Engineering Mathematics, 7 th edition, edition, ISBN 9781137031204 Programme 10: Sequences
Test exercise 10 1 (a) Find the next two terms terms and and form form of the general general term term for the following following sequence: 1, 3, 5, 7, . . .
Solution: Solution: Step 1: 1: The numbers increase by 2 at a time so the next two numbers in the sequence are 7 + 2 = 9 and 9 + 2 = 11 Step 2: 2: The form of the general term is obtained by recognizing that this is an arithmetic sequence: a + nd where the first term is a = 1 and the common difference is d = = 2. Step 3: 3: The general form is then: 1 + 2n 2n
Engineering Mathematics seventh Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (b) Find the next two terms and form form of the general term for for the following sequence:
Solution: Solution: Step 1: 1: The numbers increase by 3 at a time so the next two numbers in the sequence are 5 + 3 = 8 and 8 + 3 = 11 Step 2: The form of the general term is obtained by recognizing that this is an arithmetic sequence: a + nd where the first term is
and the common difference is d = = 3.
Step 3: The general form is then:
Engineering Mathematics seventh Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (b) Find the next two terms and form form of the general term for for the following sequence:
Solution: Solution: Step 1: 1: The numbers increase by 3 at a time so the next two numbers in the sequence are 5 + 3 = 8 and 8 + 3 = 11 Step 2: The form of the general term is obtained by recognizing that this is an arithmetic sequence: a + nd where the first term is
and the common difference is d = = 3.
Step 3: The general form is then:
Engineering Mathematics seventh Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (c) Find the next next two terms and form of the general general term term for the following sequence: 16, 8, 4, 2, . . . Solution: Solution: Step 1: The numbers decrease by a half at a time so the next two numbers in the sequence are and Step 2: The form of the general term is obtained by recognizing that this is a geometric sequence:
where the first term is
and the common ratio is
.
Step 3: The general form is then:
Engineering Mathematics seventh Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (d) Find the next two terms and form of the general term for the following sequence:
Solution: Step 1: Each number in the sequence is obtained by multiplying the previous number by minus 2 so the next two numbers in the sequence are and Step 2: The form of the general term is obtained by recognizing that this is a geometric sequence:
where the first term is
and the common ratio is
.
Step 3: The general form is then:
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (e) Find the next two terms and form of the general term for the following sequence:
Solution: Step 1: By inspection it is seen that each number in the sequence is obtained by adding the three previous numbers so the next two numbers in the sequence are and Step 2: The form of the general term is obtained by recognizing that this is a recursive sequence of order three and so needs three starting values: where
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (f) Find the next two terms and form of the general term for the following sequence:
Solution: Step 1: By inspection it is seen that each number in the sequence is obtained by adding the two previous numbers and multiplying that sum by 2 so the next two numbers in the sequence are and Step 2: The form of the general term is obtained by recognizing that this is a recursive sequence of order two and so needs two starting values: where
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
3 (a) Find the recursive description corresponding to the following prescription for the output of a sequence:
Solution: Step 1:
Step 2: Therefore the recursive form is: where
(found from the prescription)
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
3 (b) Find the recursive description corresponding to the following prescription for the output of a sequence:
Solution: Step 1:
Step 2: Therefore the recursive form is: where
(found from the prescription)
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
3 (c) Find the recursive description corresponding to the following prescription for the output of a sequence:
Solution: Step 1:
Step 2: Therefore the recursive form is: where
(found from the prescription)
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(a) Evaluate
Solution: Step 1:
Step 2:
Step 3:
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(b) Evaluate
Solution: Step 1: because 5 > 1
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(c) Evaluate
Solution: Step 1:
Step 2:
because 0 < 0.999 < 1
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(d) Evaluate
Solution: Step 1:
Step 2:
Because the sequence just oscillates between
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(e) Evaluate
Solution: Step 1:
Step 2:
Step 3:
Step 4: So
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(f) Evaluate
Solution: Step 1:
Step 2:
Step 3: So
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
Further problems 10 1 (a) Find the next two terms and form of the general term for the following sequence: 1, 6, 11, 16, . . . Solution: Step 1: The numbers increase by 5 at a time so the next two numbers in the sequence are 16 + 5 = 21 and 21 + 5 = 26 Step 2: The form of the general term is obtained by recognizing that this is an arithmetic sequence: a + nd where the first term is a = 1 and the common difference is d = 5. Step 3: The general form is then: 1 + 5n
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (b) Find the next two terms and form of the general term for the following sequence:
Solution: Step 1: The numbers increase by 2 at a time so the next two numbers in the sequence are
Step 2: The form of the general term is obtained by recognizing that this is an arithmetic sequence: a + nd where the first term is
and the common difference is d = 2.
Step 3: The general form is then:
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (c) Find the next two terms and form of the general term for the following sequence: 10, 1, 0.1, 0.01, . . . Solution: Step 1: Each term is one tenth the previous term so the next two numbers in the sequence are
Step 2: The form of the general term is obtained by recognizing that this is a geometric sequence:
where the first term is 0
and the common ratio is
.
Step 3: The general form is then:
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (d) Each term is one tenth the previous term so the next two numbers in the sequence are 1234.5, 123.45, 12.345, . . . Solution: Step 1: Each number in the sequence is obtained by dividing the previous number by 10 so the next two numbers in the sequence are 1.2345 and 0.12345 Step 2: The form of the general term is obtained by recognizing that this is a geometric sequence:
where the first term is
and the common ratio is
.
