Geometric Progressions and Sum to Infinity

Find terms and sums of a geometric progression, decide whether it has a sum to infinity and find it, and solve problems linking a GP with an AP.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 32 real past-paper questions below

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Add or multiply?

Fold a sheet of paper in half, then in half again. Each fold doubles its thickness: 1, 2, 4, 8 layers. You multiply by the same number every time.

A list like this is a geometric progression, or GP. The number you multiply by is the common ratio, written $r$ . In an arithmetic progression (AP) you add the same number $d$ instead.

AP: $5, 8, 11, 14, \dots$

Subtract neighbours: always 3. So $d = 3$ .

GP: $5, 10, 20, 40, \dots$

Divide neighbours: always 2. So $r = 2$ .

Rule

To test a list: subtract neighbours. Same answer each time? It is an AP. Divide neighbours. Same answer each time? It is a GP.

Any term divided by the term before it gives the ratio:

Formula

r = \frac{u _{2}}{u _{1}}

$r$ — the common ratio. It can be a fraction or negative.
$u_{1}, u_{2}$ — the 1st and 2nd terms. Any term ÷ the term before it works too.

To reach the 2nd term you multiply the first term by $r$ once. To reach the 3rd term, twice. So the nth term needs $r$ exactly $n - 1$ times.

Formula

u_{n} = a r^{n - 1}

$a$ — the first term.
$r$ — the common ratio.
$n$ — the position of the term: 1st, 2nd, 3rd and so on.

Worked example

1
A GP starts $48, - 24, 12, \dots$ Find the common ratio and the 8th term.
2
$r = \frac{- 24}{48} = - \frac{1}{2}$ — Divide the 2nd term by the 1st. The signs swap each time, so r is negative.
3
$u_{8} = 48 (- \frac{1}{2})^{7}$ — The 8th term needs r seven times: 8 − 1 = 7.
4
$(- \frac{1}{2})^{7} = - \frac{1}{128}$ — An odd power of a negative number stays negative.
5
$u_{8} = - \frac{48}{128} = - \frac{3}{8}$ — Multiply and cancel by 16.

Rule

0606 and 4037 print $u_{n} = a r^{n - 1}$ in the List of formulas. 4PM1 does not print it, so learn it.

Common mistake

Writing the 8th term as $a r^{8}$ . The first term has no $r$ at all, so the 8th term has only 7 of them: $a r^{7}$ .

Key idea

AP: add the same $d$ . GP: multiply by the same $r$ . One subtraction or one division tells you which.

Find a and r from two terms

A very common question gives you two terms and asks for $a$ and $r$ . Write each term with $u_{n} = a r^{n - 1}$ . Then divide one equation by the other, and $a$ cancels.

Worked example

1
The 2nd term of a GP is 24 and the 5th term is $- 81$ . Find the common ratio and the first term.
2
$a r = 24$ — 2nd term: r appears once.
3
$a r^{4} = - 81$ — 5th term: r appears four times.
4
$\frac{a r ^{4}}{a r} = \frac{- 81}{24}$ — Divide the second equation by the first. The a cancels.
5
$r^{3} = - \frac{27}{8}$ — Simplify: divide top and bottom by 3.
6
$r = - \frac{3}{2}$ — Cube root: 3 × 3 × 3 = 27 and 2 × 2 × 2 = 8. A cube root of a negative is negative.
7
$a = 24 \div (- \frac{3}{2}) = - 16$ — Put r back into ar = 24.

When the two positions are an even distance apart, you get $r^{2}$ or $r^{4}$ . Then there are two possible ratios, one positive and one negative.

