What is 4.5: Geometric sequences and series?

A geometric sequence has constant ratio r and nth term u_n=ar^n-1, where a is the first term.

How many marks are 4.5 questions worth?

4.5 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 4.5 appeared in?

We have not yet indexed a verified past-paper appearance for 4.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

4.5

Understand and work with geometric sequences and series, including the formulae for the nth term and the sum of a finite geometric series; the sum to infinity of a convergent geometric series, including the use of |r| < 1; modulus notation.

Draft — not yet indexed

Geometric sequences and series

Worked answers and methods for 4.5 on Edexcel A-level Maths 9MA0.

Explanation

A geometric sequence has constant ratio $r$ and nth term $u_n=ar^{n-1}$ , where $a$ is the first term.
For $r\ne1$ , subtracting $rS_n$ from $S_n$ proves $S_n=a(1-r^n)/(1-r)$ ; an equivalent form may be more convenient when $r>1$ .
A geometric series has a finite sum to infinity only when $|r|<1$ , in which case $S_\infty=a/(1-r)$ ; for a finite-sum threshold, isolate $r^n$ and use logarithms.
A negative ratio makes term signs alternate, but convergence still depends on its modulus.
A common error is to use $r<1$ instead of $|r|<1$ , which would wrongly accept ratios below $-1$ .

Worked example

Find the exact sum to infinity of $12-3+0.75-\cdots$ .

1.The common ratio is $r=-3/12=-1/4$ , and $|r|=1/4<1$ , so the series converges.
2.Its sum is $S_\infty=12/[1-(-1/4)]=12/(5/4)=48/5$ .

Answer: $\dfrac{48}{5}$

Common mistakes

Don't use $ar^n$ rather than $ar^{n-1}$ for the nth term of a geometric sequence.
Don't apply the sum-to-infinity formula without checking that the common ratio has modulus less than one.

Exam tip

State the common ratio and verify convergence before calculating an infinite geometric sum.

Worked practice

Tier 1 · Easy

Find the

8

th term of the geometric sequence

3,6,12,\ldots

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$384$	2
Notes
The first term is $a=3$ and the common ratio is $r=2$ . Hence $u_8=3\times2^7=384$ .

(2 marks)

Tier 2 · Standard

A geometric sequence has positive common ratio. Its third term is

18

and its sixth term is

486

. Find the first term, the common ratio and the sum of the first five terms.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	First term $2$ Common ratio $3$ $S_5=242$	5
Notes
With first term $a$ and ratio $r>0$ , $ar^2=18$ and $ar^5=486$ . Dividing gives $r^3=27$ , so $r=3$ . Then $a=18/3^2=2$ . Therefore $S_5=2(3^5-1)/(3-1)=2(242)/2=242$ .

(5 marks)

Tier 3 · Hard

A geometric series has first term

800

and common ratio

0.9

. Find the least value of

n

for which the sum of the first

n

terms exceeds

7000

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$n=20$	5
Notes
The finite sum is $S_n=800(1-0.9^n)/(1-0.9)=8000(1-0.9^n)$ . The condition $S_n>7000$ gives $0.9^n<0.125$ . Taking logarithms and accounting for $\ln(0.9)<0$ gives $n>\ln(0.125)/\ln(0.9)=19.73\ldots$ . Therefore the least integer is $n=20$ .

(5 marks)

Tier 1 · Easy

The three positive numbers

12

k

and

27

, in that order, are consecutive terms of a geometric sequence. Find

k

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$k=18$	2
Notes
For three consecutive geometric terms, the square of the middle term is the product of its neighbours. Thus $k^2=12\times27=324$ . Since $k$ is positive, $k=18$ .

(2 marks)

Tier 2 · Standard

A convergent geometric series satisfies

S_\infty=20

and

S_2=\dfrac{84}{5}

. Its ratio

r

is positive. Determine

r

and the initial term

a

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$a=12$ $r=\dfrac25$	4
Notes
From the sum to infinity, $a/(1-r)=20$ , so $a=20(1-r)$ . Also $a(1+r)=84/5$ . Substitution gives $20(1-r^2)=84/5$ , hence $r^2=4/25$ . The ratio is positive, so $r=2/5$ , and then $a=20(3/5)=12$ .

