Chapter 9
Sequences and Series Solutions
SECTION - A
Objective Type Questions (One option is correct) 1.
If ge gene nera rall ter term m of a seq seque uenc nce e is is n(n + 1)(2n 1)(2n + 1), then its 5 th term is (1) 110
(2) 155
(3) 330
(4) 420
Sol. Answer (3) 1)(2n + 1) Here, an = n(n + 1)(2n
a5 = 5 × 6 × 11 = 330 2.
If ge gene nera rall ter term m of of an an A. A.P P. is is 2n 2n + 5, then its common difference is (1) 2
(2) 3
(3) 5
(4) 7
Sol. Answer (1) Here, an = 2n + 5
Alternate method:
a1 = 7
an = 2n + 5
a2 = 9 Thus, d = = a2 – a1 = 2 3.
an –
1
= 2( 2(n n – 1) + 5 = 2n 2n + 3
d = = an – an – d
1
=2
If 6 th and 12 th term of an A.P. are 13 and 25 respectively, then its 20 th term is (1) 37
(2) 39
Sol. Answer (3) Here a6 = a + 5d = = 13 and an d a12 = a + 11 11d d = = 25 = 12 6d = ⇒
d = d =2
and a = 3 Hence a20 = a + 19d 19 d = = 3 + 38 = 41
(3) 41
(4) 43
2
4.
Sequences and Series
Solution of Assignment (Set-2)
If a, b, c , d , e, f are are in A.P., then e – c is is equal to (1)) d (1 d – – c
(2) 2(d – – c )
(3) 2(c – – a)
(4) c – b
Sol. Answer (2) Let D be the common difference, then c c = = a + 2D d = = a + 3D e = a + 4D Clearly, e – c = = 2D = 2( 2(d d – – c ) 5.
The m th term of an A.P. is n and n th term is m. Its Its p p th term is (1)) m – n + p (1
(2)) n + p – m (2
(3)) m + n – p (3
(4)) m + n + p (4
Sol. Answer (3) Given, am = n an = m
⇒
a + (m – 1) 1)d d = = n
a + (n – 1) 1)d d = =m
⇒
…(i) …(ii)
Subtracting (ii) from (i), we get (m – n)d = = n – m ⇒
d = = –1
and an d a = n + m – 1 = (m (m + n – 1) p – 1) 1)d d a p = a + ( p – = m + n – 1 + 1 – p = m + n – p 6.
If 8 th term of an A.P. is 15, then the sum of first 15 terms is (1) 180
(2) 210
(3) 225
(4) 240
Sol. Answer (3) Given, a8 = 15
S 15 =
=
⇒
a + 7d 7 d = = 15
15 1)d 2a (15 1) 2 15 2
2a 14d
= 15(a 15(a + 7d ) = 15 × 15 = 225 7.
If a1, a2, a3, … are in A.P. such that a1 + a5 + a10 + a15 + a20 + a24 = 225, then a1 + a2 + a3 + …+ a24 is equal to (1) 600
(2) 900
(3) 1200
(4) 1800
Solution of Assignment (Set-2)
Sequences and Series
3
Sol. Answer (2) 3(a 3( a1 + a24) = 225
a1 + a24 = 75
Required sum =
8.
24 2
(a1 a24 ) 900
The first and last terms of an A.P A.P.. are 1 and 7. If If the sum sum of its its terms is 36, then then the number of terms will be (1) 6
(2) 7
(3) 8
(4) 9
Sol. Answer (4) Given, a = 1, l = = 7 and S n = 36
∵
n
S n =
2
n
36 =
n=
9.
2
a l
1 7
72 8
9
An A.P. has has comm common on dif diffe feren rence ce 2, 2, sum sum of firs firstt n terms 49 and 7 th term 13. The value of n is equal to (1) 5
(2) 6
(3) 7
(4) 8
Sol. Answer (3) Given d = = 2, S n = 49, a7 = 13
a + 6d = = 13 a=1
Also, 49 =
49 =
n
2
n
2a n 1 d
2
2 n 1 2
9 = n(1 + n – 1) n2 = 49 n=7 10.. If th 10 the e sum sum of fi firs rstt n terms of an A.P. is 2 2n n 2 + 5n, then its n th term is (1) 3n – 5
(2) 4n – 3
(3) 4n + 3
(4) 3n + 5
4
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer (3) S n = 2n2 + 5n ∵
S 1 = a = 7
2 a + d S 2 = 8 + 10 = 18 = 2a ⇒
d = d =4
T n = a + (n – 1) – 1)d d = 7 + (n (n – 1)4 = 4n 4n + 3 11.
The sum of all all two two digit digit odd odd numb numbers ers is (1) 2375
(2) 2475
(3) 2560
(4) 4920
Sol. Answer (2) Sequence of two digit odd numbers : 11, 13, 15, … 99 No. of terms n =
99 11 2
Required sum =
45 2
1
45
11 99
= 45 × 55 = 2475 12. The number number of numbe numbers rs lying betwee between n 81 and 1792 1792 which which are divisible divisible by by 17 is (1) 105
(2) 107
(3) 109
(4) 101
Sol. Answer (4) Numbers divisible by 17 and lying between 81 and 1792 are 85, 102, 119, …, 1785 ( n – 1) × 17 1785 = 85 + (n (n – 1) × 17 1700 = (n
100 = n – 1 n = 101 13. Three numbers numbers are in A.P A.P.. such that that their sum is 24 and sum of their squares squares is 200. 200. The numbers numbers are (1) 2, 2, 8, 14
(2) 4, 4, 8, 12
Sol. Answer (3) (a – d ) + a + (a + d ) = 24 ⇒
a=8
Also, (8 – d )2 + 8 2 + (8 + d )2 = 200 ⇒
64 – 16d 16d + + d 2 + 64 + 64 + d 2 + 16 16d d = = 200
⇒
2d 2 + 192 = 200
⇒
2d 2 = 8
⇒
d = = ±2
Hence numbers are 6, 8, 10
(3) 6, 6 , 8, 10
(4) 5, 5, 8, 11
Solution of Assignment (Set-2)
Sequences and Series
5
14. The sum to p p terms terms of an A.P. is q and the sum to q terms is p p.. The sum to p to p + q terms is (1) – ( p + p + q )
(2) 0
(3) p – q
(4)) p + q (4
(3) 37
(4) 38
Sol. Answer (1) p
2 q
2
[2a ( p 1)d ] q
... (i)
[2a (q 1)d ] p
... (ii)
Equation (i) – Equation (ii)
⇒
a( p q )
⇒
a
15. If
d
2
d
2
[( [( p 2 q 2 ) ( p q )] q p
( p q 1) 1
3 5 7 to
n
term s
5 8 11 to 10 term s
(1) 35
= 7, then n is equal to (2) 36
Sol. Answer (1) n
2 3 n 1 2 2 10 2 5 9 3 2
n
2
7
6 2n 2 = 7 × 5 × 37
2n2 + 4n = 2590
n2 + 2n – 1295 = 0 n2 + 37 37n n – 35 35n n – 1295 = 0 35( n + 37) = 0 n(n + 37) – 35(n
(n + 37)(n 37)(n – 35) = 0 n = 35 (n –37) 16. If n A.M.’s are inserted between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value of n is (1) 4
(2) 6
Sol. Answer (2)
17 3 n n 1
3
Given,
17 3 n 1
3
3 1
(3) 8
(4) 9
6
Sequences and Series
42 42 9 n 1
3
Solution of Assignment (Set-2)
14n n
1
51 + 9n 9n = 17 17n n + 3 8n = 48 n=6 17. Four numbers numbers are in A.P A.P.. The sum of of first and last is 8 and and the product product of both middle middle terms is 15. 15. The least least among the four numbers is (1) 1
(2) 2
(3) 3
(4) 4
Sol. Answer (1) Let, a – 3d , a – d , a + d , a + 3d be be the four numbers. Given, a – 3d + + a + 3d = = 2a = 8
a=4 Also, a2 – d 2 = 15
16 – d 2 = 15 d 2 = 1 d = = ±1. Hence numbers are 1, 3, 5, 7. 18. If the sum sum of three numbers in A.P is 27 and and the product product of first and last is 77, then the the numbers are (1)) (1
7 51 , , 22 2 4
(2) 6, 9, 12
(3) 7, 10, 11
(4) 7, 9, 11
Sol. Answer (4) Let a – d , a, a + d be be the numbers then 3a 3a = 27
a=9 and an d a2 – d 2 = 77
81 – d 2 = 77 d = = ±2 Hence numbers are 7, 9, 11. 19. If n A.M.’s are inserted between 2 and 38 such that the sum of the resulting series obtained is 200, then the value of n is (1) 6
(2) 8
(3) 10
Sol. Answer (2) Sum of n A.M.s. =
a b 2
n
a b 2
Sum of series = n
(n + 2)(a 2)(a + b) = 400
n + 2 = n=8
400 40
= 10
a b
= 200
(4) 12
Solution of Assignment (Set-2)
Sequences and Series
7
20. If S n denotes the sum to n terms of an A.P. whose first term is a and common difference is d , then S n +
3
– 3S n +
2
+ 3S n +
– S n is equal to
1
(1) – d
(2) – a
(3) 0
(4) 2a
Sol. Answer (3) S n +
3
– 3S n +
2
+ 3S n +
– S n
1
n3
2a n 3 1 d 3 2
=
( n 2) (n 1) n 2a n 2 1 d 3 2a n 11 d 2a n 1 d 2 2 2
3(n 2)(n 1) 3n(n 1) n(n 1) n 3 3 n 6 3n 3 n ( n 2)(n 3) 3( d 2 2 2 2 2 2 2 2
2a
=
=0
21.. If thr 21 three ee pos positi itive ve nu numb mbers ers a, b, c are are in A.P. and
(1)) a = b = c (1
(2) 2b = 3a + c
1 a
2
,
1 b
2
,
1 2
c
also in A.P., then
(3)) b 2 = (3
ac
(4) 2c = 2b + a
8
Sol. Answer (1) 2
b
2
1
a
2
1
2
c
2 2
(a c )
(a c) 2 2ac 2
a c
2
(a + c )2 = 4ac , –2ac –2 ac
a = c and and (a ( a + c )2 –2 –2ac ac
22. If S n denotes the sum of first n terms of an A.P., and S 2n = 3S n then (1) 4
(2) 5
(3) 6
S 3n S n
is equal to
(4) 7
Sol. Answer (3) 2n 3n [2a ( 2n 1)d ] = [2a (n 1)d ] 2 2
2a = (n + 1) 1)d d 23. Let S n denotes the sum to n terms of an A.P. whose first term is a. If the common difference d is is given by d = S n – kS n – 1 + S n – 2 , then k is equal to (1) 1 Sol. Answer (2)
(2) 2
(3) 3
(4) 4
8
Sequences and Series
Solution of Assignment (Set-2)
24.. If the first 24 first,, second second and last last term term of an A.P A.P.. are a, b and 2a 2a respectively, then its sum is
(1)) (1
ab
ab
(2)) (2
2(b a)
b a
(3)) (3
3ab 2(b a)
(4)) (4
2ab
a
b
Sol. Answer (3) d = d =b–a Let n be the number of terms then 2a 2a = a + (n – 1)(b 1)(b – a) a ba
= n – 1
a
n=
ba
b
+ 1 =
ba
b
Sum =
2(b a )
(a 2a ) =
3ab 2(b a)
25. If A1, A2 are two A.M.’s between two positive numbers a and b, then (2 A1 – A2)(2 A2 – A1) is equal to ab
(1)) a + b (1
(2)) ab (2
(3)) (3
ab
a
(4)) (4
b
Sol. Answer (2) A1 = a +
A2 = a +
b a 3
b a 3
2
Now, (2 A1 – A2)(2 A2 – A1) = 2a
2(b a ) 2(b a ) 4(b a ) b a 2 a a a 3 3 3 3
= a(a + b – a) = ab 26.. The fou 26 fourth rth term of the G.P .P.. 4, 4, –2, 1, … is (1)) (1
1
(2)) (2
2
1 2
Sol. Answer (1) 3
4 1 1 a4 = ar 3 = (4) 2 8 2
(3) – 1
(4)
1 4
Solution of Assignment (Set-2)
Sequences and Series
9
512
27.. Whic 27 Which h term term of the the G.P G.P.. 18, – 12, 8, 8, … is (1) 7 th
729
?
(2) 9 th
(3) 11th
(4) 13 th
Sol. Answer (2) an = ar n –
512 729
1
2 18 3
2 3
n
1
n
1
512 729 18
256 81 9 9
2 3
8
n – 1 = 8 n=9 28. The third term of a G.P G.P.. is 3. The product product of its its first five terms terms is (1) 81
(2) 243
(3) 729
(4) 343
Sol. Answer (2) ar 2 = 3 Now, a·ar ·ar 2·ar 3·ar 4 = a5r 10 = (ar ( ar 2)5 = 3 5 = 243 29. If (2 p p)) th term of a G.P. is q 2 and (2q (2q) th term is is p p 2, then ( p + p + q) th term is (1)) pq (1
(2)) (2
1 2
2
p q
2
(3)) p 2q 2 (3
(4)) (4
1 4
3
p q
3
Sol. Answer (1) ar 2 p – 1 = q2
... (i)
ar 2q – 1 = p2
... (ii)
Multiply (i) and (ii), ar p +
q – 1
= pq
30. Three numbers numbers whose product product is 512 are in G.P G.P.. If 8 is added to the first and 6 to the the second, the number number will be in A.P. The numbers are (1) 2, 2, 8, 32
(2) 8, 8, 8, 8
Sol. Answer (3) Given,
a
�a �ar
r
512
a3 = 512 a=8 Now,
8 r
+ 8r +
14, 8r are in A.P. 8, 14,
1 1 r = 28
8
(3) 4, 4, 8, 16
(4) 2, 2, 8, 14
10
Sequences and Series
2
8r
8 8 r r
Solution of Assignment (Set-2)
= 28
8( 8(r r 2 + r + + 1) = 28r 28 r 2r 2 – 5r + + 2 = 0 2r 2 – 4r – – r + + 2 = 0 – 2) –(r –( r – – 2) = 0 2r (r –
2r r = = 1 or r r = =2 1
r r = =
,2
2
Hence numbers are 4, 8, 16. 31.. If first 31 first and and eighth eighth terms terms of of a G.P G.P.. are are x x –4 and and x x 52 and its second term is x t , then t is is equal to (1) 2
(2) 3
(3) 4
(4) 13
Sol. Answer (3) a = = x x –4 and ar 7 = = x x 52
( x x –4)r 7 = = x x 52
r 7 = = x x 56 r = = x 8 ar = = x 4 = x t t t = 4
32. The sum to n terms of the G.P. 1 +
(1)) (1
1
(2)) (2
n
2
1 n
2
1 2
1
1 4
is (3)) (3
1
1 n
2
(4)) (4
2 1
1 n
2
Sol. Answer (4)
1 2 n
11
S n =
1
1
2 1
1 n
2
2
33. The n th term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then its first term is (1) 1
(2) 3
Sol. Answer (1) ar n – 1 = 128 and
ar
n
1) = 255 r 1
(
a r
n
1 r
�r
1
a
= 255
(3) 5
(4) 7
Solution of Assignment (Set-2)
1 28 28 2 a 1
Sequences and Series
11
= 255
256 – 255 = a a=1 a=1 34. The sum of three numbers numbers in a G.P G.P.. is 26 and the sum of products taken taken two at a time is 156, 156, then the numbers are (1) 1, 1 , 5, 25
(2) 2, 2, 6, 18
(3) 1, 1 , 4, 16
(4) 2, 2 , 8, 16
(3)) c 2 – b 2 (3
(4) 0
Sol. Answer (2) a + ar + + ar 2 = 26
... (i)
a2r (1 (1 + r + + r 2) = 156
... (ii)
Squaring (i) and divide by (ii), 1
r = = 3,
3
and a = 2, 18 ba
35. If a, b, c are are in G.P., then
b c
(1)) ab (1
b a
is equal to
b c
(2)) ac (2
Sol. Answer (4) Since a, b, c are are in G.P. b = ar , c = ar 2
a b a = b c bc
b
Now,
=
( 1) a(r 1) ar (1 r ) ar (1 r ) a r
1 r
1 r
=0
36. If x , 2 x + + 2, 3 x + + 3 are the first three terms of G.P., then the fourth term is (1) – 27
(2) 27
(3) – 13.5
(4) 13.5
Sol. Answer (3) (2 x + + 2)2 = = x x (3 (3 x + + 3) ⇒
x 2 + 5 x + + 4 = 0
⇒
x = = –1, –4
But x But x –1, –1, x x = = –4 37. If g 1, g 2, g 3 are three geometric means between two positive numbers a and b, then g 1g 3 is equal to (1)) g 2 (1 Sol. Answer (3) g 1, g 2, g 3 are also in G.P.