Worked example

1
A GP has a positive common ratio. Its 3rd term is 36 and its 5th term is 16. Find $r$ and $a$ .
2
$a r^{2} = 36, a r^{4} = 16$ — Write both terms with the formula.
3
$r^{2} = \frac{16}{36} = \frac{4}{9}$ — Divide: the a cancels.
4
$r = \pm \frac{2}{3}$ — Square root gives two answers.
5
$r = \frac{2}{3}$ — The question says r is positive, so reject the negative one.
6
$a = 36 \div \frac{4}{9} = 81$ — Put r² = 4/9 into ar² = 36.

Common mistake

Dropping the negative root when the question does not rule it out. $r^{2} = \frac{4}{9}$ gives $r = \frac{2}{3}$ or $r = - \frac{2}{3}$ . Read the question for a hint such as "positive ratio" or "has a sum to infinity".

Common mistake

Subtracting the equations instead of dividing. In a GP the terms are multiplied, so divide to make $a$ cancel.

Key idea

Two terms, two equations, divide. The $a$ disappears and a power of $r$ is left.

Three terms with an unknown

Sometimes three terms in a row contain an unknown, $x$ . In a GP, every term divided by the one before gives the same $r$ . So the two ratios must be equal.

Formula

\frac{u _{2}}{u _{1}} = \frac{u _{3}}{u _{2}}

$u_{1}$ — the first of the three terms.
$u_{2}$ — the middle term. It appears twice.
$u_{3}$ — the third term.

Cross-multiply and it becomes $u_{2}^{2} = u_{1} u_{3}$ : the middle term squared equals the outer two multiplied.

Worked example

1
The first three terms of a GP are $x - 1$ , $x + 3$ and $3 x + 1$ . Find the possible values of $x$ and the common ratio for each.
2
$\frac{x + 3}{x - 1} = \frac{3 x + 1}{x + 3}$ — The two ratios are equal.
3
$(x + 3)^{2} = (x - 1) (3 x + 1)$ — Cross-multiply.
4
$x^{2} + 6 x + 9 = 3 x^{2} - 2 x - 1$ — Expand both sides.
5
$2 x^{2} - 8 x - 10 = 0$ — Move everything to one side.
6
$x^{2} - 4 x - 5 = 0$ — Divide by 2.
7
$(x - 5) (x + 1) = 0$ — Factorise: −5 × 1 = −5 and −5 + 1 = −4.
8
$x = 5 : 4, 8, 16, r = 2$ — Put x = 5 into the three terms and divide.
9
$x = - 1 : - 2, 2, - 2, r = - 1$ — Put x = −1 in. Both values give a real GP.

Common mistake

Using $u_{2} - u_{1} = u_{3} - u_{2}$ . That is the test for an AP. A GP needs equal ratios, not equal differences.

Common mistake

Stopping at the values of $x$ . Questions often go on to ask for $r$ or a sum, so put each $x$ back into the terms.

Key idea

Equal ratios, cross-multiply, solve the quadratic, then check each answer in the three terms.

Sum of the first n terms

Adding terms one by one is slow. There is a neat trick: multiply the whole sum by $r$ , then subtract. Almost every term cancels.

$S_{n}$ $= a + a r + a r^{2} + \dots$ $+ a r^{n - 1}$ — the sum we want.

$r S_{n}$ $= a r + a r^{2} + \dots + a r^{n - 1}$ $+ a r^{n}$ — every term times r.

$S_{n} - r S_{n} = a - a r^{n}$ — subtract. All the middle terms cancel.

$S_{n} (1 - r) = a (1 - r^{n})$ — take out the common factors.

$S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ — divide by 1 − r.

Formula

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

$a$ — the first term.
$r$ — the common ratio. It must not be 1, or you divide by 0.
$n$ — how many terms you add.

Rule

Every board prints this formula: the 0606 and 4037 List of formulas, and the 4PM1 formulae sheet. When $r > 1$ , the same formula is neater as $\frac{a ( r ^{n} - 1 )}{r - 1}$ .