Step 3: The general form is then:
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (e) Find the next two terms and form of the general term for the following sequence:
Solution: Step 1: By inspection it is seen that each number in the sequence is obtained by subtracting the previous term from the one before that so the next two numbers in the sequence are and Step 2: The form of the general term is obtained by recognizing that this is a recursive sequence of order two and so needs two starting values: where
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
1 (f) Find the next two terms and form of the general term for the following sequence:
Solution: Step 1: By inspection it is seen that each number in the sequence is obtained by adding the two previous numbers so the next two numbers in the sequence are and Step 2: The form of the general term is obtained by recognizing that this is a recursive sequence of order two and so needs two starting values: where
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
3
(a) Solve the difference equation:
Solution: Step 1: Assuming
So that
then substituting into the difference equation gives:
.
Step 2: The general form of the solution to the difference equation is then:
Step 3: Applying the given conditions
: [1] [2]
From [1]
and substituting into [2] gives
.
Step 4: Therefore
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
3
(b) Solve the difference equation:
Solution: Step 1: Assuming
then substituting into the difference equation gives:
So that
.
Step 2: The general form of the solution to the difference equation is then:
Step 3: Applying the given conditions
: [1] [2]
From
[1]
and
substituting
into
[2]
gives
and therefore Step 4: Therefore
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
3
(c) Solve the difference equation:
Solution: Step 1: Assuming
then substituting into the difference equation gives:
So that
.
Step 2: The general form of the solution to the difference equation is then:
Step 3: Applying the given conditions
: [1] [2]
From
[1]
and
substituting
into
[2]
gives
and therefore Step 4: Therefore
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(a) Does the sequence
converge or diverge?
Solution: Step 1:
Step 2: Sequence diverges
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(b) Does the sequence
converge or diverge?
Solution: Step 1:
Step 2: Sequence diverges
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(c) Does the sequence
converge or diverge?
Solution: Step 1: Assume that the sequence converges to A. Then:
That is: so that Step 2: Sequence converges to
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(d) Does the sequence
converge or diverge?
Solution: Step 1: Since
then
and so sequence converges
Step 2: Sequence converges to 0
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
5
(e) Does the sequence
converge or diverge?
Solution: Step 1:
Each term of the sequence is 10 times greater than the previous term. Step 2: Sequence diverges
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
7 The first, third and sixth terms of an arithmetic sequence form three successive terms of a geometric sequence. If the first term of both the arithmetic and geometric sequence is 8, find the second, third and fourth terms and general term of the geometric sequences. Solution: Step 1: The terms 8, 8 + 2d , 8 + 5d are the first, third and sixth terms of the arithmetic sequence in question and are three successive terms of the corresponding geometric sequence. Step 2: So:
Step 3: That is: and so Step 4: Therefore: and so The common ratio is then
.
Step 5: The second, third and fourth terms of the geometric sequence are then:
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
9
For what value of
does
differ from 2 by 0.02 to 2 dp?
Solution: Step 1: When n = 0 the sequence term has the value
and:
Step 2: So the terms of the sequence are increasing but are always less than 2. So we need to know the value of n corresponding to a term value of 1.975, this being the first value that is equal to 1.98 to 2 dp: Step 3: That is
that is
Step 4: This gives: so that
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
11 If three numbers in geometric sequence are respectively subtracted from three numbers also in geometric sequence leaving remainders that are also in geometric sequence show that all three geometric sequences have the same common ratio. Solution: Step 1: Let three sequential numbers from two geometric sequences be given as: and Step 2: Then it is given that geometric sequence. That is if
also form three terms of another
then
and
.
Step 3: Furthermore: and Step 4: Therefore that is Step 5: So that:
Step 6: That is: so Step 7: Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
Therefore: and so That is,
so all three common ratios are the same.
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
13 If x , 5 x , 6 x + 9 form three successive terms of an arithmetic sequence, find the next four terms. Solution: Step 1: Let Step 2: Therefore:
Step 3: That is:
Step 4: The three given terms of the arithmetic sequence are then 3, 15, 27 and the next four are: 39, 51, 63, 75
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
15
The twelfth and nineteenth terms of an harmonic sequence are respectively and
. Find the fourth term.
Solution: Step 1: The reciprocals of an harmonic sequence form an arithmetic sequence so the twelfth and nineteenth terms of the corresponding arithmetic sequence are: 5 and
respectively.
Step 2: That is:
Step 3: Therefore: and so
and
Step 4: The fourth term of the arithmetic sequence is then
Step 5: The fourth term of the harmonic sequence is then
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
17
If
form three successive terms of an arithmetic sequence
show that sequence.
also form three successive terms of another arithmetic
Solution: Step 1: Let:
Then:
Step 2: That is:
Step 3: So that:
Step 4: And hence: that is Step 5: Therefore if: then
Step 6: Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
That is: form three successive terms of an arithmetic sequence
Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204
19
Show that the difference equation
can be derived
from the coupled difference equations:
Find
and
given that
and
.
Solution: Step 1: and so Step 2: As a consequence:
Step 3: Since then
, that is
Step 4: Assume
and substitute into the difference equation to obtain:
That is:
Step 5: Therefore: and so Step 6: Applying the conditions: Engineering Mathematics seventh edition © K.A. Stroud and Dexter J. Booth 2013, published by Palgrave Macmillan, ISBN: 9781137031204