g 1g 3 = g 22
(2) 2g 2
(3)) g 22 (3
(4)) g 23 (4
12
Sequences and Series
Solution of Assignment (Set-2)
38. The fifth term term of a G.P G.P.. is 32 and common common ratio is 2, then the the sum of first 14 terms terms of the G.P G.P.. is (1) 16388
(2) 32667
(3) 32766
(4) 64342
Sol. Answer (3) a × 24 = 32
a=2 2(r 14 1) 2( 214 1) S 14 = = 32766 1 r 1 39. If the sum of first first three numbers numbers in G.P G.P.. is 21 and their product product is 216, 216, then the numbers numbers are (1) 3, 3 , 6, 12
(2) 5, 5 , 7, 9
(3) 6, 6 , 2, 213
(4) 6, 6 , 12, 24
Sol. Answer (1) a r
·a·ar = = 216
a3 = 216 a=6 Also,
6
+ 6 + 6r 6r = = 21
r
r 2 1 = 15 r
6
5r = =0 2r 2 + 2 – 5r
2r 2 – 4r – – r + + 2 = 0 2r (r – – 2) – (r (r – – 2) = 0 r r = =
1 2
,2
Hence numbers are 3, 6, 12 40. If x , y , z are are three geometric means between 6 and 54, then z is is equal to (1)) (1
9 3
(2) 18
(3)
18 3
(4) 27
Sol. Answer (3) 3
b z = 3rd G.M. = ar 3 = a a
n
1
3
54 4 = 6 = 6
6 3 3
18
3
41. If thr three ee dis distin tinct ct nu numb mbers ers a, b, c are are in A.P. and b – a, c – – b, a are in G.P., then a : b : c is is equal to (1) 2 : 3 : 4 Sol. Answer (3) b – a = c – b 2b = a + c
(2) 1 : 2 : 4
(3) 1 : 2 : 3
(4) 1 : 3 : 5
Solution of Assignment (Set-2)
Sequences and Series
13
Also, (c (c – – b)2 = a(b – a)
c c – – b= a a + b = c c
a
a+
= c
2
3a + c = 2c 3a = c
a
1
c
3
2b = a + 3 a = 4 a a
1
=
b
2
Hence required ratio = 1 : 2 : 3 42. Three numbers numbers form an increasing increasing G.P G.P.. If the middle term is doubled, then the numbers numbers are in A.P. A.P. The common common ratio of the G.P. is (1)) (1
2
(2)) (2
3
2
3
(3)) (3
(4)) (4
3 2
3
2
Sol. Answer (2) 4ar = = a + ar 2
r
2
3
For an increasing G.P., r = = 2
3
43. If a, b, c form a G.P. with common ratio r such that 0 < r < 1, and if a,
3 2
b
, – 4c form an A.P., then r is
equal to
(1)) (1
2
1
(2)) (2
2
Sol. Answer (4) 3ar = = a – ar 2
r = = –1,
1 4
But 0 < r < < 1, r =
1 4
3
1
(3)) (3
3
(4)) (4
1 4
14
Sequences and Series
Solution of Assignment (Set-2)
44. If x , y , z are are in A.P.; ax , by , cz are are in G.P. and a
(1)) (1
b
b
a
(2)) (2
a
c
1 1 1 , , are in A.P., then
a
b
c
z
c a
x
a
(3)) (3
c
z
is equal to
x
c
b
(4)) (4
b
c
c a
Sol. Answer (2) b2y 2 = ac ac × × z
4a2c 2 ( x z )2 . 4 (a c )2
( x z )2
x z
z
(a c )2
xz
ac
ac
z
x
a c
c a
45. If the second, third and sixth sixth terms of an A.P. A.P. are distinct and form consecutive consecutive terms of a G.P G.P., ., then the common ratio of the G.P. is 1
(1)) (1
(2) 1
2
(3) 2
(4) 3
Sol. Answer (4) a + d , a + 2d , a + 5d are are in G.P.
a 5d
a 2d
a 2d
a d
a2 + 6ad + + 5d 2 = a2 + 4d 2 + 4ad 2 ad = = 0 d 2 + 2ad
d (d d + + 2a 2 a) = 0 d 0, d d = = –2 –2a a Now, a + d = = –a a + 2d = = –3 –3a a a + 5d = = –9 –9a a common ratio = 3 1
46. If disti distinct nct posi positiv tive e numbe numbers rs a, b , c are in G.P. and
a b
,
1 c a
,
1 b c
are in A.P., then the value of
a + 4b + c is is equal to (1) – 3
(2) 0
Sol. Answer (2) 2 ar
2
a
1
a
ar
1
ar
ar 2
(3) 3
(4) 4
Solution of Assignment (Set-2)
2 1 r 1
Sequences and Series
1 r
2 r + + 1 = –2r –2 r r 2 + 2r
ar 2 + 4ar + + a = 0 or a + 4b + c = 0 47. Th The e sum sum 1 + 3 + 32 + … + 3 n is equal to n
n + 1
(1) 3
(2)) (2
3
1
1
2
(3) 3n + 1 – 1
(4) 3n
Sol. Answer (2) 1 + 3 + 3 2 + … 3n
S n =
3
n
1
1
2
48.. Th 48 The e sum sum of th the e ser serie ies s 12 + 1 + 22 + 2 + 32 + 3 + … + n2 + n is equal to ( 1)( n 2) 3
n n
(1)) (1
(2)) (2
( 1)(3n 1)
n n
2
(3)) (3
(2n 1)(3n 1) 6
n
n(n 1) (4) (4) 3
Sol. Answer (1) (1 + 2 + 3 + … + n) + (12 + 22 + … + n2) ( 1) 2
n n
=
( 1) 1) 2n 1 1 3 2
n n
=
( 1) 3 2n 1 2 3
n n
=
( 1)( n 2) 3
n n
=
1
49.
2
( 1)( 2n 1) 6
n n
(1)) (1
1 4
1
1
8
16
to is equal to
1
(2) 1
2
Sol. Answer (2) 1 a
S =
1 r
2 1
1 2
1
(3) 2
(4) 3
2
15
16
Sequences and Series
Solution of Assignment (Set-2)
50.. The sum 50 sum of the serie series s 3 + 7 + 13 + 21 21 + 31 + … to n terms is equal to
(1)) (1
(
n n
2
2n 3) 3) 3
(2)) (2
(
n n
2
3n 5) 5) 3
(
n n
(3)) (3
2
2n 5) 5) 3
(4)) (4
(
n n
2
2n 2) 2) 3
Sol. Answer (2) Let S n = 3 + 7 + 13 + 21 + … + an – 1 + an Again S n = 3 + 7 + 13 + … + an – 2 + an –
1
…(i) + an
…(ii)
Subtracting (ii) from (i), 0 = 3 + [4 + 6 + 8 + … + ( an – an – 1 )] – an
an = 3 +
(n 1) [2 × 4 + (n ( n – 1 – 1) × 2] 2 ( n 1) (8 + 2n 2n – 4) 2
an = 3 +
=3+
(n 1) 2
(2n (2 n + 4)
= 3 + n2 + n – 2 = n2 + n + 1
S n =
=
=
∑ (n
2
n 1)
( 1)(2n 1) n( n 1) n 6 2
n n
(
n n
2
3n 5) 5) 3
51.. If th 51 the e sum sum of fi firs rstt n terms of an A.P. is an2 + bn and nth term is An is An + B then (1)) A = 2a (1
(2)) A = a (2
(3)) a = 2 A (3
(4)) A = 3a (4
Sol. Answer (1) Sn
an 2 bn ,
S1
a b A B
…(i)
S2
4a 2b (A B ) (2A B )
…(ii)
Tn
An
B
On solving equation (i) & (ii) we get, A get, A = = 2a 52. If tenth term of an A.P A.P.. is 19 and sum of first fifteen fifteen terms is 225 then fifth fifth term of A.P A.P. is (1) 5 Sol. Answer (3)
(2) 6
(3) 9
(4) 11
Solution of Assignment (Set-2)
Sequences and Series
17
Let first term is a and common difference is d , then T10
a 9d 19
S15
2a a
15 [2a (15 1) 1) d ] 225 2
14d
… (i)
30
… (ii)
9d 19
On solving equation (i) & (ii) we get, a = 1 and d = = 2 then
T5
a (5 1)d 1 4 2 9
53.. The max 53 maximum imum sum of the the serie series s 100 100 + 98 98 + 96 96 + .... is (1) 2500
(2) 2550
(3) 2050
(4) 2555
Sol. Answer (2) For maximum sum, Tn
a (n 1) d 0
100 ( n 1) ( 2) 0
n
Then,
1 50 ⇒ n 51 S n
51 2
5 0 2550 100 0 51 50
54. I f 1 , 2 , 3 , ......., n are in A.P. whose common difference is d , then sind sin d [sec [ sec 1 sec 2 + sec 2 sec 3 + ........ + sec n–1sec n] = (1) sec n – sec 1
(2) sin n – sin 1
(3) cos n – cos 1
Sol. Answer (4)
2 – 1 = 3 – 2 = …………… = n – n–1 = d sin d [sec [sec 1 sec 2 + sec 2 sec 3 + ……………+ sec n–1 sec n] sin d
=
cos 1 cos 2
sin d cos 2 cos 3
...............
sin d cos n cos n –1
sin(3 2 ) sin(2 1) sin(n n1) = cos cos cos cos ............... cos cos 2 1 2 3 n n –1
=
sin 2 cos 1 cos 2 sin 1 cos 2 cos 1
sin 3 cos 2
cos 3 sin 2 ............... cos 2 cos 3
(tan 2 tan 1) + (tan 3 tan 2) + ………+ (tan n tan n–1) tan n – tan 1, remaining terms are cancelled out
(4) tan n – tan 1
18
Sequences and Series
Solution of Assignment (Set-2)
55. Consider that 10 arithmetic arithmeti c means are inserted betwee between n 3 an and d 7 an and d the their ir su sum m is is a. Again consider that the sum of five numbers in A.P. is 30 and the value of middle terms is b. Then a + b equals (1) 16
(2) 56
(3) 46
(4) 36
Sol. Answer (2) Let,
A1 A2 A3
then
A1 A2
,
,
,....... .,
A10 are
10 A.M.’s between 3 and 7
3 7 A3 ....... A10 10 50 2
Now sum of five terms of an A.P. b
2d b d b b d b 2d 30
5b
30
b= 6 a
b 50 6 56
56. Let t r denote the r th th term of an A.P. Also suppose
t m
1 n
and
t n
1 m
, (m n), for some positive integers
m and n, then which of the following is necessarily a root of the equation ? ( l+ m –2 n ) x 2 + ( m + n –2 l ) x + ( n + l –2 –2 m )= 0 (1)) t n (1
(2)) t m (2
(3)) t m + n (3
(4)) t mn (4
Sol. Answer (4) Let first terms of A.P. = a Common difference = d t m =
t n =
1 n
1 m
= a + (m 1) 1)d d
…(i)
= a + (n ( n 1) 1)d d
…(ii)
by (i) and (ii) a
1 mn
, d
1 mn
t mn = a + (mn 1) 1)d d =
1 mn
(mn 1)
1 mn
1 mn 1 mn
1
Also x Also x =1, =1, is the root of the equation, hence root is t mn. 57. The sum of the first 100 terms terms common to the series 17, 21, 25, 29, 29, 33, ...... and and 16, 21, 26, 31, 36..... is (1) 10 100101
(2) 11 111000
(3) 10 101100
(4) 11 110010
Solution of Assignment (Set-2)
Sequences and Series
19
Sol. Answer (3) Common difference of first A.P. = d 1 = 4 Common difference of second A.P. = d 2 = 5 L.C.M. of (d ( d 1, d 2) = 20 First common terms of both A.P. is 21 Hence the A.P. formed by common terms is 21, 21 + 20, 21 + 2 × 20, 21 + 3 × 20, …… Sum of 100 terms
100 2 21 (100 1) 20 2
50 42 99 20 50 42 1980 50 2022 = 101100 58. If the sixth term term of a GP be 2, then the the product product of first first eleven terms is (1) 1024
(2) 2047
(3) 2048
(4) 1023
Sol. Answer (3) Let G.P. is a, ar , ar 2, ar 3 ………… T 6 = ar 5 = 2
… (i)
P = a.ar.ar 2 ………… (ar (ar 9) (ar (ar 10) = a11r 1
+ 2 + 3 + ………. 10
1011
=a
11
r
2
= a11r 55 = (ar ( ar 5)11 = 211 = 2048 by (i) 100
th
59. Let an be the n term of the G.P. G.P. of positive positi ve numbers. Let Le t
∑ n
100
a2 n
and
1
∑
a2 n 1
such that , then the
n 1
common ratio is
(1)) (1
(2)) (2
(3)) (3
Sol. Answer (1) Let G. G.P. P. = a, ar , ar 2 ……, where ‘r ‘ r ’ is the common ratio
= a2 + a4 + …… + a200 = ar ar + + ar 3 + ar 5 + ……+ ar 199 ar
=
(1 (r 2 )100 ) 1 r 2
ar
(1 r 20 0 )
(1 r 2 )
…(i)
(4)) (4
20
Sequences and Series
Solution of Assignment (Set-2)
Similarly, = a1 + a3 + a5 + …………… + a199 = a + ar 2 + ar 4 + …………… + ar 198
(1 (r 2 )100 )
a
=
(1 r 2 )
(1 r 200 )
a
…(ii)
(1 r 2 )
By (i) and (ii)
r 2
3
2 x 2 x 60. The se seri rie es ....... will have a definite sum when x 3 x 3 x 3 2 x
(1) –1 < x < < 3
(2) 0 < x < < 1
(3) x = = 0
Sol. Answer (1) For definite sum –1<
2 x x
3
Case (i)
<1
2 x x
3
<1
2 x x
–1<0
3
2 x ( x 3) <0 x 3
3 ( x 3)( x 3) <0 0 � x 3 ( x 3)2 x
�
�
�
� �
�
�
�
( x x – – 3) ( x x + + 3) < 0 x ( (– – 3, 3) Case (ii)
2 x x
3
…(i)
1 ⇒
2 x x
3
1 0
x 3 0 x 3
2 x
3 x 3 x
3
0
⇒
( x 1)( x 3) ( x 3)2
( x 1) 0 ( x 3)
�
� � �
��
0
or ( x + + 1) ( x x + + 3) > 0 x
( , 3) ( 1, )
…(ii)
By (i) and (ii) the common interval of x is is (– 1, 3)
(4) x = = –3
Solution of Assignment (Set-2)
Sequences and Series
21
61. Four geom geometric etric mean means s are inser inserted ted betwee between n the numb numbers ers 211 – 1 and 2 11 + 1. The product of these geometric means is (1) 222 – 211 + 1
(2) 244 – 223 + 1
(3) 244 – 222 + 1
(4) 222 – 212 + 1
Sol. Answer (2) The product of Geometric mean
=
( 211 1)(211 1)
4
= 222 1)
2
244 2.222 1 = = 244 223 + 1
62. The va valu lue e of of x in in (–, ) which satisfies the equation
(1)) (1
(2)) (2
3
1|cos x ||cos2 x ||cos3 x |.......
8
(3)) (3
3
2 3
43 is
(4) Al Alll of of thes these e
Sol. Answer (4) 1 |co |cos s x||cos |cos2 x |...... ......... ...... ..... ..
8
=
1|cos |cos x| |co |cos s x| 2 |cos |cos x |3 .... ...... .... ..
8
c os x | 1 | co
x
3
,
8
1 1|cos |cos x |
3
4
82
1 2
1 2
| co cos x |
cos x
=
1 2
3
,
�
3
,–
�
3
, hence only four solutions exist
63. If one geo geome metri tric c mean mean G and two A.Ms p p and q be inserted between two given quantities then (2 p – q)(2 )(2q q – – p p)) equals (2)) G2 (2
(1)) G (1 Sol. Answer (2)
Let the numbers are a and b.
G
2
ab
…(i)
Also a, p, q, b b be in A.P.
2 p p = = a + q (2 p p – – q) = a and 2q = p + b (2 (2q q – p p)) = b
(2 p p – – q) (2q (2 q – p p)) = ab ab = = G2 by (i)
(3) 2G2
(4) 2G
22
Sequences and Series
Solution of Assignment (Set-2)
64. The sum of three three numbers in G.P G.P.. is 56. If we substract substract 1, 7, 21 from these numbers numbers in that order, order, we obtain an A.P. the three numbers are (1) 10, 18, 26
(2) 8, 16, 32
(3) 9, 16, 23
(4) 5, 8, 11
Sol. Answer (2) Let the numbers are a, ar , ar 2 a + ar + + ar 2 = 56
…(i)
a – 1, ar – – 7, ar 2 – 21 are in A.P. 2(ar ar – – 7) = (a ( a – 1) + ((ar ar 2 – 21) 2( 2ar – – 14 = a + ar 2 – 22
…(ii)
By (i) and (ii) 2ar 14 = (56 ar ) 22 3ar = = 56 22 + 14 3ar = = 48 ar = = 16
…(iii)
By (i), (iii) a + 16 + ar 2 = 56 a + ar 2 = 40 Again ar = = 16 a =
16 r
16
.r
2
16 r
40
r
solving, we get r = = 2, r r = =
1 2
Case (i) if r = = 2, then a = 8 then numbers are 8, 16, 32. Case (ii) if r = =
1 2
then a = 32
then numbers are 32, 16, 8. 65. L e t a , b be the roots of the equation x 2 – 4 x x + + k 1 = 0 a n d c , d the roots of the equation x 2 – 36 x + k 2 = 0. If a < b < c c < < d and and a, b, c , d are in GP, then the product k 1k 2 equals (1) 81
(2) 729
Sol. Answer (2) Let common ratio of G.P. is r b = ar , c c = = ar 2, d d = = ar 3 x 2 4 x x + + k 1 = 0
(3) 256
(4) 64
Solution of Assignment (Set-2)
Sequences and Series
By the properties of roots a + b = 4 a + ar = = 4
…(i)
ab = k 1 a(ar ) = k 1
a2r = k 1
…(ii)
Similarly, x 2 36 x x + + k 2 = 0 ar 2 + ar 3 = 36
…(iii)
(ar 2) × (ar 3) = k 2
…(iv)
By (i) and (iii), (1 r )
a 2
(1 r )
r a
4 1 1 ⇒ ⇒ r 3 2 36 r 9
Case (i) if r = = 3, then by (i) a = 1 k 1 = 3, k 2 = 243
k 1 k 2 = 243 × 3 = 729 By a < b < c < < d we we find that r is is positive hence only one case is possible. That is given by r r = =3 66. If a, b, c are are three distinct real numbers and they are in a G.P. If a + b + c = xb, then (1)) x < (1 < –1 or x or x > > 3
(2) x –3 or x or x 2
(3) x –4 or x or x 3
(4) x 2 or x or x 5
(3) 99.2100
(4) 99.2100 + 1
Sol. Answer (1)
–
+
Let common ratio = r = r
–1
+ 3
b = ar , c c = = ar 2 a + b + c = xb
a + + ar ar + + ar 2 = x (ar )
1 + r + + r 2 = xr r 2 + r (1 (1 x ) + 1 = 0 Because r is is real, hence Discriminant > 0
(1 x )2 4 × 1 × (1) > 0 ((1 x ) 2) (1 x + + 2) > 0 –
+
1 x ) ( x + + 3) > 0
–1
+ 3
( x + + 1) ( x x 3) > 0
x ( , 1) (3, ) 67.. Th 67 The e sum sum of th the e ser serie ies s 1 + 2.2 + 3.2 2 + 4.23 + 5.24 + ......+ 100.2 99 is (1) 99.2100 – 1
(2) 10 100.2100
23
24
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer (4) S = = 1 + 2.2 + 3.2 2 + 4.23 + ……… + 100.2 99
…(i)
2S = = 2 + 2.2 2 + 3.23 + ……… + 99.2 99 + 100.2100
…(ii)
Subtracting (ii) from (i) S = = 1 + (2.2 2) + (3.22 2.2 2) + ……… + (100.2 99 99.299) 100.2100 S = 1 + 2 + 2 2 + 23 + ……… + 2 99 100.2100 S = 100
2
S = = S
1
2 1
100.2100
S = = 2100 1 100.2101 S S = = 99.2100 + 1
68. If 3
1 4
1
(3 p )
42
then p equals equals (3 2 p ) ..... to 8 , then p
(1) 1
(2) 5
(3) 7
Sol. Answer (4) 3
1 4
( 3 p)
Let S 3
S
4
1 4
2
( 3 2 p) .......... = to = 8
1 1 (3 p ) 2 (3 2p ) .......... 4 4
1 1 (3) 2 (3 p ) .......... 4 4
Subtracting (ii) from (i) 3S 1 1 . .............. 3 ( p ) 2 ( p ) .. 4 4 4 3S 4
3 p
1 4
1 4
1 3S 3 p 4 1 4 1 4 3S 4
3
2
...............
p 3
But S = 8
38 4
p 3
3
p 3
= 6 3 = 3 p = 9
…(i)
…(ii)
(4) 9
Solution of Assignment (Set-2) 1
1
69. The pro rod duc uctt
2 4.4 8.8
1 16
Sequences and Series
.... to
(1) 4
equal to
(2)
(3) 8
2
(4) 2
Sol. Answer (4) 1
1
1
1
4
8
16 16
32
2 4 8
2
1
2
3
4
8
16
32
.2 .2
24
...............