Worked example

1
Find the sum of the first 8 terms of $3 + 6 + 12 + \dots$
2
$a = 3, r = 2, n = 8$ — Read the values. 6 ÷ 3 = 2.
3
$S_{8} = \frac{3 ( 2 ^{8} - 1 )}{2 - 1}$ — r is bigger than 1, so use the neater form.
4
$= 3 \times 255 = 765$ — 2⁸ = 256, so 256 − 1 = 255.

Harder questions ask for the least number of terms to pass a target. Write an inequality, then use logs to bring $n$ down from the power.

Worked example

1
A GP is $5 + 4 + 3.2 + \dots$ Find the least number of terms for the sum to be more than 24.
2
$a = 5, r = 0.8$ — 4 ÷ 5 = 0.8.
3
$\frac{5 ( 1 - 0. 8 ^{n} )}{1 - 0.8} > 24$ — Set the sum formula greater than the target.
4
$25 (1 - 0. 8^{n}) > 24$ — 5 ÷ 0.2 = 25.
5
$0. 8^{n} < 0.04$ — Divide by 25, then rearrange: 1 − 0.96 = 0.04.
6
$n l g 0.8 < l g 0.04$ — Take lg of both sides. The power law brings n down.
7
$n > \frac{l g 0.04}{l g 0.8} = 14.4 \dots$ — lg 0.8 is negative. Dividing by a negative flips the sign.
8
$n = 15$ — n must be a whole number, so round up. Check: S₁₄ ≈ 23.90, S₁₅ ≈ 24.12.

Common mistake

Not flipping the inequality. When $r$ is between 0 and 1, $l g r$ is negative. Dividing by it turns $<$ into $>$ .

Rule

Log questions like this are on calculator papers: 0606 Paper 2 and 4PM1. Show the line with the logs, because it earns a method mark.

Key idea

Multiply by $r$ and subtract: the middle terms cancel. That one idea gives the whole sum formula.

Sum to infinity

Look at $8 + 4 + 2 + 1 + \frac{1}{2} + \dots$ Each term is half the one before. The terms shrink towards 0, so the total never passes 16. It gets closer and closer to 16.

Key idea

About 2,500 years ago the Greek thinker Zeno of Elea argued that you can never cross a room: first you walk half the way, then half of what is left, and so on forever. But $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ adds up to exactly 1. You do arrive.

This only happens when $r$ is between −1 and 1. Then $r^{n}$ shrinks towards 0 as $n$ grows, and the sum formula becomes $\frac{a ( 1 - 0 )}{1 - r}$ .

Rule

A GP has a sum to infinity only when $- 1 < r < 1$ , written $∣ r ∣ < 1$ . Such a GP is called convergent.

Formula

S_{\infty} = \frac{a}{1 - r}

$a$ — the first term.
$r$ — the common ratio, which must be between −1 and 1.

Rule

Every board prints this formula, with the condition $∣ r ∣ < 1$ next to it. You must still check it and write it down in your answer.

Worked example

1
A GP starts $36, - 12, 4, \dots$ Explain why it has a sum to infinity, and find it.
2
$r = \frac{- 12}{36} = - \frac{1}{3}$ — Divide the 2nd term by the 1st.
3
$∣ r ∣ = \frac{1}{3} < 1$ — Write the condition. This line is the explanation the examiner wants.
4
$S_{\infty} = \frac{36}{1 - ( - \frac{1}{3} )}$ — Use the formula. Keep the brackets round the negative r.
5
$= \frac{36}{\frac{4}{3}} = 27$ — 1 + 1/3 = 4/3, and 36 × 3/4 = 27.

Example

$2 + 3 + 4.5 + \dots$ has $r = 1.5$ . Since $1.5 > 1$ , the terms grow, so there is no sum to infinity.

Worked example

1
A GP has first term 10 and sum to infinity 40. Find the common ratio.
2
$\frac{10}{1 - r} = 40$ — Put the facts into the formula.
3
$1 - r = \frac{1}{4}$ — Rearrange: 1 − r = 10 ÷ 40 = 1/4.
4
$r = \frac{3}{4}$ — And 3/4 is between −1 and 1, so the answer is valid.

