Close Arithmetic Progressions QA Test (1) Subjective Test Question 1 ( 2.0 marks) Determine whether the given sequences are arithmetic progressions or not. Give reasons to justify your answers. (a) 1, −5, −11, −17, −23 … (1 mark) (b) 62, 99, 133, 166, 185 … (1 mark) Solution: (a) Difference between the second term and the first term = −5 − 1 = − 6
Difference between the third term and the second term = −11 − (−5) = −11 + 5 = −6 Difference between the fourth term and the third term = −17 − (−11) = −17 + 11 = −6 Difference between the fifth term and the fourth term = −23 − (−17) = −23 + 17 = −6 The difference between the consecutive terms of the given sequence is constant. Thus, this sequence is an arithmetic progression with the common difference −6.
(b) Difference between the second term and the first term = 99 − 62 = 37 Difference between the third term and the second term = 133 − 99 = 34 Difference between the fourth term and the third term = 166 − 133 = 33 Difference between the fifth term and the fourth term = 185 − 166 = 19 The given sequence does not have a common difference. Thus, this sequence is not an arithmetic progression. Question 2 ( 2.0 marks) How many terms of the arithmetic progression 54, 50, 46 … should be taken so that their sum is zero? Solution: Here, first term, a = 54
Common difference, Let the sum of the first n terms of the given sequence be zero, i.e., S n = 0
Now, using the formula
, we get
⇒
0 = n × [108 + (n −1) × (−4)]
⇒
0 = 108 + (n −1) × (−4) [Since n ≠ 0]
⇒
−4n + 4 = −108
⇒
4n = 108 + 4
⇒
4n = 112
Hence, 28 terms of the given sequence should be taken so that their sum is zero. Question 3 ( 2.0 marks) Find the sum of the first 22 terms of an AP whose nth term is given as an = 2n + 3. Solution: The nth term of the AP is given as
For n = 1, a1 = 2 × 1 + 3 = 5 First term, a = 5
∴
For n = 2, a2 = 2 × 2 + 3 = 7
an
= 2n + 3
Common difference = a2 − a1 = 7 − 5 = 2
∴
Number of terms, n = 22 Using the formula
⇒
S 22
= 11 × (10 + 21 × 2)
⇒
S 22
= 11 × (10 + 42)
⇒
S 22
= 11 × 52
⇒
S 22
= 572
, we get
Hence, the sum of the first 22 terms of the AP is 572. Question 4 ( 2.0 marks) Find the number of integers between 100 and 500 that are divisible by 6. Solution: We can observe that 102 is the lowest integer greater than 100 that is divisible by 6 and 498 is the largest integer lesser than 500 that is divisible by 6.
The integers between 100 and 500 that are divisible by 6 form an AP as 102, 108, 114 … 492, 498
In order to determine the number of integers between 100 and 500 that are divisible by 6, we need to find the number of terms in the arithmetic progression mentioned above. The first term (a1) of this AP is 102 and the common difference (d ) is 6. Let the number of terms in the AP be n. The nth term of the AP is 498. Using the formula an = a1 + (n − 1) × d , 498 = 102 + (n − 1) × 6 ⇒
498 − 102 = 6n − 6
⇒
396 + 6 = 6n
⇒
402 = 6n
⇒
n
= 67
Hence, sixty-seven integers between 100 and 500 are divisible by 6. Question 5 ( 2.0 marks) Does 30 belong to the AP −52, −48, −44, −40 …? Solution:
First term, a = −52 Common difference, d = −48 − (−52) = −48 + 52 = 4 Let 30 be the nth term of the given AP. Using the formula an = a1 + (n − 1) × d , we have 30 = −52 + (n − 1) × 4 ⇒
30+ 52 = 4n − 4
⇒
82 = 4n − 4
⇒
82 + 4 = 4n
⇒
86 = 4n
⇒
n
=
Now, n cannot be fraction. It must be a natural number. Hence, 30 does not belong to the given AP. Question 6 ( 2.0 marks) Write the first four terms of an AP whose 7 th term is 15 and 11th term in −5. Solution:
Let the first term of the given AP be common difference be d .