4
1
Let
16
2
3
8
16
2
1
P
P
.2
32
1
8
8
..........
2
4
4
.... ...... .... .... .... .... ....
2 16
3 16
3 32
4 32
...............
…(i)
..............
…(ii)
Subtracting (ii) from (i) P
P
2
1 4
1 P 2
1
8
1 16 16
.............. upto
1
4
1
1 2
4 1
1 2
2
P = 1
1
2
4
1
1
8 16
..........
2P 21 2
70. Let S denotes denotes the infinite sum 2 + 5 x + 9x2 + 14x3 + 20x4 + ..., where | x x |<1. |<1. Then S equals equals (1)) (1
2 x (1 x )
3
(2)) (2
2 x (1 x )
3
(3)) (3
2 x (1 x )
Sol. Answer (2) S = = 2 + 5 x + + 9 x 2 + 14 x 3 + 20 x 4 + ………….
Sx = = 2 x + + 5 x 2 + 9 x 3 + 14 x 4 + ………….
…(i) …(ii)
Subtracting (ii) from (i) S (1 (1 – x – x ) = 2 + 3 x + + 4 x 2 + 5 x 3 + 6 x 4 + ………….
…(iii)
S (1 (1 – x – x ) x x = = 2 x + + 3 x 2 + 4 x 3 + 5 x 4 + ………….
…(iv)
3
(4)) (4
2 x (1 x )3
25
26
Sequences and Series
Solution of Assignment (Set-2)
Subtracting (iv) from (iii) S (1 (1 – x – x )(1 )(1 – x – x ) = 2 + x + x + + x 2 + x 3 + …………. x
S (1 (1 – – x x )2 =
2
S (1 (1 – – x x )2 =
2 2 x x 2 x (1 – x ) (1 x )
1–
x
2 x
S = S =
(1 x )3
71. The val alu ue of of x + y + z is is 15 if a, x , y , z , b are in A.P. while the value of
1
x
1
y
1
z
5
is
3
if a, x , y , z , b are in i n H.P. H.P.
the value of a and b are (1) 9, 1
(2) 7, 4
(3) 8, 2
(4) –1, 3
Sol. Answer (1) We know that the sum of n A.M. between two quantities are equal to n times their single mean.
a b x x + + y y + + z z = =3 2
15 =
3 2
(a + b) a + b = 10
…(i)
Similarly a, x, y, z, b b are in H.P.
1 1 1 1 1 , , , , are in A.P.
a x y z b 1
Again
5
3
x
10 9
1
y
1
z
31
1 b 2 a
31
1 2 a b
ab ab
10 ab
ab ab = = 9 By (i) and (ii) we get that a(10 – a) = 9
a2– 10a 10 a + 9 = 0 (a – 1) (a (a – 9) = 0 a = 1, 9 if
a = 1 then b = 9 and if a = 9, then b = 1
Hence numbers are 9, 1
…(ii)
Solution of Assignment (Set-2)
Sequences and Series
27
72. If x , y , z are are positive reals satisfying 4 xy + + 6yz + + 8zx = 9, then the greatest possible value of the product xyz is (1) 0.125
(2) 0.375
(3) 0.500
(4) 0.625
Sol. Answer (2) Using A.M. G.M. 1 4 xy 6 yz 8 zx (4 xy )(6 yz)(8zx ) 3 3
1
9 ( 4 4 2 2 3 x 2 y 2 z 2 ) 3 3
27 192 x 2y 2z 2 x 2y 2z 2
27 192
xyz 0.375 73. If a + b + c = 3 and a > 0, b > 0, c > > 0, then the greatest value of a2b3c 2 is 9
3 .2
(1)) (1
7
4
10
(2)) (2
7
3
.2
7
4
8
(3)) (3
7
3 .2 7
4
7
7
(4)) (4
3 .2 7
7
Sol. Answer (2) A G a 2
a
b
2
3
b 3
b 3
c 2
7
7
3 7 10
3 .2
2
a
2
2
.
b
3
3
3
10
2
a b c 3
2
7
2 3 2
2
.
c 2
2
a 2b 2c 3
7
2
1 2
4
7
2
c
3
a b c
3 .2
4
7
7
10
Hence the maximum value is
3 .2
4
7
7
.
74.. Th 74 The e sum sum to 100 100 term terms s of the the seri series es 1.2.3. + 2.3.4. + 3.4.5. +...+ n(n + 1)(n 1)(n + 2) +... is integral multiple of (1) 2525 Sol. Answer (1) T n = n(n + 1 ) (n (n + 2) = n3 + 3n2 + 2n 2n
(2) 2526
(3) 2527
(4) 2528
4
28
Sequences and Series
Solution of Assignment (Set-2)
S n = T n = n3 + 3n 2 + 2n 2
2n( n 1) n( n 1)(2n 1) n(n 1 3 = 2 6 2
=
( 1) n 2 n 3
n n
4 100 101 102 103
S 100 =
4
= 2525 × Integer
75.. Th 75 The e sum sum of th the e firs firstt n terms of the series 1 2 + 2.22 + 32 + 2.42 + ..... is
( 1)2 when n is even. Then the sum 2
n n
if n is odd, is (
n n
(1)) (1
2
1)
(2)) (2
3
n(n + 1)2
(3)) (3
n
2
(n 1) 2
(4)) (4
(
n n
2
1)
2
Sol. Answer (3) If n is odd then n – 1 is even then Sn
Sn
(n 1)n 2 , t n2 1 2 n
S
n
1 t n
(n 1)n 2 (n 1)n2 n2 2 2
76. If x , y , z and and w are are non-zero real numbers and x 2 + 5y 2 + 5z 5 z 2 + 4w 4 w 2 – 4 xy xy – – 4yz 4 yz – – 4zw = = 0, then x , y , z , w are in (1) A.P.
(2) A.G.P.
(3) H.P.
(4) G.P.
(3) H.P.
(4) A.G.P.
Sol. Answer (4) The given expression can be written as ( x x – – 2y )2 + (y – – 2z)2 + (z – – 2w )2 = 0 x = = 2y , y = = 2z, z = = 2w x y
y
z
z w
2
Hence x, Hence x, y, z, w are are in G.P. 77. If x 18 = y 21 = z 28, then, 3log y x , 3logz y , 7logx z are are in (1) A.P.
(2) G.P.
Sol. Answer (1) x 18 = y 21 = z 28 log x = = 21 log y = = 28 log z 18 log x
…(i)
Solution of Assignment (Set-2)
3 logy x 3
and 3 log
z
21 7 3 18 2 log y log x
y
7 logx z 7
Sequences and Series
3
(by (i))
28 3 4 21 log z
log y
(by (i))
18 9 7 28 2 log x log z
Hence the number becomes
(by (i)) 7 2
, 4,
9 2
, which are in A.P.
78.. Th 78 The e coe coeff ffic icie ient nt of x101 in the expansion of (1 – x – x )(1 )(1 – 2 x )(1 )(1 – 22 x ) ... (1 – 2101 x ) is (1) 24950 – 2 5050
(2) 25051 – 2 5152
(3) 24950 – 2 5051
Sol. Answer (4) (1 – x – x ) (1 – 2 x ) (1 – 2 2 x ) (1 – 23 x ) ………… (1 – 2 101 x ) 1 1 2 1 2 3 101 = ( 1)( x 1)( 2) x ( 2) x 2 ( 2) x 3 .......( 2 ) x 101 2 2 2 2 1 1 1 = ( 1)102 2.22.23.2 4.........2101( x 1) x x 2 .......... x 101 2 2 2 1 1 1 ........ .....1 ..101 01 = 21 2 3..... .( x – 1) x x 2 .......... x 101 2 2 2 101102
= 2
2
.( x102 – 1
1 1 1 1 2 3 ......... 101 x 101 ..........) 2 2 2 2
Coefficient of x of x 101 101102
= 2
2
(–1) 1
1 1 1 2 ............... 101 2 2 2
1 102 1 102 1 1 2 2 101 51 101 51 2 (– (–1) ( 1) = 2 1 1 1 1 2 2 2102 1 2102
10151 . = 2
2 1
= 210151 . 2 101(2102 1) = 2 5050 – 25152
(4) 25050 – 2 5152
29
30
Sequences and Series
Solution of Assignment (Set-2)
79. If a, b, c are are in A.P., p A.P., p,, q, r are are in H.P. and ap ap,, bq bq,, cr are are in G.P., then
(1)) (1
p r
r
c
p
a
a
(2)) (2
c
p q
q p
c a
a c
(3)) (3
p r
r p
b c
c
(4)) (4
b
p q
q p
c a
a c
Sol. Answer (1) 2b
q
…(i)
a c 2 pr
…(ii)
p r
b 2 q 2 (ap )(cr )
…(iii)
By (i), (ii), (iii) 2
2
a c 2pr 2 p r (ap)(cr )
(a c )2
( p r )2
ac
a
c
c
pr
a
p r
r p
80. Let C C be be a circle with centre P 0 and and AB AB be be a diameter of C . Suppose P 1 is the midpoint of line segment P 0 B, P 2 the midpoint of line segment P 1 B and so on. Let C 1, C 2, C 3 be circles with diameters P 0P 1, P 1P 2, P 2P 3, ... respectively. Suppose the circles C 1, C 2, C 3 ... are all shaded. The ratio of the area of the unshaded portion of C to to that of the original circle is (1) 8 : 9
(2) 9 : 10
(3) 10 : 11
(4) 11 : 12
Sol. Answer (4) Let the radius of largest circle is r AB = 2r 2 r
P 0P 1 =
r
2
, P 0C 1 =
r
4
A
2
( r 2 ) r Area of circle C 1 = 4 42
P 0
2
r 2 r Similarly area of circle C 2 = 2 8 8
2
2
r r Area of the circle = C 3 = 2 and so on. 16 16
C 1
C 2
P 1 P 2
B
Solution of Assignment (Set-2)
Sequences and Series
The sum of all areas is 1 1 1 2 2 ............... 2 8 16 4
2 = r
1 1 r 2 2 2 16 16 S = r r 3 1 1 12 4 4 2 S 1 = Area of unshaded portion = r
11
S 1
r 2
12
r 2
r 2
r 2 12
11r
2
12
11 12
81. If 1, log9(31 – x + 2) and log 3(4.3x – 1) are in A.P., then x is is (1) log34
(2) 1 – log34
(3) log30.25
(4) lo log43
Sol. Answer (2) We have, 2log9(31 –
x
+ 2) = 1 + log 3(4.3 x – 1)
⇒
log3(31 –
⇒
31 –
⇒
3 + 2.3x = 12.(3x )2 – 3.3x
⇒
12.(3x )2 – 5.3x – 3 = 0
⇒
12y 12 y 2 – 5y 5 y – – 3 = 0, let 3x = y
⇒
12y 12 y 2 – 9y + + 4y – – 3 = 0
⇒
3y (4y (4y – – 3) + 1(4y 1(4y – – 3) = 0
⇒
y
3
⇒
x
x
+ 2) = log 33(4.3 x – 1)
+ 2 = 12.3
1
3
3
4
,
x
x
–3
3 4
x log log 3 = log 3/4
log3 3 – log3 4
⇒
x
⇒
1–log3 4
82. If b – c , 2b 2 b – x and and b – a are in HP. Then (1) AP
(2) GP
a
x , 2
b
x and 2
(3) HP
x are in 2
c
(4) None of these
31
32
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer (2) b – c , 2b – x , b – a are in HP 1 ⇒
c
b
1
,
2b x
2 ⇒
2b
x
1
b
1
,
c
ba
are in AP
1
b
a
⇒
2(b 2( b – c )(b )(b – a) = (2b (2b – x )(2b )(2b – a – c )
⇒
2(b 2( b2 – ab – bc bc + + ac ) = 4b2 – 2ab – 2bc bc – – 2bx bx + + ax ax + + cx 2ac ac = = 2b 2 b2 – 2bx 2 bx + + ax ax + + cx
⇒
x
(a c ) b 2 xb
⇒
2ac
⇒
ac
⇒
x x x a 2 c 2 b 2
⇒
a
2
x
2
x
2
(a c )
, b
x
2
x
2
4
, c
2
x
b 2 xb
x
2
4
2
are in GP.
83. Let log23 = , then log64729 is (1)) (1
(2)) (2
(3) 3
3
Sol. Answer (1) log2 3 = 6
6 log64729 = log(2 6 ) (3 ) =
84.
n
∑ n
C 0
n
C 1
........
n!
1
n
C n
6
log2 3 =
is equal to
(1)) e2 (1
(2)) e2 + 1 (2
(3)) e2 – 1 (3
(4) e–2
Sol. Answer (3)
∑ n
1
n
C0
n
n
C1 C2 n!
n
..........
Cn
n
2
∑ n!
n
1
=
2 1!
2
2
2!
3
2
3!
......... e2 1
(4) 2
Solution of Assignment (Set-2)
85.
∑ n
(ln x )n n
0
Sequences and Series
is equal to
!
(1) ln x
(2)) x (2
(3)) (3
1 ln x
(4)) (4
1 x
Sol. Answer (2)
n
(ln x ) ∑ n! 0
= 1
n
(ln x ) (ln x )2 .......... = 1! 2!
∑
86. The su sum of of
n
n
C 2 .
n
3
x .
equals
e
(1)) (1
ln x
2
n!
2
e
(2)) e2 (2
2
(3)) e3 (3
(4)) (4
1
e
2
Sol. Answer (4)
n
∑ n
n
C 2 .
3
2
=
n!
2
∑ n
1 3 2 ∑ = 2 2 (n 2)! n
n
( 1) 3n 2 . 2! n( n 1)( n 2)!
n n
2
1 3
0
2 0!
4
87.. Th 87 The e coe coeff ffic icie ient nt of of x x in
1
3
1!
3
2!
.... =
1 3 x x e
3 e
2
2
equals
x
12
(1)) (1
2
24
(2)) (2
15
25
(3)) (3
25 24
(4)) (4
15 12
Sol. Answer (3) (1 3 x x 2 ) e
x
= (1 3 x
(1 3 x x 2 )e
2 x
Coefficient of
x
x x 2 x 3 x 4 ) 1 ......... 1! 2! 3! 4! 4 x
1
4!
3 3!
1 2!
25 24
88.. Th 88 The e sum sum of th the e ser serie ies s 2 1!
4 3!
6 5!
(1)) e + 1 (1 Sol. Answer (3)
...... to equals (2) e – 1
(3) e
(4)) e–1 (4
3
33
34
Sequences and Series 2
4
1!
6
3!
Solution of Assignment (Set-2)
......
5!
2r 2r 1 1 1 1 (2r 1)! (2r 1)! (2r 2)! (2r 1)!
t r
S
∑ t =
1 1 ∑ (2r 2)! ∑ (2r 1)! 1 1
r
r
1
r
1
= 1
2!
r
1 4!
1 1 1 .......... 1! 3 ! 5 !
.........
=e
89.
2 1!
24
2!
246
3!
2468 4!
(1)) e (1
...... to equals
(2) 2e
(3) 3e
(4) 3e – 2
Sol. Answer (3) 2 1!
t r
2 4
2!
3!
2 4 6 ....2r
2 46
r !
t r
8
4!
......
2(1 2 3 ......r ) r ( r 1) r! r !
1 r 1 2 1 2 (r 1)! ( r 1)! ( r 2 )! ( r 1)!
1 1 2∑ 1)! 1 (r 2)! 1 (r 1)
∑ t ∑ r
1
r
S 1
=
2 46
r
S
r
1 1!
r
1 2!
1 1 2 1 ..... 1! 2!
.....
= e + 2e 2 e = 3e 1
90.
2! 1 1!
1 4! 1
3!
1 6! 1
......... ..........
5!
1 e
(1)) (1
1 e 1 e
(2)) (2
1 e
1 1 e e
(3)) (3
Sol. Answer (2) 1 2! 1 1!
1 4! 1 3!