(4 marks)

Tier 3 · Hard

A convergent geometric series has positive first term

a

and common ratio

r

. The sum of its first two terms is

12

, and the sum of its first four terms is

15

. Find all possible pairs

(a,r)

and show that both series have the same sum to infinity.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$(a,r)=(8,\dfrac12)$ or $(24,-\dfrac12)$ Both sums to infinity are $16$	6
Notes
Since $S_4=S_2(1+r^2)$ , $15=12(1+r^2)$ , giving $r^2=1/4$ and hence $r=\pm1/2$ ; both satisfy $\|r\|<1$ . From $a(1+r)=12$ , $r=1/2$ gives $a=8$ , while $r=-1/2$ gives $a=24$ . The sums to infinity are respectively $8/(1-1/2)=16$ and $24/[1-(-1/2)]=16$ .

(6 marks)

Tier 2 · Standard

Express the recurring decimal

0.234234234\ldots

as a fraction in its simplest form by modelling the repeating blocks as a geometric series.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\dfrac{26}{111}$	3
Notes
The repeating blocks give the geometric series $234/1000+234/1000^2+\cdots$ , with ratio $r=1/1000$ . Since $\|r\|<1$ , its sum to infinity is $(234/1000)/(1-1/1000)=234/999=26/111$ .

(3 marks)

Tier 3 · Hard

The infinite geometric series is

(2x-1)+(2x-1)\left(\dfrac{x}{3}\right)+(2x-1)\left(\dfrac{x}{3}\right)^2+\cdots

. Find the set of real values of

x

for which the series converges and its sum lies strictly between

4

and

10

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\dfrac32<x<\dfrac{33}{16}$	5
Notes
Convergence requires $\|x/3\|<1$ , so $-3<x<3$ . On this interval, $3-x>0$ and the sum is $S_\infty=(2x-1)/(1-x/3)=3(2x-1)/(3-x)$ . Solving $S_\infty>4$ gives $x>3/2$ , while $S_\infty<10$ gives $x<33/16$ . Both bounds lie inside the convergence interval, so $3/2<x<33/16$ .

(5 marks)

Tier 3 · Hard

For the first six terms of a geometric sequence, the sum of the terms in odd-numbered positions is

42

and the sum of the terms in even-numbered positions is

21

. Find the first term and the common ratio. Hence find the sum of the first six terms and the sum to infinity.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	First term $32$ ; common ratio $\dfrac12$ $S_6=63$ $S_\infty=64$	6
Notes
If the first term is $a$ and ratio is $r$ , the odd-position sum is $a(1+r^2+r^4)=42$ . The even-position sum is $ar(1+r^2+r^4)=21$ , so division gives $r=1/2$ . Hence $a(1+1/4+1/16)=42$ , so $a(21/16)=42$ and $a=32$ . The first six terms have total $42+21=63$ . Since $\|r\|<1$ , the sum to infinity is $32/(1-1/2)=64$ .

(6 marks)

Q10

Tier 3 · Hard

10.

For a real constant

k

, consider the geometric series with first term

5

and common ratio

q=k^2-4k+2

. Determine all values of

k

for which the series converges, and find its sum in terms of

k

on this set. Classify what happens at every boundary value of the convergence set, and explain why the series diverges for all remaining values of

k

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$2-\sqrt3<k<1$ or $3<k<2+\sqrt3$ $S_\infty=\dfrac{5}{3-(k-2)^2}$ At $k=1$ or $k=3$ , the partial sums alternate between $5$ and $0$ ; at $k=2\pm\sqrt3$ , the partial sums are $5,10,15,\ldots$	6
Notes
Since $q=(k-2)^2-2$ , convergence requires $-1<(k-2)^2-2<1$ . This is equivalent to $1<(k-2)^2<3$ , giving $2-\sqrt3<k<1$ or $3<k<2+\sqrt3$ . On this set, $S_\infty=5/(1-q)=5/[3-(k-2)^2]$ . At $k=1$ or $k=3$ , $q=-1$ , so the partial sums alternate between $5$ and $0$ and have no limit. At $k=2\pm\sqrt3$ , $q=1$ , so every term is $5$ and the partial sums grow as $5,10,15,\ldots$ . For every other value of $k$ , $\|q\|>1$ , so the terms do not tend to zero and the series diverges.

(6 marks)

Verified exam appearances

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