a
and the
Using the formula an = a1 + (n − 1) × d , the 7th term of the AP can be written as: a7 = a ⇒
+ (7 − 1) × d
15 = a + 6d … (1)
Using the formula an = a1 + (n − 1) × d , the 11th term of the AP can be written as: a11 = a ⇒
+ (11 − 1) × d
−5 = a + 10d … (2)
On subtracting equation (2) from equation (1), we get 15 − (−5) = a + 6d − (a + 10d ) ⇒
15 + 5 = a + 6d − a − 10d
⇒
20 = −4d
⇒
d =
−5
Thus, the common difference of the AP is −5. Now, using d = −5 in equation (1), we have
15 = a + 6d ⇒
15 = a + 6 × (−5)
⇒
15 = a − 30
⇒
a
= 15 + 30 = 45
Hence, the first term of the AP is 45. Therefore, the first four terms of the given AP are: a1
= 45
a2
= 45 + (−5) = 40
a3
= 45 + 2 × (−5) = 35
a4
= 45 + 3 × (−5) = 30
Question 7 ( 3.0 marks) What is the sum of all 3-digit numbers that leave 2 as remainder when divided by 3? Solution: The smallest 3-digit number that leaves 2 as remainder when divided by 3 is 101.
The next number that will leave 2 as remainder when divided by 3 is 101 + 3 = 104
The next number that will leave 2 as remainder when divided by 3 is 104 + 3 = 107 Hence, the numbers forms an AP with the first term (a) as 101 and the common difference ( d ) as 3. The largest 3-digit number that leaves 2 as remainder when divided by 3 is 998. Hence, the last term (l ) of the AP = 998 Let the number of terms in the AP be n. Thus, l will be the nth term of the AP. Using the formula an = a1 + (n − 1) × d , we get 998 = 101 + (n − 1) × 3 ⇒
998 − 101 = 3n − 3
⇒
897 + 3 = 3n
⇒
3n = 900
⇒
n
= 300
Thus, the number of terms in this AP is 300. Now, using the formula
, we get
⇒
S 300
= 150 × 1099
⇒
S 300
= 164850
Thus, the sum of all 3-digit numbers that leave 2 as remainder when divided by 3 is 164850. Question 8 ( 3.0 marks) Classify each of the following arithmetic progressions as finite or infinite. Give reasons to justify your answers.
(a) Weights of the students of a class when arranged in ascending order: 40 kg, 42 kg, 44 kg … 56 kg, 58 kg, 60 kg (1 mark) (b) The taxi fare in a city is such that one is charged Rs.15 for the first kilometre and Rs.10 for every subsequent kilometre. The series formed by the fare to be paid after every kilometre. (1 mark) (c) Rupam planted two trees in her garden when he was 20 years old. She increases the number of trees he plants every year by two. She plants trees up to 45 years of age. (1 mark) Solution:
(a) Since the number of students in a class is a finite number, the number of terms in the sequence of their weights will also be finite. Hence, the series formed by the weights of the students of a class will be a finite AP. (b) The series formed by the fare to be paid after every kilometre will have infinite terms as there is no limit to the distance that can be travelled in a taxi ride. Hence, this series will be an infinite AP. (c) Rupam plants tress for next 25 years (45 − 20). Hence, the number of terms in this series will also be 26. Thus, it is a finite AP. Question 9 ( 3.0 marks) If the 12th and 19th terms of an AP are respectively 16 and 32, then show that the 5 th term of this AP is zero. Solution: Let the first term of the AP be difference be d .
a
and let its common
th
We know that the n term of an AP is given by an = a + (n − 1)d 12th term of the AP = a + (12 − 1)d = 16
∴
⇒
a
+ 11d = 16 ... (1)
19th term of the AP = a + (19 − 1)d = 32
∴
⇒
a
+ 18d = 32 ... (2)
Multiplying equation (1) with 2 and comparing it with equation (2), we get 2(a + 11d ) = a + 18d ⇒
2a + 22d = a + 18d
⇒
2a + 22d − a − 18d = 0
⇒
a
+ 4d = 0
Now, the 5th term is given by a + (5 − 1)d = a + 4d , and we know that its value is 0. Thus, the 5th term of the AP is 0. Question 10 ( 3.0 marks) The angles of a quadrilateral form an AP with a common difference of 20°. Find the angles. Solution: A quadrilateral has four angles. It is given that these angles form an AP.
Let the four angles be
a
− 2d , a − d , a, and a + d .
We know that the sum of the angles of a quadrilateral is 360°. ∴
(a − 2d ) + (a − d ) + (a) + (a + d ) = 360°
⇒
4a − 2d = 360°
It is given that the common difference of the AP is 20°. d = 20º 4a − 2 × 20º = 360° 4a − 40º = 360° 4a = 360° + 40º 4a = 400º ⇒
a
= 100°
Thus, the four angles are 100º − 2 × 20°, 100º− 20°, 100º, and 100º+ 20°, i.e., 60°, 80°, 100°, and 120°.