1 6! 1 5!
e
..........
...........
e 1 2 e
1
e 1 2
e
=
2
2e 1 e
2
1
(e 1) 1)2 = (e 1)(e 1)
1 e 1 e
1 e 1 e
(4)) (4
Solution of Assignment (Set-2)
Sequences and Series
91.. Th 91 The e sum sum of th the e ser serie ies s
(a b)(a b )
(a b )(a b )(a 2 b 2 ) 2!
(a b )(a b)(a 4 b 4 a 2 b 2 ) ...... to 3! is equal to (1)) (1
e
a
2
eb
2
(2)) (2
e
a
2
b2
(3)) (3
e
a
2
b2
(4)) (4
e
a
2
eb
2
Sol. Answer (4)
(a b )(a b )(a2 b2 ) (a b)(a b) ......... 2! 2
=
a
=
e
a
2
b 2
a
4
b4 2!
a
6
b6
.......
3!
2
2
2 a4 a6 2 b 4 b6 ....... ....... a b = – 2! 3! 2! 3!
2
1 (eb 1) ea e b
2
3
1 a b a b 1 a b 92. The 92. The sum sum of the the ser serie ies s ...... will be 3 a a 2 a
b a
a b
(2)) loge (2
(1) logeab
(3)) loge (3
(4) logeab
Sol. Answer (3) 2
3
1 a b a b 1 a b 2 2 3 ........... = log a a
e
1
93.. Th 93 The e sum sum of se seri ries es
1 .2
1
2.3
(1) 2 loge2
1 3 .4
1 4 .5
a b 1 a log
1.2
1 2.3 1
=
1
=
1 2
2
(2) loge2 – 1
1 2
1 3.4
1 3
1 1 2 3
4 = loge e
1
4.5 1 3
1 4
(3) loge2
......... 1 4
1 4
1
.........
5
2) 2 lo log 2 1 = 1 2(1 log 2)
........
e
b a
log
a e
b
.......... is
Sol. Answer (4) 1
e
e
4 e
(4) loge
35
36
Sequences and Series
94.
2
n 1 is equal to (n 1)!
n
∑ n
Solution of Assignment (Set-2)
0
(1) 2e
(2) 2e + 1
(3) 2e – 1
(4) 1 – 2e
Sol. Answer (3)
∑ n
0
95. If
2
n 1 (n 1)!
n
x
1
1
y = = 1
1
3. 4 1
2. 3
n
1
1. 2
( 1) 1 ∑ (n 1)! (n 1)! = 0 n n
4. 5
5. 6
1 ∑ (n 1)! 0
n
1
∑ (n 1)! =
n
e
e 1 2e 1
0
...... , and
1
6. 7
...... , then
(1)) x = (1 = y
(2)) x > (2 > y
(3)) x < (3 < y
(4)) xy = (4 = 1
Sol. Answer (1) x
1 1.2
1
3.4 1
1
x
y 1
2
1 2.3
...........
5.6
1
1
3
1 4.5
1 4
1 5
1
1
6
..........
.........
6.7
1 1 1 1 1 1 ........... 2 3 4 5 6 7
y 1
=
1
1
2
1 3
1 4
1
.............
5
x = = y . 3 5 y y y .... 3 5 3 5 96. If x 2y = = 2 x – – y and and x |x | < 1, then x x x ...... 3 5
(1) 1
(2) –1
Sol. Answer (3)
1 y 1 y 3 5 3 5 1 x x x ............ log x 1 x 3 5 y
y
3
5
y
loge
............
e
2
x y
2x y
y
2 x 2
x
1
(3) 2
(4) –2
Solution of Assignment (Set-2)
1 y 1 2 x 2 x 1 2 x
1 y 1 x
2
1
1 x 1 x
1 y 1 y
x
1 2x x
x
2
Sequences and Series
2
2
1
1 2x x
2
1
( x 1)2 2
x
1
(1 x )2 2
x
1
2
2 1 y 1 x log 1 y 1 x 2 = 1 x 1 x log log 1 x 1 x
loge
e
e
e
97. Th The e num numbe bers rs lo log g180 12, log216012, log2592012 are in (1) A.P. (2) G.P .P.. (3) H.P. (4)) None of (4 of the abov above e progres progression sions s Sol. Answer (3) As log12 180 log12 25920 = log log12 ((180) 180) (259 (25920 20)) )) = 2 log log12 (216 (2160) 0) Hence the numbers, log12 180 180, log log12 2160 2160 and log12 259 25920 are in A.P.
log180 12, log2160 12, lo log25920 12 are in H.P.
98.
1 1 1 2 is equal to; 2 x 1 3 2 x 13 5 2 x 15
x x 1
x 1 x
(1)) log (1
(2)) log (2
Sol. Answer (2)
∵
3 5 x x 1 x x 2 1 x 3 5
log
2
1
2 x 1
1
3 2 x
1
3
1
5 2 x 1
5
1 1 2 x 1 l og 2 x 2 log x 1 log 1 2 x x 1 2 x 1
(3) log(2 x + + 1)
1 2 x 1
(4) log
37
38
Sequences and Series e
99. If
x
1 x
Solution of Assignment (Set-2)
B0 B1x B2 x 2 B
1
(1)) (1
n
n
Bn – Bn –1 equals;
1
(2)) (2
n!
x , then
(3)) (3
n 1 !
1 n!
1
n 1 !
(4) 1
Sol. Answer (1) e
x
1 x
⇒
B0 B1x B2 x 2 B
n
n
e
x
1 x
x 1 1!
1
x
B0 B1x B2 x 2 B
n
x
2
2!
n
x
n!
1 x x
2
n
x
x 3 x B0 B1x B2 x 2 B n
n
n
x
On equating coefficients of x n and and x x n – 1 on both sides, We get;
and
1 n!
1
1
n ! 1 ! 1
n 1 ! n 2 ! Bn
⇒
B
n
1
1 2!
1 2!
1 1!
1 1!
1 B
n
1 B
n
1
1 n!
100. 10 0. The valu values es of of x such such that log 1/2 x > > log1/3 x , is/are (1)) x ( (1 (0 0, 1)
(2) x [0, 1)
(3) x (0, 1]
(4) x [0, 1]
Sol. Answer (1) Using change of base formula; log1/ 3 x
log1/2 x 1 log1/ 3 .log1/ 2 x 2 log1/2 1 / 3
Now, the inequality can be considered as; 1 Now, log1/ 2 x 1 log1/ 3 0 2
ve
log1/2 x 0
1
…(1)
But original inequality is meaningful when x > 0
…(2)
⇒ x
x (0, 1) 101. 10 1. If the sum sum of first first n terms of an A.P. is cn2, then the sum of squares of these n terms is
(4n 2 1)c 2 6
n
(1)) (1
(4n 2 1)c 2 3
n
(2)) (2
(4n 2 1)c 2 3
n
(3)) (3
n
(4)) (4
[IIT-JEE 2009]
(4n 2 1)c 2 6
Solution of Assignment (Set-2)
Sequences and Series
39
Sol. Answer (3) S n = cn2 S n – 1 = c (n –1)2 t n = c [n2 – (n – 1)2] = c (2n (2n – 1) n
∑ t
So now
2
n
= c 2[12 + 32 + ........ + (2n (2 n – 1)2]
1
n
= c 2[12 + 22 + ........ + (2n (2n)2 – (22 + 42 + ........ + (2n (2 n)2)] 2
=
c
=
c
=
c
=
c
=
c
=
2
2
2
2
(2n )(2n 1)( 4n 1) – 22 (1 22 ........ n 2 ) 6 (n )(n 1)(2n 1) (2n )(2n 1)(4n 1) –4 6 6
(2n 1) 3 n( 4n 1) – 2n (n 1) (2n 1) 2 2 ( 4 n n – 2n – 2n ) 3 (2n 1) (2n 2 – n ) 3
2
(2n 1)(2n – 1) 3
2
(4n 2 – 1) 1) 3
c n
c n
=
102 10 2. Let Let a1, a2, a3, .... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which [IIT-JEE 2012] an < 0 is (1) 22
(2) 23
Sol. Answer (4) 1
As,
a20
1 25
1 5
19 d
1 1 25 5
19d
d
1
Also,
a
n
4 25 19
1 4 (n 1) 0 5 25 19
4 (n 1) 1 25 19 5
n
99 4
The least integral value of n is 25
(3) 24
(4) 25
40
Sequences and Series
Solution of Assignment (Set-2)
103 10 3. Le Lett bi > 1 for i = 1, 2, ..., 101. Suppose loge b1, log e b2 , ..., loge b101 are in Arithmetic Progression (A.P.) with the common difference log e2. Suppose a1, a2, ..., a101 are in A.P. such that a1 = b1 and a51 = b51. [JEE(Advanced)-2016] If t t = = b1 + b2 + ... + b51 and s = a1 + a2 + ... + a51, then (1)) s > t and a101 > b101 (1
(2)) s > t and a101 < b101 (2
(3)) s < t and a101 > b101 (3
(4)) s < t and a101 < b101 (4
Sol. Answer (2) b1, b2, b3, ..., b101 are in G.P. with common ratio = 2 Let a1 = b1 = a Given, a51 = b51
t
s
a 50d
a 250
a(251 1)
51
51
a1 a51
2
a
2
1 250
So, s > t
50
a101
a101
a 2(a 2
b101
So,
a 100d and
a101
Also,
b101
a)
100
a2
a 2100
251 1 2100
1
b101 SECTION - B
Objective Type Type Questions (More than one options are correct) 1.
If the the Arit Arithme hmetic tic mea mean n of of two two posit positive ive num number bers s a and b ( a > b) is twice their geometric mean, then a : b equals b equals 2
(1)) (1
3: 2
(2)) (2
3
7 4 3:1
(3)) (3
Sol. Answer (1, 2, 3) If A.M. = A, G.M. = G a
Then
b
A
A
2
G2
A
A
2
G2
or
A
A
2
G2
A
A
2
G2
But A But A = 2G a
b
2
3
2
3
a
also if a
then
b
b
or
2
3
2
3
2 3 2 3
2
3
2
3
(2 (2 3 ) (2 (2 3 )
,
74 3 1
1 74 3
1: 7 4 3
(4)) (4
2:
3
Solution of Assignment (Set-2)
2.
Sequences and Series
If th the fir firs st an and th the e (2 (2n n + 1)th terms of an A.P., a G.P. and an H.P. of positive terms are equal and their (n + 1)th terms are a, b and c respectively, respectively, then (1)) a = b = c (1
(4)) ac = (4 = b2
(3)) a b c (3
(2)) a + c = 2b (2
Sol. Answer (3, 4) If total term = (2n (2 n + 1) then middle term = (n ( n + 1)th Let first term of all progressions is A A and and last term is B. then A B
For A.P., a =
2
For G.P., b =
For H.P., c =
AB 2 AB A B
are A.M., G.M. and H.M. of A A and and B respectively. a, b, c are Hence, ac ac = = b2 and a b c 3.
41
If cos( – ), cos, cos( + ) are in H.P. and cos 1, then the angle cannot lie in the (1) I quadrant
(2) II quadrant
(3) III quadrant
(4) IV quadrant
Sol. Answer (1, 4)
2 cos( ) cos( ) cos( ) cos( )
cos =
cos 2 cos 2
cos =
2cos 2cos cos cos
2 cos2 cos = cos 2 + cos 2
2 cos2 cos = 2 cos2 + 2 cos2 – 2
2 cos2 (1 – cos ) = 2 sin2
cos
sin
2
sin
cos
2
2
2
2
cos
2 2 cos 2 cos 0 2 2
1 2
1 2
2
cos
cos
1 2
cos
2
1 2
3 2
Hence does not lie in I and IV quadrant.
4
2
3 4
2
2
2 sin2 4 sin2 cos 2 2 2 2
2
0⇒
2
n ⇒ 2n
42
4.
Sequences and Series
Solution of Assignment (Set-2)
If x , y , z are are in A.P. and x and x 2, y 2, z 2 are in H.P., then which of the following is correct? (2)
(1)) x , y , z ar (1 are all equal
x
2
, y , z a arre in G.P.
(3)
x
2
, y , z are in G.P.
(4) Eit ith her (1) or (2 (2)
Sol. Answer (4) 2y y = = x x + + z 2
y
2
… (i)
2
2 x z
x
2
… (ii)
z 2
By (i) and (ii)
( x z)2 2 x2 z 2 2 4 ( x z 2 )
( x 2 z2 2 xz)( x2 z2 ) 8 x2 z 2 ( x 2 z2 )2 2 xz( x2 z2 ) 8 x2 z 2 0
( x 2 z2 )2 4 xz( x2 z2 ) 2 xz( x2 ( x 2 z2 )( x 2 z2 4 xz) 2 xz( x2
2 z
2 z
) 8 x2 z 2 0
4 xz ) 0
( x 2 z2 2 xz)( x2 z2 4 xz ) 0 ( x z)2 (( x z)2 2 xz ) 0
x
4y
x
z
2
or ( x z)2 2 xz
2xz
z
2
or
y
xz 2
x
Hence x = = y = = z or or , y, z are in G.P. 2
5.
If a, b, c are are in H.P., then a
(1)) (1
(3)) (3
b
b
c a
a
b
,
b
2 2
,
c a
, c
b
2
c
b
,
a
b c
arre in G.P. a
Sol. Answer (1, 2, 3, 4) If a, b, c are are in H.P.
arre in H.P. a
(2)
(4)
2
b
1
b
a b
c
a
1
b
c
b ,
c a
c ,
a
b
are in H.P.
Solution of Assignment (Set-2) 1 1 1 , , are in A.P.
Then
(1)) (1
Sequences and Series
a b c
1 1 1 , , are in A.P.
a b c
abc a
a
abc
b c
a
c
b
a ,
b c
a
c
are in A.P.
2,
a
b c c
b c are in A.P.
c
a b
,
2 are in A.P.
c
b ,
b c c
c
b
a
or
ab
a b
,
a
a ,
b
2,
a
or
b c
,
a
are in H.P.
b c
Hence (1) is correct (2)) (2
2
b
2
b
1
ba
=
1
b
c
2b a c (b a)(b c )
2(b 2( b2 – bc bc – – ab + ac ) = 2b2 – ab – bc
2b2 – 2bc bc – – 2ab + 2ac ac = = 2b2 – ab – bc
– bc bc – – ab = 2ac
b =
2ac a
c
, hence a, b, c are are in H.P.
2
b b b (3)) a c (3 2 2 2
b
2
4
ac
b=
(4)) (4
ab 2
2ac a
c
bc 2
b
2
4
hence a, b, c are are in H.P.
1 1 1 , , are in A.P.
a b c
abc abc a a
,
b
,
b c c
are in A.P.
43
44
Sequences and Series
6.
abc a bc c ,
a a
a b c
1,
a b
b a
,
c
b ,
bc c
a
b
a
1,
a
b c c
1, are in A.P.
are in A.P.
c ,
Solution of Assignment (Set-2)
b
, are in H.P.
The sum of of the first ten terms of an A.P A.P.. equals 155 155 and the the sum of the first two two terms of of a G.P G.P.. equals 9. The first term of the A.P. is equal to the common ratio of the G.P. and the common difference of the A.P. is equal to the first term of G.P.. Given that the common difference of the A.P. is less than unity, which of the following is correct?
(1)) (1
25 2
25
is i s a term of the G.P.
(2)
is i s a term of the A.P.
(4)
25
(3)) (3
3
3
25 2
Sol. Answer (2, 4) Let A. P. is a, a + d , a + 2d , …………… S 10 =
10 2
[2a [2 a + 9d 9 d ] = 155
= 31 2a + 9d =
…(i)
Let G. G.P. P. is A, AR, AR 2, …… …… AR AR n –1 ……… A + AR = = 9
…(ii)
a = R
…(iii)
A = d
…(iv)
and it is given then d < < 1 A + Aa Aa = = 9
…(v)
By (i) and (ii), (iii), (iv) A
31 9 A A 9 2
A + + 18 = 0 9 A2 – 33 A 9 A2 – 27 A – 6 A A + + 18 = 0 3 A A(3A (3A – 9) – 2(3 A A – – 9) = 0
A = 3, A =
2 3
is a term of the G.P.
is a term of the A.P.
Solution of Assignment (Set-2)
Sequences and Series
45
if A if A = 3, then d = = 3, R R = = 2 not possible as d d < < 1. 2
if A =
3
, then d d = =
2 3
, R R = =
25 2
Terms Term s of G.P. G.P. = A A,, AR , AR 2 2 25 6 25 25 , , = 3 3 6
Terms Ter ms of A.P A.P.. = a, a + d , a + 2d 2 d =
Hence
and
7.
25 2
25 3
25 25 25 2 25 4 , , 2 2 3 2 6
is a term of G.P.
is a term of A.P.
In an equilate equilateral ral triangle triangle of side side 3 cm, a circle circle is inscri inscribed bed in which which again again an equila equilatera terall triangle triangle is inscrib inscribed ed and so on. This continues for an infinite number of times. Then (1)) Sum of (1 of areas areas of all circle circles s (in c cm m2) is
3
(2)) Sum of (2 of areas areas of of all circl circles es (in (in cm2) is (3)) Sum of areas of all equilat (3 equilateral eral triangles triangles (in cm2) is
3 3
(4)) Sum of areas of all equilat (4 equilateral eral triangles triangles (in cm2) is
3
Sol. Answer (2, 3)
A
3 (3)2 4
The area of ABC = = A1 =
P
r 1 = The radius of circle C1 = inradius of ABC 3
=
S
33
4 3 3 3
3
2
4
C 1 Q
9
9 2
3
B
2
Let the area of circle C 1 is B1 where B1 =
3 2
2
Again, let the side of equilateral triangle PQR = = a1 Circle C 1 is the circumcircle of PQR . r 1 =
a1 2sin60
T 1
C 2 T 2
R C
46
Sequences and Series 3
a1
=
2
⇒
3
2
a1
Solution of Assignment (Set-2)
3
2
2 3
Area of equilateral triangle PQR = = A2 =
4
3 2
2
Again the radius of C 2 is the in radius of PQR , Let that is r 2. 3
r 2 =
S
4
3 2
3 2
2
3
3 2
4
9 4
9 4
3
r 2 =
4
and so on Sum of areas of equilateral triangles = A1 + A2 + A3 + …………… 3
=
=
=
4
2 3 2 3 2 ���� 3 . . . . . . . . . . . . . . . 2 4
3 4
3 4
32 1
1 4
1
9
1
1
1 16
3
4
...............
9
4 3
3
4
Sum of areas of all circles = B1 + B2 + …………… upto = r 12 + r 22 + ……………….
=
=
3
3 2
2
................
3 4
2
1 1 ................ 4 16
1 1 = 3 4 3 cm2 1 3 1 4
3
cm2
Solution of Assignment (Set-2) 143
8.
Let
N
10
1
9
Sequences and Series
. Then which of the following is true?
(1)) N is (1 is prime (2)) N (2 N =(1+10 =(1+1012+1024+...+ 10132)(1+10+102+...+1010) (3)) N (3 N =(1+10 =(1+1011+1022+...+ 10132)(1+10+102+...+1010) (4)) N (4 N =(1+10 =(1+1013+1026+...+ 10130)(1+10+102+...+1012) Sol. Answer (3, 4) (1)) N is (1 is not a prime because it is divisible by 9. (2) (1 + 10 10 + 102 + 103 + ..……… + 10 132) (1 + 10 + 10 2 + ..………1010)
10133 1 1011 1 10143 1 = 12 9 10 1 102 1 Hence (2) is false. (3) (1 + 1011 + 10 22 + 10 33 + …………+10132) (1 + 10 + 10 2 + ………+1010)
(1011)13 1 1011 1 10143 1 1011 1 = 10 1 1011 1 10 1 1011 1
143
=
10
1
9
N
Hence option (3) is true. (4)) (1 10 (4 1013 1026 .......... 10130 )(1 10 10 102 ............ 1012 )
(1013 )11 1 1013 1 10143 1 N = 9 1013 1 10 1 Hence (4) is true 9.
/2 If logx a, ax /2 and logb x are are in GP, then x then x is is equal to
(1) loga(logba)
(2) loga(logea) – loga(logeb)
(3) – loga(logab)
(4) loga(logeb) – loga(logea)
Sol. Answer (1, 2)
47
48
Sequences and Series
logx a,
a
a
x 2
x 2
2
Solution of Assignment (Set-2)
, logb x are in GP
logx a logb x .
ax = logba
loga (logb a )
x
Now, loga (loge a) loga (loge b) 10. The sum sum of the squares squares of of three distinct real numbers numbers which which are are in GP is S 2, if their sum is S , then (1) 1 < 2 < 3
1
(2)
3
2 1
(3) 1 < < 3
Sol. Answer (1, 2) Let the numbers are,
a ,
a, ar
r
Now, we have, a
2
2
r
Let,
a2 a2r 2 S 2 and
1
r
x
,
x
2
as
r
a r
a ar S
1
r
2 1 r 2 a 2 S 2 r
a
2
a
2
[ x 2 2] a 2 S 2
a
2
[ x 2 1] S 2
…(I)
ax + + a = S
a2( x x + + 1)2 = 2S 2
…(II)
Dividing (I) by (II) we get, 2
x
1
( x 1)2
1
2
1 1 x 1 2 x
1 x 1 1 2 2 x 1 x 1 1 x
x
1
2
2 1
(4)
1 3
1
Solution of Assignment (Set-2)
r
1
r
Sequences and Series
2 1 2 1
(r 2 1)( 2 1) r ( 2 1) ( 2 1)r 2 r ( 2 1) ( 2 1) 0
D0
( 2 1)2 4( 2 1)2 0 2 2 2 2 1 2( 1) 1 2 2 0
(3 2 1)( 2 3) 1 3
2 3
1
2 But 1,
1 3
11.
3
, 3
2 1 ,
1
2 3 3
2
The Th e equ equat atio ion n (lo (log gx 10) – (logx 10) – 6 log x 10 = 0 is satisfied by a value of x of x given given by (1)) (1
3
10
(2)) (2
1 10
Sol. Answer (1, 2) Let logx 10 = y y 3 – y 2 – 6y = = 0 y (y ( y 2 – y – – 6) = 0
y (y ( y – – 3) (y ( y + + 2) = 0
= 0, y = = 3, y y = =– 2 y =
logx 10 = 0, 3, – 2 1
log10 x = = ,
log10 x x = = 1
1 3
3
,–
1
,
2 1
x = x = 103 ,10 – 2
1 2
(but is not defined)
(3)) (3
10
(4)) (4
1 3
10
49
50
Sequences and Series
Solution of Assignment (Set-2)
k ( k 1)
4n
12. Let
Sn
2
∑ ( 1)
2
k
[JEE(Advanced) 2013]
. Then S n can take value(s)
k 1
(1) 1056
(2) 1088
(3) 1120
(4) 1332
Sol. Answer (1, 4) k ( k 1)
4n
Sn
∑ ( 1)
2
2
k
k 1
Sn
–12 – 22 32 42 52 62 ........ ( 4n 3)2 ( 4n 2)2 (4n 1)2 (4n )2
S n
(32 1 ) ( 42 22 ) (72 52 ) (82 6 2 ) (112 92 ) (122 10 2 ) ...... (4 (4n 1)2 ( 4n 3)2 ( 4n)2 ( 4 n 2)2
S n = 2(1 3) 3 ) 2( 4 2) 2(7 5) 2(8 6) ...... 2( 4 n 1 4 n 3) 2( 4 n 4 n 2) S n = 2[1 2 3 ..... 4 n ]
2 4n( 4n 1) 2
From A From A,, 4n 4 n(4 (4n n + 1) = 1056 4n2 + n = 264 4n2 + n – 264 = 0 n=8 From B, 4n 4 n(4 (4n n + 1) = 1088
(Not poss possible) ible)
From C , 4n 4 n(4 (4n n + 1) = 112 1120 0
(Not poss possible) ible)
From D, 4n 4 n(4 (4n n + 1) = 1332 n=9
SECTION - C
Linked Comprehension Type Questions Comprehension-I The ubiquitous AM–GM inequality has many applications. It almost crops up in unlikely situations and the solutions using AM–GM are truly elegant. Recall that for n positive reals ai , i = 1, 2, ..., n, the AM–GM inequality tells n
∑
ai
1
n
1
n
1
n ai
The special case in which the inequality inequalit y turns into equality help solves many problems where at fir st we seem to have not enough information to arrive at the answer. 1.
If the equation x 4 – 4 x 3 + ax 2 + bx + + 1 = 0 has four positive roots, then the value of (1) 5
(2) –5
(3) 3
|a||b| a
(4) –3
b
is
Solution of Assignment (Set-2)
Sequences and Series
51
Sol. Answer (1) Let the roots are x are x 1, x 2, x 3, x 4 and all are positive By the properties of roots x 1 + x 2 + x 3 + x 4 = 4
…(i)
x 1 x 2 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 + x 3 x 4 = a
…(ii)
x 1 x 2 x 3 + x 1 x 2 x 4 + x 2 x 3 x 4 + x 3 x 4 x 1 = – b
…(iii)
x 1 x 2 x 3 x 4 = 1
…(iv)
By (i) and (iv) A.M. of x of x 1, x 2, x 3, x 4 = G.M. of x of x 1, x 2, x 3, x 4 Hence the roots are equal
x 1 = x 2 = x 3 = x 4 = 1
by (i)
a = 6, and b = – 4 | a | | b | 6 4 10 5 ab 64 2 2.
The Th e num numbe berr of of or orde dere red d pa pair irs s ( x x , y ) of real numbers satisfying the equation x 8 + 6 = 8 | xy| – xy| – y 8 is (1) 10
(2) 8
(3) 4
(4) 2
Sol. Answer (3) x
8
y 8 6 8xy
Here we use A.M. G.M. for
8
x y
8
,
and 6.
8
y 8 1 1 1 1 1 1 ( x 8 .y 8.1.1 .1.1.1 .1.1 .1.1 .1.1 .1))1/ 8 8
8
y 2 6 8 | xy |
x
x
… (ii)
By (i) and (ii) A.M. = G.M. number are equal Hence,
8 x
1
x
1
8
1
y
1
y
Hence total 4 order pairs (1, 1), (1, –1), (–1, 1), (–1, –1) are formed. 3.
If a, b, c are are positive integers satisfying a b
c
(1) Is
b c a
85 4
c a
b
3 2
, then the value of
(2) Is
17 4
abc
1
abc 65
(3) Is
8
(4)) Can (4 Can’t ’t be be dete determi rmined ned
52
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer (3) a bc
b c
a
c a
b
3 2
a b c 1 1 1 c a ab bc
1 (a b c ) ab
3 2
3
a bc
1 1 9 2 bc c a
bc
a bc c
a
a
b c a b
9 2
…(i)
Now let us take three numbers a + b, b + c , c c + + a and using A.M H.M. (a b) ( b c ) ( c a) 3
2(a b c ) 3
3 1 1 1 a b b c c a
3 1 1 1 a b b c c a
1 1 9 1 (a b c ) a b b c c a 2
…(ii)
By (i) and (ii) We find that A.M = H.M. Hence numbers are equal this is possible only when a = b = c Let a = b = c c = = 2
Then
abc
1 abc
8
1 8
65 8
Hence only (3) is correct
Comprehension-II One useful way of defining sequences is by a recursion relation. Many recurrence relations can be transformed to some known sequences, say GP or sometimes nth term can be found by algebraic jugglery. 1.
If an – an–1 = 1 for every positive integer greater than 1, then a1 + a2 + a3 + ....a ....a100 equals (1) 5000 . a1
(2) 5050 . a1
Sol. Answer (2) an –an –
1
=1
a2 = 2, a3 = 3, …….. an = n a1 + a2 + …. a100 = 1 + 2 + 3 + …..100 = 5050
(3) 5051 . a1
(4) 5052 . a2
Solution of Assignment (Set-2)
2.
Sequences and Series
Some chocolat chocolates es are distribu distributed ted between between 25 children children in such a way that first first child child gets 5 chocolates chocolates,, second second child child th th gets 7 chocolates and in general, n child gets n chocolates more than the ( n – 1) child. The total number of chocolates distributed is (1) 3250
(2) 2525
(3) 2750
(4) 3025
Sol. Answer (4) S = 5 + 7 + 10 + 14 + …………. T n
… (i)
Again S = = 5 + 7 + 10 + …. T n–1 + T n
… (ii)
By subtracting (ii) from (i) O = 5 + 2 + 3 + 4 + …… ( T n – T n–1) – T n T n = 5 + 2 + 3 + 4 + …… ( T n – T n–1) T n = 5 + [2 + 3 + 4 + ……. ( n – 1) terms]
n 1 (n 2) 2
= 5
S
1
T n
2
n
2
n8 2
( n 2 n 81) =
1 n(n 1)(2n 1) n( n 1) 8n 2 6 2
At n = 25 S
3.
53
1 25 25 26 51
2
6
25 26 2
8 25 3025
Let {an} be a sequence defined by a1 = 1 a1 + 2a2 + 3a 3 a3 + ... + (n (n – 1)a 1) an–1 = n2an, n 2 The value of a786 is 2
(1)) (1
1
(2)) (2
789
2
(3)) (3
393
Sol. Answer (4) a1 = 1 and a1 + 2a 2 a2 + 3a 3 a3 + ……………(n ……………(n – 1) an – 1 = n2an
…(i)
using (i) at n = 2, a1 = 4a 4 a2 a2 =
1 4
at n = 3, a1 + 2a2 = 9a3
1+2×
1 4
Similarly a4 =
an =
1+
= 9a3 1 8
, and so on.
1 2n
a786 =
1 2 786
1 1572
1 2
= 9a3 a3 =
1 6
393
1
(4)) (4
1572
54
Sequences and Series
Solution of Assignment (Set-2)
Comprehension-III If A,, G and H are If A are respectively arithmetic, geometric and harmonic means between a and b both being unequal and positive, then A
G
H
ab
a + b = 2 A
2
G2 = ab
ab 2ab
a
G2 = AH
b
On the basis of above information answer the following questions. 1.
If the the geom geometric etric and harm harmonic onic mea means ns of two two numb numbers ers are 16 and and
12
4 5
, then the ratio of one number to the
other is (1) 1 : 4
(2) 2 : 3
(3) 1 : 2
(4) 2 : 1
Sol. Answer (1) Since,
2
ab
16 and
256 ab
2ab ab
64 5
64 5
2 256 5
ab
ab
40 a = 8, b = 32
64
a : b = 1 : 4 2.
The sum sum of AM and GM of two positive positive numbers equal to the differe difference nce between between the numbers. numbers. The numbers numbers are in the ratio (1) 1 : 3
(2) 1 : 6
Sol. Answer (3) Since,
ab 2
2 ab
ab
ab
a 3b
4ab = a2 – 6ab + 9b2 10 ab + 9b2 = 0 a2 – 10ab
a2 – 9ab – ab + 9b2 = 0 a(a – 9b) – b(a – 9b) = 0 (a – b) (a ( a – 9b) = 0 a = 9b a : b = 9 : 1 (as a b)
(3) 9 : 1
(4) 1 : 12
Solution of Assignment (Set-2)
3.
Sequences and Series
55
The Th e num numbe bers rs wh whos ose e A. A.M. M. an and d G.M G.M.. are are A A and and G is
(1)) (1
A ( A
2
G2 )
(2)) (2
A
A
2
G2
(3)) (3
A
A
2
G2
(4)) (4
A
A
2
G2
2
Sol. Answer (3) Let two numbers be a and b ab
AM
and,
2
GM
A
G
ab
Now, (a (a – b)2 = (a + b)2 – 4ab 4 ab = 4 A2 – 4G2 = 4( A2 – G2)
a – b = 2
A
2
G2
A
2
G2
Now, a + b = 2 A a – b = 2 Solving, we get 2
G2
2
G2
a
A
A
b
A
A
and
Comprehension-IV While solving logarithmic equations or logarithmic inequalities care must be taken to ensure that the value of the variable obtained do indeed satisfy the given equation. Often the solution consists in transforming the original equation to a form which can be solved with ease. But in bargain the process of solution brings with it extrareous solutions, because the transformations carried carr ied out are not always equivalent. In what follows one must verify that the values of variables obtained indeed satisfy original equation or inequation. x
1.
How ma many ny solu solutio tions ns in in real real numb numbers ers doe does s the the equat equation ion (1) None
(2) Exactly one
Sol. Answer (3) x
3
x
2
8
x
2
x2
3x
=6
1
3x x
3 2 x2
2.3
31 x
Taking log on both sides with base 2,
3 8 2 x
x
6 have?
(3) Exactly two
(4) Infinite
56
Sequences and Series
Solution of Assignment (Set-2)
3x log2 3 1 (1 x ) lo x2
2(x 2(x 1) = (1 – x) log 2 3 x2
x – – 1 = 0,
2 x2
log log2 3
2 1 x = = 1, x 2, x 2 log 2 3 x 2 log3 2 2 x 2[log3 2 1] 2[log 3 2 log 3 3] 3]
x 2l 2log3
4 log3 9 3 2
= 1, x log3 22 2 x =
x = = 1, x log3
1 36
Equation has exactly two solutions. 2.
The Th e sol solut utio ion n of of the the in ineq equa uali lity ty x
log 1 ( x 2 x 1) 0 10
is given by (1) – < < x x < < 2
(2) – < < x x < < –1
Sol. Answer (2) x log
10 1
( x x 2 + x + + 1) > 0
log 10 ( x 2 + x + + 1) < 0 x log 2
1 here x here x + x + + 1 = x 2 2
Now again x log log 10 ( x 2 + x + + 1) < 0 here two cases arise
3 4
> 0, for all x all x R
(3) –3 –3 < x < < 1
(4) 1 < x < < 3
Solution of Assignment (Set-2)
Sequences and Series
(i) x > > 0
57
…(i)
and log10 ( x 2 + x + + 1) < 0
x 2 + x + + 1 < 1 x2 + x x < < 0 x ( x x + + 1) < 0 –
+
or x ( (– –1, 0)
…(ii)
+
–1
0
by (i) and (ii), we find that no value of x exists exists –
+
(ii) x (ii) x < < 0, and log 10 ( x 2 + x + + 1) > 0
+
–1
0
x < < 0, 0, x x 2 + x + + 1 > 10 0 and x 2 + x + + 1 > 1
x < x <0
x < < 0 and x 2 + x + + 1 > 1 x 2 + x x > >0 x < < 0 and x and x (– , – 1) (0, )
x (– , – 1) Hence option (2) is correct.
3.
2 Let S be be the set of all solutions solutions x x in in real numbers of the equation (log5 x ) log5 x
(1) 126
(2)
151 25
Sol. Answer (2) We have,
5
(log5 x )2 + log5x = 1 x = 1 (log5 x )2 + log5x 5 – log5x x = 1 1 (log5 x )2 + log 5 x – log 5 x = 1 5 x
1 1 – (log5 x ) + (1 log x ) (log 5) 5) 1 = 1 5 2
x
Let log5 x = = y y 2 +
1 1 y
–
1 1
y
y 2 +
1 1 y
–
=1
1
y y 1
=1
(3)) (3
131 25
5 x
1. Then
(4) 131
∑
x
x S
equals
58
Sequences and Series
Solution of Assignment (Set-2)
y 2 (1 + y ) + 1 – y = = (y + + 1) y 2 + y 3 + 1 – y = = y + + 1 y 3 – 2y + + y 2 = 0 y 3 + y 2 – 2y = = 0 y (y ( y 2 + y – – 2) = 0 y (y ( y + + 2) (y (y – – 1) = 0 y = = 0, y = = – 2, y y = =1 = 0, – 2, 1 log5 x = = 50, 5 –2, 5 x =
x = = 1,
1 25
,5
Sum of value of x = = 1 +
1 25
+5=6+
1 25
=
151 25
Comprehension-V Let V denote the sum of the first r terms of an arithmetic progression (A.P.) whose first term is r and the [IIT-JEE 2007] common difference is (2r (2 r – – 1). Let T = V +1 – V – 2 and Q = T +1 – T for r = = 1, 2,..... r
r
1.
r
r
r
n
(3)) (3
1 12 1 2
( 1)(3n 2 n 1)
n n
(2)) (2
(2n 2 n 1)
(4)) (4
n
1 12
( 1)(3n 2 n 2)
n n
1 ( 2n 3 2n 3) 3
T is always r
(1) An odd number 3.
r
The sum V 1 + V 2 + ..... + V is (1)) (1
2.
r
(2) An even number
(3) A prime number
(4) A composite number
Which Wh ich on one e of th the e follo followin wing g is a corre correct ct sta statem temen ent? t? (1)) Q1, Q2, Q3,.... are in A.P. with common difference 5 (1 (2)) Q1, Q2, Q3,.... are in A.P. with common difference 6 (2 (3)) Q1, Q2, Q3,.... are in A.P. with common difference 11 (3 (4)) Q1 = Q2 = Q3 = .... (4
Solution of Comprehension-V 1.
Answer (2)
2.
Answer (4)
3.
Answer (2) V turns out be r
r
2
(2r 2 r 1)
V 1 = 1 V 2 = 7 wrong choices can be eliminated using these two values. T = (r + 1) (3r (3r – – 1), obviously a composite number Q = T difference 6. r
r
r
+1
– T = 6r + + 5, which is an P with common r
Solution of Assignment (Set-2)
Sequences and Series
59
Comprehension-VI Let A1, G1, H 1 denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For n 2, let An – 1 and H n – 1 have arithmetic, geometric and harmonic means as An, Gn, H n [IIT-JEE 2007] respectively. 1.
Which Whi ch one one of of the the follo followin wing g stat statem emen ents ts is is corre correct ct? ? (1)) G1 > G2 > G3 > ... (1
(2) G1 < G2 < G3 < ...
(3)) G1 = G2 = G3 = ... (3
(4) G1 < G3 < G5 < ... and G2 > G4 > G6 > ...
Sol. Answer (3) 2.
Which Whi ch one one of of the the follo followin wing g stat statem emen ents ts is is corre correct ct? ? (1)) A1 > A2 > A3 > ... (1
(2) A1 < A2 < A3 < ...
(3)) A1 > A3 > A5 > ... and A2 < A4 < A6 < ... (3
(4) A1 < A3 < A5 < ... and A2 > A4 > A6 > ...
Sol. Answer (1) 3.
Which Whi ch one one of of the the follo followin wing g stat statem emen ents ts is is corre correct ct? ? (1)) H 1 > H 2 > H 3 > ... (1
(2) H 1 < H 2 < H 3 < ...
(3)) H 1 > H 3 > H 5 > ... and H 2 < H 4 < H 6 < ... (3
(4) H 1 < H 3 < H 5 < ... and H 2 > H 4 > H 6 > ...
Sol. Answer (2)
SECTION - D
Assertion-Reason Type Questions 1.
STATEM EMEN ENT T-1 : If a, b, c are are distinct positive reals in G.P., then log a n, logb n, logc n are in H.P., n N and STATEMENT-2 : The reciprocals of terms of Arithmetic progression with non-zero terms, form an H.P.
Sol. Answer (4) Statement 1 is false for n = 1 But statement 2 is clearly true.
2.
STA ST ATE TEMEN MENT T-1 : The The sum sum of of recipr reciproc ocals als of of first first n terms of the series
1
1 3
1 5
1 7
1 9
.... is n2.
and STATEMENT-2 : A sequence is said to be H.P. if the reciprocals of its terms are in A.P. Sol. Answer (2) Sum of reciprocals = 1 + 3 + 5 + …………… upto n terms =
n
2
1 1 (n 1) 2 n 2
Hence statement 1 is true and the statement 2 is clearly true.
60
3.
Sequences and Series
Solution of Assignment (Set-2)
STA ST ATEM TEMENT ENT-1 -1 : For thre three e posit positive ive une unequa quall quant quantitie ities s a, b, c are are in H.P., we must have a2008 + c 2008 > 2b 2 b2008. and STATEMENT-2 : A.M. G.M. H.M. for positive numbers.
Sol. Answer (1) Using A.M G.M. a
2008
c 2008
2
But
ac
a
2008
ac
2008
> b
c 2008 2
ac
2008
2b 2008
Hence statement 1 is clearly true statement (2) is also true and explains statement 1. 4.
STA ST ATEMEN TEMENTT-1 1 : The sum sum of 100 100 arithmet arithmetic ic means means between between two two given given numbers numbers 1000 1000 and and 3016 is 2008 200800. 00. and STATEMENT-2 : The sum of n arithmetic means between two given numbers is nth power of their single A.M.
Sol. Answer (3) 100
Sum =
2
1000 3016 3016 1000
=
401600 2
200800
hence statement-1 is true statement (2) is false because sum = n(single A.M.) 5.
STATEM EMEN ENT T-1 : If a, b, c , d are are positive and distinct numbers in H.P., then a + d > > b + c . and a
STATEMENT-2 : If a, b, c , d are are in H.P., then
d
ad
b
c
bc
.
Sol. Answer (2) ac
2 1 a
,
b
b,
1 b
,
1 c
ad ad
d 2
,
1
c
in AP
d
b
c
d > > b + c c [[∵ AM > HM] a + d
1
b
1
a
1
d
1
c
1
a
1
d
1
b
1
c
bc
Statement 1 and statement 2 both are true but statement 2 is not correct reason for statement 1
Solution of Assignment (Set-2)
Sequences and Series n
6.
n
STATEM EMEN ENT T-1 : Fo Forr n N , n > 1, 2 > 1 + n,
2
61
1
2
.
and STATEMENT-2 : A.M. of distinct positive number is greater than G.M. Sol. Answer (1) Statement 2 is obviously true 2
1 2 2 . . . . .. 2
n
1
n
n
2
n(n 1) 1 n 2 2
n
7.
n
2
1 n.2
1,2,2 2.......2n 1
1n
1
n
[AM > GM]
1 2 n.2 n
1
2
STA ST ATE TEM MEN ENT T-1 : The The su sum m of of n terms of two arithmetic progressions are in A.P. in the ratio (7n (7 n + 1) : (4n (4n + 17), then the ratio of their nth terms is 7 : 4. and STATEMENT-2 : If S n = ax 2 + bx bx + + c , then T n = S n – S n – 1 .
Sol. Answer (4) S n Sn '
n(7n 1) 7n 1 1) 4 n 17 n( 4n 17)
Sn
Tn Tn
8.
'
(7n 2 n ) Sn
Sn
S n 1
' S 'n 1
Sn '
( 4n 2 17n ) (4
7(2n 1) 1 14n 6 4(2n 1) 17 8n 13
STA ST ATEMENT TEMENT-1 -1 : If the the angle of a convex convex polygon polygon are are in A.P A.P.. 120°, 125°, 130°.... 130°....., ., then it has 16 sides. and STATEMENT-2 : The sum of the angles of a polygon of x sides sides is (n ( n – 2) 180°.
Sol. Answer (4) Let n be the number of sides n
2
[240 ( n 1)5 ] ( n 2)180 n2 – 25 25n n + 144 = 0
n = 9, 16 Largest angle 120° + 15 × 5° = 195° > 180° Which is not possible in a convex polygon
62
9.
Sequences and Series
Solution of Assignment (Set-2) 3
STA ST ATEM TEMENT ENT-1 -1 : If the the infini infinite te A.G A.G.P .P.. 1,
, 2, x 2, x ..... ..... has a finite sum, then x = = 2.
and STATEMENT-2 : The infinite A.G.P. a, (a ( a + d )r , (a ( a + 2d )r 2 ........ has a finite sum only if | r | < 1. Sol. Answer (1)
3 (1 (1 d )r , 2 (1 2d )r 2
1
r
2
r
2
2
r
r
x
2 3
3r
20
1,
3
2
3
d
r
r
3
1
1 Taking
3
d
3
1
3
1
1
2
53 3 3 (1 3d )r 3 1 ( 3 1) ( 3 1)3 ( 3 1)( 4 2 3 ) 2 2
(5 3 3 )( 3 1)(2 3 ) (3 3 5)(3 3 5) 27 25 2 10.. ST 10 STA ATE TEME MENT NT-1 -1 : If If ax = by = c z ; where x , y , z are unequal positive numbers and a, b, c are in G.P., then x 3 + z 3 > 2y 3. and STATEMENT-2 : If a, b, c are are in H.P. a3 + c 3 2 2b b3, where a, b, c are are positive real numbers. Sol. Answer (1) Given ax = by = c z = k
∵
∵
a, b, c are are in GP
2
y
Let
1
x
1
z
b a
3
z 3 ( 2
log a b c
k1 x c b
,
b
k
k1 y
1 y 1 x
Hence x Hence x , y , z are are in H.P.
y is is H.M. of x and and z
x
11.
a
3
xz )
log b c a
xz
y
y 3 x 3 + z 3 > 2y 3
log c a
b
STATEMENT-1 : a a b b c c = 1. and STATEMENT-2 : ab + c bc + a c a + b = 1.
,
c
k 1 z
k 1 z 1 y
Solution of Assignment (Set-2)
Sequences and Series
63
Sol. Answer (2) log a b
– c
=
log b c
–a
=
log c a
= K (Let) (Let)
–b
log a = K (b ( b – c ) log b = K (c ( c – – a) log c c = = K (a ( a – b) Also, a, b, c > 0 Statement-1 : Let y y = = aa bb c c log y y = = a log a + b log b + c log log c = aK aK ((b – c ) + b . K (c c – – a) + c c .. K K ((a – b) = 0
y = = 1, hence statement 1 is true. Statement-2 : Let y = ab + c . bc + a . c a + b log y y = = (b + c ) log a + (c c + + a) log b + (a + b) log c = (b + c ) . K K ((b – c ) + (c c + + a) K K .. (c c – – a) + (a + b) K . (a – b) = K K [[b2 – c 2 + c 2 – a2 + a2 – b2] = 0
y = = 1 Hence statement-2 is also true and explains statement-1. 12.. ST 12 STA ATE TEME MENT NT-1 -1 : log log416 is a rational number. and STATEMENT-2 : log23 is an irrational number. Sol. Answer (2) log4 16 = log4 4 2 = 2 log4 4 = 2, a rational number but log 2 3 is an irrational number. Because we may prove this by following method.
Let log2 3 is a rational number
log2 3 =
p q
3 =
p q
, where p where p and and q are integers and q 0.
p
2
q
3 q = 2 p, which is not possible for integers p p and and q, hence log2 3 is not a rational
number. log23 is an irrational number. 13. ST STA ATEMEN TEMENT-1 T-1:: Number Number of solutio solution n of of log| log x |x | = ex is two. and STATEMENT-2: If log303 = a, log305 = b than log308 = 3(1 – a – b).
64
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer (4) ex = log| x x | has only one solution y
x
=e
y
x
y
= log|x |
14.. Suppose 14 Suppose four four disti distinct nct posit positive ive numb numbers ers a1, a2, a3, a4 are in G.P. Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and [IIT-JEE 2008] b4 = b3 + a4. STATEMENT-1 STA TEMENT-1 : The numbers b1, b2, b3, b4 are neither in A.P. nor in G.P. and STATEMENT-2 STA TEMENT-2 : The numbers b1, b2, b3, b4 are in H.P. Sol. Answer (3) Let a1 = 1, a2 = 2, a3 = 4, a4 = 8
b1 = 1, b2 = 3, b3 = 7, b4 = 15 Clearly, b1, b2, b3, b4 are not in HP. Statement-2 is false.
SECTION - E
Matrix-Match Type Questions 1.
Match the foll llo owing Column-I
(A) If
1 1 1 , , ar are in H.P., then
a
b
c
Column-II
(p) loga x , logb x , logc x are are in H.P. (Assume that log functions are defined)
(B) If a, b, c ar are in A.P., then
(q)
(C) If a, b, c ar are in G.P., then
(r)
(D)) If sum of roots (D roots of the quadratic quadratic ax 2 + bx + + c = = 0
(s)
(abc 0) equals to the sum of squares of their reciprocals then
a
c a
b
,
2 2 a
,
b
ab
b
, c
b ,
c
ac
bc
b
2
are in A.P.
are in A.P.
bc ba ca ,
ca
,
cb
ab
are in A.P.
Solution of Assignment (Set-2)
Sequences and Series
Sol. An Sol. Answer swer A(q, s), B(q, s), C(p), D(r) (A)
1
If
a
1
,
1
,
b
a, b, c are in A.P. b
Subtract
a
are in H.P.
c
2
from each term
b b
b
2 2
2
, , c
are in A.P.
(B) a, b, c are in A.P. Divide each term by abc
1 bc
,
1
1
,
are in A.P.
ca a b
multiply each term by (ab ( ab + bc bc + + ca ca))
(C)
ab bc
ca
bc ab ac ac bc
bc ca
ca
bc ,
ab ab
ab
ca
1,
ca ca ,
ab
bc ca
,
ca
bc
1,
ab ac bc
ab ,
bc
ca
ab
ab ca ca
ab
1 are in A.P.
are in A.P.
If a, b, c are are in G.P. then log a, log b, log c are in A.P.
or
1 1 1 , , are in H.P. log a log b log c
log x log x log x , , are in H.P. log a log b log c
loga x , logb x , logc x are are in A.P. (D)
1
2
1
2
, where , are the roots
2 2
( )
( )2 ( )2 2 ( )2 b
2
–2
c
b a a a 2 c a 2
are in A.P.
65
66
Sequences and Series
Solution of Assignment (Set-2)
2
b – 2ac
b – a
a
2
c
2
a
2
2
b – 2ac
2
c
– bc 2 = b2a – 2a2c 2
2
2a c b a bc
2
2.
2
a b
b c
c a b ,
,
a b c
c a
are in A.P.
Matc Ma tch h the equ equat ation ions s wit with h the their ir so solut lution ions s Column-I (Equations)
1 1 .2 .3 .4
(A)
x
(B)
x
(C)) (C
x
(D)) (D
x
Column-II (Solutions)
1 2 .3 .4 . 5
... to 2
2 3 1 1.3 1.3.5 1.3.5.7 ... to 20
(q) 80
3 1 .2 .4
(r) 36
4 2 .3 .5
5 3 .4 .6
... to 58
2 1 3 1 4 1 2 . 3 ... to 20 5.7 3 1.3 3 3.5 3
Sol. Answer A(r), B(s), C(p), D(q)
(A)
x
1 1 .... 2 1.2. 3.4 4 2.3. 2.3.4. 4.5 5 1.2.3.
(B)
(p) 72
x
1 1 1 1 .... 2 2.3.4 3.4.5 3 1.2.3 2.3.4
x
1 1.2.3 2 3 1.2. 3
x
x
18 2 36 36
2 3 1 .... 20 1.3 1.3. 1.3.5 5 1.3. 1.3.5. 5.7 7
4 6 2 .... 20 2 1 .3 .3 1.3.5 1.3.5.7
x
(s) 40
Solution of Assignment (Set-2)
Sequences and Series
1 1 1 1 1 1 .... 20 1.3 1. 1.3.5 1. 2 1.3 1.3.5 1.3.5.7
x
x
2
(1) 20
x = = 40
(C)) (C
T n
(n 2) (n 2)2 n(n 1)( n 3) n( n 1)( n 2)( n 3)
=
( 1) 3n 4 n(n 1)( n 2)( n 3)
=
1 3 4 (n 2)(n 3) (n 1)( n 2)(n 3) n( n 1)( n 2)(n 3)
n n
Hence,
Sn
K
At n = 1, k =
S n
S
x
(D)) (D
29 36
1 3 4 [Using the properties of series] n 3 2(n 2)(n 3) 3( 3(n 1)(n 2)(n 3)
29 36
1 3 4 n3 2( 2(n 2)(n 3) 3( 3(n 1)(n 2)(n 3)
29 36
29 36 58 72
x
T n
1 n 1 . (2n 1)(2n 1) 3
n
3 1 1 1 = . 4 (2n 1) 4(2n 1) 3n
=
S n
1 1 1 1 1 3 1 1 . 1 . . = 4 2n 1 3 (2n 1) 1) 3 4 2n 1 2n 1 3 1
n
1 4 x
4
[1 0]
1 4
20 x = = 80
n
n
67
68
3.
Sequences and Series
Solution of Assignment (Set-2)
Matc Ma tch h the the se seri ries es wi with th th thei eirr sum sum Column-I
Column-II
(A)) 1.2. (A 1.2.3.4 3.4 + 2.3. 2.3.4.5 4.5 + ... ... + n(n + 1)(n 1)(n + 2)(n 2)(n + 3) 3) + ... to 11 terms
(B)
13 1
13 2 3
1 3
...
13 23 ... n 3 1 3 ... (2n 1)
... t to o 16 terms
(C) 1.3 + 2.4 2.4 + ... ... + n(n + 2) + ... to 30 terms ( 1)
n n
(D) 1 + 3 + ... +
2
+ ... to 60 terms
Sol. Answer A(s), B(r), C(p), D(q) (A) T n = n(n + 1) (n (n + 2) (n (n + 3)
Sn
n(n
1)(n 2)(n 3)(n 4) C 5
At n = 1 1.2.3.4
1 2 3 4
5
5
C = = 0
S 11
C
11 1 2 13 14 15 5
72072
13 23 ...............n 3 (B) T n = 1 3 5 .... ...... .... .... .... .... ...( .(2 2 n – 1)
n(n 1) 2
=
n
T n =
1 4
S 16 =
=
=
=
2
2
n 1
1 (n 1)2 4
2
1
22 32 42 ...............172 4 1
12 22 ...............172 1 4
1 17 17 18 35
4 1 4
6
1785 1
1
1784 4
446
(p) 10385
(q) 37 37820 (r) 446 (s) 72072
Solution of Assignment (Set-2)
Sequences and Series
(C)) T n = n(n + 2) = n2 + 2n (C S n = n2 + 2n ( 1)(2n 1) n(n 1) 6
n n
S n =
30 31 61
S 30 =
6
30 31
= 5 × 31 × 61 + 30 × 31 = 10385
(D)) T n = (D
( 1) 1 2 2
n n
n
2
n)
1
∑ n2 ∑ n 2
S n =
S n =
1 n( n 1)(2n 1) n(n 1) 2 6 2
n n
S 60
1 n 2 6
60 61 62 6
37820 4.
Match Mat ch the the follo following wing ent entries ries of colum column n I with with tho those se of of colum column n II. II. Column I (A)) If the root (A roots s of the the equatio equation n x 3 – 9 x 2 + 23 x – – 15 = 0
Column II (p)
2
are in AP, then their common difference is (B)) Num (B Number ber of of value values s of x of x for for which [ x x ], ], sgn x sgn x , { x x } { x x 0 0}}
(q) 5
are in AP, is (C) If a, b, c ar are in AP, then the straight line
(r) 1
ax + + by + + c = = 0 will always passed through a fixed point whose ordinate and abscissae are (D)) An infinite G.P (D G.P.. has the first term x term x an and sum 5, then
2
C1 P 0
(s) 2
x may may be (t) – 2
69
70
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer A(p, s, t), B(p, s), C(r, t), D(p, q, r, r, s) (A) The roots of the given given equation equation are 1, 1, 3, 5 Hence, common difference is = 2 (B)) Obv (B Obviou iously sly,, for x for x = = 2, [ x x ], ], sgn x sgn x and and { x x } Will be in A.P. Hence there are two values of x . (C) Sin Since ce,, a, b, c are in A.P.
2 b = a + c , then straight line ax + by + c = 0 will pass through (1, – 2) because it satisfy the condition a – 2b + c = = 0 or 2b 2 b = a + c (D) We have, have, sum sum of infinite infinite G.P
S
a , | r | 1 1 r , | r | 1
S
1 r
r
x 1 r
x
5
5
1
S
x
5
exists only when |r | r | < 1
5 x 5
1
x < <5 – 5 < 5 – x x < <0 – 10 < – x
0 < x x < < 10 5.
Match Mat ch the the follo followin wing g entrie entries s of col column umn I with with those those of colum column n II Column I
Column II
(A) Le Lett, S 1, S 2, .... be squares such that for each
(p) 2
n 1, the length of a side of S n equals the length of a diagonal of S n + 1 . If the length of a side of S 1 is 10 cm. The value of n for which area of S n is less than 1 sq.cm is 12
2
(B)) If the first (B first two two terms terms of a HP are
5
and
13
respectively. If the largest term is T , then [T [T ] is less than or equal to
(q) 4
Solution of Assignment (Set-2)
Sequences and Series
1 1 log2.5 ...... 2 3 3
(C) Th The e valu value e of (0.16)
is i s less than or
(r) 6
1 1 (3 d ) (3 2d ) ..... , then d is is greater 4 42
(s) 8
equal to (D) If 3
than or equal to (t) 9 Sol. Answer A(s, t), B(p, q, r, s, t), C(q, r, r, s, t), D(p, q, r, r, s, t) (A) We hav ave, e, Length of side of S n = Length of diagonal of S n + 1 = Length of side of S n
S n 1
Length of side of
(Length of side of S n + 1 )
2
2 for all n 1
Sides of S 1, S 2, ………, S n form a GP with common ratio
10
S n
Side of
2
1
n
1
10 n
2
S n
Area of
1
2
100
n
2
1
Now, S n < 1 100
n
2
1 2n – 1 > 100
1
n 8 (C) Le Let, t,
S
1 3
1 3
2
.......
1 3
1
1
Let
y
(D)) (D
0.4
1
3
2 log
0.4 log
4
2 1 2
–1
4
1 4
1 1 4
d d = =9
1
0 .4
4
0.4
d 1
2
(.16)
y
1
log2.5
3
2
8 [using
S
a
1 r
dr
(1 r )2
]
1 2
and first term 10
71
72
Sequences and Series
Solution of Assignment (Set-2) SECTION - F
Integer Answer Type Questions 1.
If H.M H.M.. of tw two o num numbe bers rs is is 4, 4, the then n A. A.M. M. 'A' and G.M. 'G' 'G' satisfy the relation 2 A + G2 = 27, then modulus of difference of these two numbers is
Sol. Answer (3) Let that numbers be a, b 2ab
Then
ab
A
a
4
b 2
,G
a b 2
2
ab
ab
2
27
a + b + ab = 27 a + b + 2( 2(a a + b) = 27 a + b = 9, ab = 18 (a – b)2 = (a ( a + b)2 – 9ab 9 ab = = 81 – 72 a–b=9 a–b=3 When a – b = 3 when a – b = –3 a + b = 9 , a = 3, b = 6 a = 6, b = 3 |a – b| = 3 2.
Let a1, a2, ......a ......a10 be in AP and n1, n2, ......n ......n10 be in HP, if a1 = n1 = 2 and a10 = n10 = 3 then a4a7 is ____
Sol. Answer (6) Let d the the common difference of the AP, then a10 = 3
a1 + 9d = = 3
a4 = a1 + 3d 3 d 3 3.
2 + 9d = = 3
1 9
2
1 3
1
Then
n10
1 n1
9D
d
7 3
1
Let D be the common difference of
n
,
1
1 n2
, .. . .. .
1 n
10
1 9
Solution of Assignment (Set-2)
1 n7
1 3
3.
2
9D
1
1 n1
1 2
Sequences and Series
9D
1
1
3
2
D
1 54
6D
1
6.
92 18
a4 .a7
54
1 2
1 9
7 18
7 3
18 7
6
If an be the nth term of an AP and if a7 = 15, then the value of the common difference that would make a2a7a12 greatest is ________.
Sol. Answer (0) Let d be be the common difference of the A.P. then, a2 a7 a12 = (15 – 5d 5 d ) 15 (15 + 5d 5d ) = 375 (9 – d 2) The value of a2 a7 a12 will be greatest when d = = 0 4.
73
If a, b, c , d are are in GP, then the value of (a ( a – c )2 + (b – c )2 + (b – d )2 – (a – d )2 is ________.
Sol. Answer (0) Let r be be the common ratio of G.P.
b = ar , c c = = ar 2, d d = = ar 3 ( b – d )2 – (a – d )2 (a – c )2 + (b – c )2 + (b = (a ( a – ar 2)2 + (ar – – ar 2)2 + (ar – – ar 3)2 – (a ( a – ar 3)2 = a2(1 – r 2)2 + a2 r 2(1 – r )2 + a2 r 2(1 – r 2)2 – a2(1 – r 3)2 = a2(1 – r )2 + [(1 + r )2 + r 2 + r 2 (1 + r )2 – (1 + r r + + r 2)2] = a2(1 – r )2 + [1 + r 2 + 2r 2 r + + r 2 + r 2 (1 + 2r 2r + + r 2) – (1 + r 2 + r 4 + 2r 2 r + + 2r 2 r 3 + 2r 2)] = a2(1 – r )2 + [1 + r 2 + 2r + + r 2 + r 2 + 2r 3 + r 4 – 1 – r 2 – r 4 – 2r – – 2r 3 – 2r 2] = a2(1 – r )2 (0) = 0
74
5.
Sequences and Series
Solution of Assignment (Set-2)
If the mth term of a HP be n and the nth term be m, then the (mn ( mn))th term is ________.
Sol. Answer (1) Let a be the first term and d be be the common difference of the corresponding A.P., then
1
n = mth term
Similarly
a (n
a (m
1)d
⇒
1)d
a (m
1)d
1 n
1 m
Solving above two equation we get,
a
1 mn
, d
1 mn
So, (mn (mn))th term of the H.P.
1 th
(mn )
1 1 mn
(mn 1).
1 1 mn
6.
term of corresponding A.P.
1
1
1 mn
1 1
1
mn
1 + logx
The Th e num numbe berr of of sol solut utio ion n of of x
________. __. = 10 x is ______
Sol. Answer (2) Taking log both sides. (1 + log x )log )log x = = (1 + log x ); x ); x > > 0
(1 + log x )(log )(log x – – 1) = 0 log x = = 1, –1
= 10, x =
1 10
Number of solution 2
7.
Let S k , k = 1, 2,.....100, denote the sum of the infinite geometric series whose first term is
1002 100 2 ∑ (k 3k 1) 1)S k is common ratio is . Then the value of k 100! k 1 1
k 1 k !
and the
[IIT-JEE 2010]
Solution of Assignment (Set-2)
Sequences and Series
75
Sol. Answer (4) We have, k 1 S k
k ! 1
1
k
1 (k 1)!
Now, (k 2 3k 1)S {(k 2)(k 1 ) 1} k
1 1 (k 3)! ( k 1)!
100
∑ | (k
2
1 1 99! 98!
3k 1)S k | 1 1 2
k 1
8.
1 (k 1)!
4
100
2
100!
1002 100 ∑ | (k 2 3k 1)S k | 4 100! k 1
Let a1, a2, a3,......, a11 be real numbers satisfying a1 = 15, 27 – 2a 2 a2 > 0 and ak = 2ak – 1 – ak – 2 for k = 3, 4, ...., 11. If
2 1
a
a22 .... a121 11
90, then the value of
a1
a2 .... a11 11
is equal to
[IIT-JEE 2010]
Sol. Answer (0) p
9.
Let a1, a2, a3, ..., a100 be an arithmetic progression with a1 = 3 and S p ∑ ai , 1 p 100 . For any integer i 1
n with 1 n 20, let m = 5n. If
S m S n
does not depend on n, then a2 is
Sol. Answer (9) We have, a1 = 3, S p = a1 + a2 + a3 + ...... + a p, 1 p 100 By the given condition, S m Sn
m n
6 (m 1) d 6 (n 1) d
k ,
which is independent of m and n
5(6 + (m ( m – 1) 1)d d ) = k (6 (6 + (n (n – 1) 1)d d ) 30 + 25nd 25nd – – 5d d = = 6k k + + nkd nkd – – kd which is independent of n
[IIT-JEE 2011]
76
Sequences and Series
Solution of Assignment (Set-2)
Hence, on comparing the coefficients of like terms, we get k = 25 30 – 5d 5d = = 150 – 25d 25 d
20 20d d = = 120 d = =6 d a2 = 9 10.. The minim 10 minimum um value value of the the sum of real real numbe numbers rs a–5, a–4, 3a 3 a–3, 1, a8 and a10 with a > 0 is
[IIT-JEE 2011]
Sol. Answer (8) For a > 0, a
5
a 4 3a 3 1 a 8 a10 8
8
a
5
a 4 a 9 1 a 8 a10
a–5 + a–4 + 3a–3 + 1 + a8 + a10 8 min( min(a a–5 + a–4 + 3a–3 + 1 + a8 + a10) = 8 11. A pa pack ck co cont ntai ains ns n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed [JEE(Advanced) 2013] cards is k , then k – 20 = Sol. Answer (5) The smallest value of n for which ( 1) 1224 2
n n
n(n + 1) > 2448
n > 49 For n = 50 ( 1) 1275 2
n n
So, k k + + (k + + 1) = 1275 – 1224 = 51 k = 25 k – 20 = 5 b
12. Let a, b, c be positive integers such that
a
is an integer. If a, b, c are in geometric progression and the
arithmetic mean of a, b, c c is is b + 2, then the value of
a
2
a 14 is a 1
[JEE(Advanced) 2014]
Solution of Assignment (Set-2)
Sequences and Series
77
Sol. Answer (4) a, ar , ar 2, 0 > 1 a
b c 3
r is is integer
2
b
a + ar + + ar 2 = 3( 3(ar ar + + 2) a + ar + + ar 2 = 3ar + + 6 ar 2 – 2ar + + a – 6 = 0 a(r – – 1)2 = 6 a = 6, r r = =2
So
a
2
a 14 = a 1
2
6
6 14 6 1
28 7
4
13. Suppose that all the terms terms of an arithmetic arithmetic progressio progression n (A.P.) (A.P.) are natural natural numbers. numbers. If the ratio of of the sum of of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 [JEE(Advanced) 2015]
and 140, then the common difference of this A.P. is Sol. Answer (9) 7 2 11 2
[2a 6d ]
[2a 10d ]
a 3d a 5d
6 11
6 7
a = 9d Also, 130 < a + 6d < < 140 130 < 15d 15d < < 140 d = =9 d
SECTION - G
Multiple True-False Type Questions 1.
STATE TEM MEN ENT T-1 : If x is is positive, then least value of
12
x
4 2
x
4 x
is 9.
STATEMENT-2 : If a2 + b2 + c 2 = 4, x 4, x 2 + y 2 + z 2 = 9 then maximum value of ax + + by + + cz is is 6. STATEMENT-3 : (1) F T F
2
3 2
2
4 3
+....... upto = 4.
2
(2) T T F
(3) T T T
(4) F F T
78
Sequences and Series
Solution of Assignment (Set-2)
Sol. Answer (3) Statement-1 : By
A.M. A.M. G.M G.M.
12 x
1 x 2
1
x
1
2
x
2
1 2 x
1 x
1 x
1 x
1 x
9
12 x
4
2 x
12 x
4 2 x
x
9
4
1
9
x
12
1
x 8 x 4
4
9
1
9
x
Statement-2 :
2 2 2 2 2 2 As, a b c x y z
ax by cz
2
ax by cz
2
0 36
ax by cz 6
Statement-3 : 3
S =
1
S
2
S
2
2
4 2
3
=
1 2
S
3
3
2
2
4
2
3
=
5
2
2
.....
4
.....
4
1 3
2
1 2
4
.....
S=2
Therefore
2
3 2
2
4 3
2
..... upto = 2 + 2 = 4
Statement –3 is true 2.
STATEM EMEN ENT T-1 : If a, b, c are are in A.P., b, c , a are in G.P. then c , a, b are in H.P. STATEMENT-2 : If a1, a2, a3,..... ,.....a a100 are in A.P. and a3 + a98 = 50 then a1 + a2 + a3 +.... +....+ + a100 = 2500. STATEMENT-3 : If a1, a2,....., ,.....,a a R then then n
(1) F T F
(2) T T F
a1
a2 a3 ........ a
n
n
a1 . a2 .... ...... ....a
n
(3) T T T
1/n
. (4) F F T
Solution of Assignment (Set-2)
Sequences and Series
Sol. Answer (2) Statement-1: b
c
a c
…(i)
2
…(ii)
ab
We have to prove
a
2bc
b c
from (1) & (2) we get
a = c required required result Statement–1 is true. Statement-2: a 3
a98 50
then,
a 1
a2 a3 ...... a100
= (a1 a 100 ) (a2 a 99 ) (a3 a 98 ) .. . ...50 terms 50 50
250 0
Statement–2 is true. Statement-3: If 3.
a1, a2 ,...., a
m
0
STATEMENT-1 : 12 – 22 + 3 2 ........ to 21 terms is 231 STATEMENT-2 : 13 – 23 + 33 – 4 3 ........ to 15 terms is 1856 STATEMENT-3 : 12 + 3 2 + 52 ........ to 8 terms is 689 (1) T T F
(2) T F F
(3) T F T
Sol. Answer (1) Statement-1: (12 – 22) + (32 – 42)…….(192 – 202) + 212 = (1 – 2) (1 +2) + (3 – 4) (3 + 4) + ……. = – (1 + 2 + 3+ ……+ 20) + 21 2 = 441 – 210 = 231 Statement-2 : 13+ 2 3 + 33 +……+ 153 – 2 (2 3 + 43 …….+ 14 3) 2
15 16 7 8 16(13 23 ..... 73 ) 120 2 16 = 2 2 = 1202 – 112 2 = 1856
2
(4) T T T
79
80
Sequences and Series
Solution of Assignment (Set-2)
Statement-3 : 12+ 2 2 + 32 ……+ 152 – (22 + 42 …….+ 142) 15 16 31 4 7 8 15 4(12 22 ..... 7 2 ) 40 31 6 6
=
= 40 (31 – 14) = 40 × 17 = 680 4.
STATEM EMEN ENT T-1 : If If lo log( x x + + z ) + log( x x – – 2y + + z ) = 2log( x x – – z ) then x then x , y , z are are in H.P. STATEMENT-2 : If p If p,, q, r in in AP and
STATEMENT-3 : If
ab
1 ab
(1) T T F
ax
px
(2) T F F
Statement-1 : log ( x + x + z ) + log ( x x – – 2y + + z ) = 2 log ( x x – – z ) x + + z ) ( x – – 2y + + z ) = ( x x – – z )2 ( x ( x x + + z )2 – ( x – – z )2 = 2y ( x x + + z ) 4 xz = = 2y ( x x + + z )
y
2 xz
x
z
Statement-2 : a x a y a z k (ay ) px qy rz
p ax
ax kx
a
y
,
a
x
,
a
y
1,
x
1
q
1
z
a
a
1,
y
,
y
1 z
ay ky
z
,
x
,
1 ab
are in A.P.
2b
1 bc
kz
1 are in A.P.
z
bc
a z
Statement-3 :
r
are in A.P.
x , y y ,, z are are in H.P.
ab
,
qy
c are in A.P. then 1 bc b
, b,
Sol. Answer (3)
ay
az rz
a,
, then x then x , y , z are are in A.P.
1 b
, c are in H.P.
(3) T F T
(4) T T T
Solution of Assignment (Set-2)
81
Sequences and Series
ab)) = 2b(1 – ab ab)) (1 – bc ) (a + b) (1 – bc ) + (b + c ) (1 – ab
5.
(1 + b2) (2abc (2 abc – – (1 (1a a + c )) )) = 0 1
b
2ac
a
c
, Hence a,
1
b
, c are in H.P.
STATEM STA TEMENT ENT-1 -1 : If the the arit arithm hmet etic ic mean mean of (b (b – c )2, (c c – – a)2 and (a (a – b)2 is same as the arithmetic mean of (b + c c – – 2a)2, (c ( c + + a – 2b 2 b)2 and (a (a + b – 2c 2 c )2 then a, b, c are are equal. STATEMENT-2 STA TEMENT-2 : If I f log x , logy logy , logz logz (( x x , y , z > > 1) are in GP then 2 x + + log(bx log(bx ), ), 3 x + + log(by log(by ), ), 4 x + + log(bz log(bz ) are in A.P. A.P. STATEMENT-3 : If n!, 3 × n! and (n (n + 1)! are in G.P. then n!, 5 × n! and (n (n + 1)! are in A.P. (1) T T F
(2) T F F
(3) T F T
(4) T T T
Sol. Answer (4) Statement-1 : (b c )2 (a b )2 (c a )2 (b c 2a )2 (c a 2b )2 (a b 2c )2 3 3
(b c 2a )2 (b c )2 (c a 2b )2 (a b )2 (a b 2c )2 (c a )2 0 a = b = c Statement-2 : If log x log x , log y , log z are are in A.P.
log (bx (bx ), ), log (by (by ), ), log (bz (bz ) are in A.P. Statement-3 : n!; 3 × n! and n + 1! are in G.P. 9(n 9( n!)2 = n! (n (n + 1)!
n + 1 = 9
n = 8
n! = 8! 5 × n! = 5 × 8!; n + 1! = 9! 9! + 8! = 5 × 9!
Hence n!, 5 × n! and n + 1! are in A.P SECTION - H
Aakash Challengers Questions 1.
The sum of n terms of the series
(1)) (1
1 x 1
k
, k 2 k x 1
n
(2)) (2
1 1 x
1 x 1
2 1 x k
2
4 1
, k 2 k x 1
n
4
............ is
x
(3)) (3
1 x 1
Sol. Answer (2) After adding
1 x
1
and subtracting it from given series, we get
k k 1
x
n
, k 2
(4)) (4
1 x 1
k k 1
x
, k 2
n
82
Sequences and Series 1
S
1
x
1
x
x
2.
1 1
1 1
x 1
1 1 x
2 1
x
2
4 1
x
4
1 1 x
K K
x
1
Solution of Assignment (Set-2)
2 1 x
2 1
x
2
4 1
2
4 1
4
4 4
1 x
............
............
x
............
4
x
K = K = 2n
The first first term term of a seque sequence nce is 1, the seco second nd is 2 and and every every term term is the the sum of of the two two precedi preceding ng terms. terms. The nth term is. 1
(1)) (1
n
2
1
1
(3)) (3
2n 1
(1 5 )n 1 (1 5 )n 1 5
(2)) (2
(1 5 )n 1 (1 5 )n 1 5
(4)) (4
Denoting the sequence by a1, a2, ….. we have an – an – 1 – an – 2 = 0 and a1 = 1, a2 = 2 Hence, an = An + Bn 2 = A2 + B2 where , are roots of
the equation x equation x 2 – – x x – – 1 = 0 eliminating A eliminating A,, B from these equations we have ( – ) an = (2 – )n –
1
– (2 – )n –
Now, + = 1, = – 1
a
n
5
1
5 1 n
2
1
( n 1 n 1)
(1 5
1
n
1
2
Sol. Answer (2)
1= A 1= A + B
1 n
5 )n 1 (1 5 )n 1
1
1 2
(1 5 )n 1 (1 5 )n 1 5
(1 5 )n 1 (1 5 )n 1 5
Solution of Assignment (Set-2)
3.
Sequences and Series
Sum of series 1 + 2 x x + + 7 x 2 + 20 x 3 + ........ up to n terms when x = = – 1 is (1)) (1
1 {4n 3 ( 3)n 1} 16
(2)) (2
1 {4n 3 ( 3)n 1} 16
(3)) (3
1 {4n 3 3n 1} 16
(4)) (4
1 {4n 3 3n 1} 16
Sol. Answer (1) The given series is of the form u 0 + u 1n + u 2n2 + ….. and let nth term satisfy the relation u n + pu n – 1 + qu n – 2 = 0 then 7 + 2 p p + + q = 0 and 20 + 7 p + 2q = 0
p = – 2, q = – 3 So, we have the relation u n – 2u 2 u n – 1 – 3u n – 2 = 0 The roots of of x x 2 – 2 x – – 3 = 0 are 3, – 1 Therefore, u n = A A.3 .3n + B(– 1)n ⇒ A
A + B = 1, 3 A – B = 2
n 1 Thus the nth term un 1x
3 4
,B
1 4
1 n [3 ( 1)n 1] x n 1 4
3 1 3n x n 1 1 ( x )n sum to n term 4 1 3 x 4 1 x 1 ( n )n 1 n n 1 lim
n
sum to n terms of the series 1 – 2 + 7 – 20 + ……. is
3 1 ( 3)n n 1 [ 4n 3 ( 3)n 1] 4 4 4 16
4.
(n 2 1) The value of ∑ is (n 2) n ! n 1
(1) 9 – e Sol. Answer (2)
(2)) (2
9 2
e
(3)) (3
9 2
e
(4) 9 + e
83
84
Sequences and Series
Let
S
n
2
Solution of Assignment (Set-2)
1
∑ n 2 n !
n
1
(n 2 1)(n 1) ∑ ( n 2) ! n 1
Now, ((n n2 + 1) (n (n + 1) = n3 + n2 + n + 1 = a + b(n + 2) + c (n + 2)(n 2)(n + 1) + d (n + 2)(n 2)(n + 1)(n 1)(n) a = – 5, b = 5, c = = – 2 , d d = = 1
1 1 1 1 5∑ 2∑ ∑ Thus, S 5 ∑ (n 2)! n 1! n! (n 1)! n 1 n 1
S
5 2) 2(e 1) e 5 e 5(e 2) 2
5e 5e 2e e
5.
9 2
25 2
10 2
e
The numbers x 1, x 2, x 3 ...... form an infinite decreasing G.P. If x If x = = 1, then the common ratio of the progression for which the expression 6 x 5 – 16 x 4 – 3 x 3 + 12 x 2 is maximum, is 1
(1)) (1
2
(2)) (2
3
(3)) (3
3
3 4
Sol. Answer (4) Let r be be the common ratio then x 2 = r , x 3 = r 2 …… Also as G.P. is infinite and decreasing, – 1 < r < < 1, r > > 0 so, 0 < r < < 1
…(i)
Now 6 x 5 – 16 x 4 – 3 x 3 + 12 x 2 = 6r 6 r 4 – 16 16r r 3 – 3r 2 + 12 12r r = = f (r ) f (r ) = 24r 3 – 48 48r r 2 – 6r + + 12 = 6[4 6[4r r 3 – 8r 2 – r + + 2] = 6 (2r (2r – – 1) (2r (2r + + 1) (r ( r – – 2)
f (r ) = 0 gives
r
1
1 , ,2 2 2
Within the domain specified by (I)
r
1 2
is the only admissible value r .
(4)) (4
1 2
Solution of Assignment (Set-2)
Sequences and Series
85
f (r ) = 6[12 6[12r r 2 – 16 16r r – – 11]
1 6[3 8 1] 0 2
f ''
So f (r ) is maximum at
r
1
2
Hence required common ratio
6.
1 2
If tot total al num numbe berr of of runs runs sc scor ored ed in n matches is
n
1 n 1 (2 n 2) where n 1, and the runs scored in the 4
k th match is given by k .2 .2n + 1 – k , where 1 k n, n is (1) 8
(2) 9
(3) 7
(4) 6
Sol. Answer (3) n
Sn
n 1 k
∑K 2
2n 1∑ K 2 k
k 1
2n 2 1
1 n
2
2n 2 2n 4
But
Sn
n
n n
2
1
2n 2 4 2n
8 n 1 n 2) (2 4
1 n 1 2 n) (2 4
…(ii)
n = 7
From (i) and (ii) n + 1 = 8
7.
…(i)
The sum S n where
( 1)n ( 1)n (1) (1) n 1! n 1!
T n
n
( 1)
n
2
n 1 is n!
( 1)n ( 1)n (2)) (2 1 n 1! n !
Sol. Answer (4)
T r
r
( 1)
r
2
r 1 r !
r 1 1 ⇒ ( 1)r r 1! ( r 1)! r !
( 1)r ( 1)r ( 1)r ( 1) r ( r 1! r 1! r 2 )! r !
( 1)n 1 ( 1)n 1 (3) (3) 1 ! ( n 1)! n
( 1)n ( 1)n 1 (4) 1 ! ( 1 ) ! n n
86
Sequences and Series
Solution of Assignment (Set-2)
( 1)r ( 1)r 1 ( 1)r 1 ( 1)r 2 ∑ ( r! r 1)! (r 1)! r 2 ! r 1 n
S n
( 1)n
8.
n
!
( 1)n 1 1 ( n 1)!
Find a three digit numb number er if its digits digits form a geomet geometric ric progre progression ssion and the digits digits of the the number number which is smaller by 400 form an A.P. is (1) 139
(2) 248
(3) 842
(4) 931
Sol. Answer (4) Let the digits be a, ar , ar 2, then the number is 100 a + 10 10ar ar + + ar 2 It is given 100a 100 a + 10 10ar ar + + ar 2 – 400 100(a 100( a – 4), ar , ar 2 are in A.P. 2ar = = ar 2 + (a – 4)
a(1 – r )2 = 4
⇒
(1 r )2
4 a
4
1 – r is is a rational number so
a
2 a
is rational
Since a > 4, a = 9 Hence
1 r
2 ⇒ r
3
1 3
So digits 9, 3, 1. Hence the number 931 9.
Let S n, n = 1, 2, 3... be the sum of infinite geometric series whose first term is n and the common ratio is 1 n
1
(1)) (1
lim
. Then
n
S1Sn
S2Sn 1 S3Sn 2 ...... Sn S1 is 2 2 2 S1 S2 ...... S n
1
(2)) (2
2
3 4
Sol. Answer (1) S n
n
1
⇒
1
n
Sn
n 1
1
S 1S n + S 2S n –
1
+ S 3S n –
2
+ …… ……S S nS 1
= 2. (n (n + 1) + 3.n 3.n + 4.(n 4.(n – 1) ……+ (n (n + 1)2
(3) 1
(4)
1 2
Solution of Assignment (Set-2) n
Sequences and Series
n
∑ (r 1)( n r 2) 6 ( n
r
2
S1
2
9 n 14 14)
1
S22 S32 ........... Sn2 22 3 2 .....( n 1) 2
(n 1)(n 2)( n 3) –1 6 (2n 2 9n 13) 6
n
2 S2Sn 1........Sn S1 n 9n 14 lim lim 2 2 2 2 n n 2n S1 S2 ......... Sn 9n 13
S1Sn
1 2
10. Let x , y (0, 1) such that there exists a positive number a( 1) satisfying log x a + logy a = 4logxy a. y
x x Then the value of is..... y Sol. logx a + logy a = 4 logxy a
log a log a 4 log a = log x log y log xy
1 1 4 = log x log y log x log y
log x log y 4 = (log x ) . (log log y ) (log x log y )
(log x x + + log y )2 = 4 log x . . log y (log
(log (log x x – – log y )2 = 0
log log x x = = log y x = = y
x y
= 1
y
11.
Let
x x 1 y = (1) = 1
a
1
x
1 x
(1)) a = 2b (1 Sol. Answer (2)
2
3
5
1 x and 2 2 3 1 x 5 1 x x
(2) a = b
b
x
2 3
x
3
1
x
5
(3) 2a = b
5
1 7
x
7
2 9
x
9
than
(4) 2a = 3b
87
88
Sequences and Series
Solution of Assignment (Set-2)
x 1 1 2 1 x x 2 1 1 x a log log 2 1 x x 2 1 x 2 1 x 2
3 x b x 3
x
x 3
5
5
x
9
3
15
x
5
1 x 3 1 x 1 log log 3 1 x 2 2 1 – x 1
1 x 1 log 2 1 x
2 x 2 x
Hence a = b
12.
1 1!
1 3 2!
x
1 3 5
3!
2 x
x
x
(1)) e (1 + x ) (1
x
(2) e (1 – x – x )
(3) xe
x
(4)) e ( x (4 x + 2)
Sol. Answer (1)
Tn
n
2
n!
x
n
1
n 2 !
1
n
n n
2
1 !
x n 1 ! 1
n
x
n
1
1
n 1 1 n
⇒ T n
1! x
n
n
x
1
2!
n
1
n
x n
1
1!
S = S = T 1 + T 2 + T 3 + ……
S
x
1!
x 1
x
2
2!
x
1!
1
2
x
2!
= xex + ex = ex ( x + x + 1) 13.. Let 2048 13 2048 arith arithmeti metic c means means be insert inserted ed betwee between n 2 22 – 1 and 2 22 + 1. Suppose the sum of these arithmetic S
means is S . Then
2
23
is ...
Sol. Sum of n, A.M.’s is n times the single A.M.
S =
2048[222 1 222 1] 2
S = = 1024 [2.222] = 1024.2 23 S 23
2
1024
Solution of Assignment (Set-2)
14. Let
S
Sol. S =
2
S
=
100
4
4
99
99
2 36
..........
4
2 99 6
...
98
4
1
1 6 100 4 4
2
2
99
4
3 4
98
2 100 6 4
..........
. Then 3S 3S equals. equals.
100 4
2S1 6S 2 S 1
1 100
4
1
4
1
10 100
4
1
4
1 4
1
4
4
100
2
4
1
S 2 × 4 =
4
99
4
4
4
10 1 00
4
1
1
– 3S 3 S 2 =
4
100
4
1
4
2
1 4
100
4
1
– 3S 2 =
1 1 4 4
3
1 – 3S 2 = 3 1 4
100
1
100
1 1 4 – 3S 2 = 3
4
98
...
99 4
…(i)
4
100 100
1 4
...............
1 1 4 1
99
3
100
.... ...... .... .... .... .... ....
99
1
4
4
Subtracting
S 2 – 4S 2 =
2
100
.... ...... .... .... .... .... ....
98
1
4
...............
98
1
1 1 4 3
3
S 2
and
1
2
99
99
1
1
..........
2
1 1 4
100
...
..........
99
1 1 4 4
1
S 1 =
4
1
1
=
2 2 6
1 4100
Where
S 1 =
26
Sequences and Series
100
100
100
100
100 100
1 100
4
100
…(ii)
10 0 4
89
90
Sequences and Series
Solution of Assignment (Set-2)
Now, S = S = 2S 1 + 6S 2
1 1 100 1 1 100 = 2 3 1 4 2 3 1 4 200 S = = 200 3S S = = 600 3
16
∑
15. The sum of
n
Sol.
tn
n
4
2
1
is
16n 4 1 1 16( 4n 2 1) 1)
1 1 ( 4n 2 1) 16 16( 4 n2 1) 1)
1 1 1 1 ( 4n 2 1) 16 32 2n 1 2n 1
16
∑ t
n
n
1 4n
4
4n 2 1
n
1
1 4 16 17 33 1 1 12376 1 16 375 16 6 33 32 33 33 1
16. Four different different integers integers form an an increasing increasing AP. One of these numbers numbers equal equal to the sum of squares squares of the other three numbers. Find the numbers. Sol. Let the numbers be a – d , a, a + d , a + 2d Since this is an increasing A.P. of integers, a, b Z and and d d > >0 Hence (a (a – d )2 + a2 + (a + d )2 = a + 2d i.e., 2 i.e., 2d d 2 – 2d + + 3a2 – a = 0
d
1 2 4 8(3a2 a) 4 1 1 2
1 2a
6a2
Since d is is positive integer 1 + 2a – 6a2 > 0 and perfect square Now 6 6a a2 – 2a – 1 < 0 2
1 1 1 a 6 36 6 0
Solution of Assignment (Set-2) 2 1 Or, a 6
7
a
6
1 6
7
6
2
6
7
Sequences and Series
since a is integer,
a = 0, then 1 + 2a 2 a – 6a2 = 1 which is a perfect square
then
1 (1 1) 1 or 0 since d > > 0 2
d
d = = 1
Hence the number – 1, 0, 1, 2
n
∑ r 17.
r
1
n
is equal to
∑ r r
4
2
1
Sol. ( x x + + 1)5 – x 5 = 5 x 4 + 10 x 3 + 10 x 2 + 5 x + + 1
[( x x + + 1)5 – x 5 ] = 5 x 4 + 10 x 3 + 10 x 2 + 5 x + + 1 n
Where stands for
∑ 1
x
( x 1)
1
5 x 4
5
n
5 x
4
10 10n 2 ( n 1)2 10n( n 1)(2n 1) 5 n( n 1) n 4 6 2
5n 4 10 n 3 10 n 2 5 n
5n( n 1) n 2 ( 1) ( 6n 3 9n 2 n 1) 6
n n
( 1)(2n 1)(3n 2 3n 1) 6 n n
x 4 x 2
3n(n 1) 1 5
5 2 5 2 n ( n 1) n( n 1)( 2 n 1) 2 3
91
92
Sequences and Series 2
Solution of Assignment (Set-2)
3
n
3 3 3 3 ...... .... .. ( 1)n and bn = 1 – an, then find the last natural number n0 such that 18. an .... 4 4 4 4 bn > an n n0.
2
Sol.
a
n
3
3 3 3 3 ...... .... .. ( 1)n .... 4 4 4 4
3 3 1 4 7
n
n
n 3 3 1 4 4
1
3 4
Since b = 1 – a and b > a n n n n n n 0
6
n 3 1 1 4 7
(– 3)n+1 < 22n – 1 When n is even, inequality is held n N and, for n is odd, inequality is held, when n 7 19. If a, b, c are are in AP and b, c , a are in GP, then show that c , a, b are in H.P. Find a : b : c Sol. a Sol. a + c = 2b
…(i)
ab = c 2
…(ii)
2bc c (a c ) by the (i) bc b c
ca ab b c
a by (ii)
c , a , b are in H.P. Eliminating of a from (i) and (ii) gives b(2 (2b b – c ) = c 2 bc – – 2b2 = 0 c 2 + bc (c c – – b) (c c + + 2 b) = 0
c b
2
∵
(b c )
a = 2b – c a b
2
c b
…(iii)
by (i)
4
= 4 : 1 : – 2 a : b : c =