Example

0606 questions sometimes use a trig ratio as $r$ . If $r = sin θ$ with $0^{\circ} < θ < 9 0^{\circ}$ , then $0 < r < 1$ , so the GP converges.

Common mistake

Using $\frac{a}{1 - r}$ when $r = 2$ or $r = - 3$ . The formula still gives a number, but it is meaningless. Check $∣ r ∣ < 1$ first.

Common mistake

Sign slips with a negative ratio: $1 - (- \frac{1}{3}) = \frac{4}{3}$ , not $\frac{2}{3}$ .

Key idea

Ratio between −1 and 1? The terms shrink and the sum settles at $\frac{a}{1 - r}$ . Otherwise there is no sum to infinity.

When an AP meets a GP

A favourite exam question links an AP and a GP. The key is to write every term in letters before you do anything else.

Steps

1
1. Write each AP term as $a + (n - 1) d$ and each GP term as $a r^{n - 1}$ . Use different letters if the first terms differ.
2. Turn each "is the same as" sentence into one equation.
3. Make one letter the subject, substitute, and solve the quadratic.
4. Use the conditions in the question (for example " $d$ is negative") to choose the answer.

Worked example

1
A GP has first term 8 and common ratio $r$ . An AP also has first term 8, and common difference $d$ , where $d < 0$ .
2
The 2nd term of the GP equals the 3rd term of the AP. The 3rd term of the GP equals the 4th term of the AP. Find $r$ and $d$ , then find the sum to infinity of the GP.
3
$8 r = 8 + 2 d$ — GP 2nd term = AP 3rd term.
4
$8 r^{2} = 8 + 3 d$ — GP 3rd term = AP 4th term.
5
$d = 4 r - 4$ — Rearrange the first equation.
6
$8 r^{2} = 8 + 3 (4 r - 4) = 12 r - 4$ — Substitute into the second equation.
7
$2 r^{2} - 3 r + 1 = 0$ — Move everything to one side and divide by 4.
8
$(2 r - 1) (r - 1) = 0$ — Factorise.
9
$r = 1 \Rightarrow d = 0$ — Reject: d must be negative.
10
$r = \frac{1}{2}, d = - 2$ — Put r = 1/2 into d = 4r − 4.
11
$S_{\infty} = \frac{8}{1 - \frac{1}{2}} = 16$ — |r| = 1/2 < 1, so the sum to infinity exists.

Worked example

1
The 1st, 3rd and 9th terms of an AP are the first three terms of a GP. The AP has $a \neq = 0$ and $d \neq = 0$ . Find the common ratio of the GP.
2
$a, a + 2 d, a + 8 d$ — Write the three AP terms.
3
$(a + 2 d)^{2} = a (a + 8 d)$ — In a GP, the middle term squared equals the outer two multiplied.
4
$a^{2} + 4 a d + 4 d^{2} = a^{2} + 8 a d$ — Expand.
5
$4 d^{2} = 4 a d \Rightarrow d = a$ — Simplify, then divide by 4d, which is allowed because d ≠ 0.
6
$a, 3 a, 9 a \Rightarrow r = 3$ — Put d = a into the three terms and divide.

Common mistake

Using the same letter for two different things, such as $a$ for both first terms when they are different numbers. Give each progression its own letters.

Key idea

Write every term in letters first. Then the sentences turn into equations and the rest is algebra you already know.

Which method? Read the question

The wording of the question tells you which tool to pick.

You see	You do
Two terms of a GP given	Write both as $a r^{n - 1}$ , then divide
Three terms with $x$ in them	Equal ratios: $u_{2}^{2} = u_{1} u_{3}$
"Sum of the first n terms"	$S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$
"Least number of terms"	Inequality, then logs; flip the sign if you divide by a negative log; round up
"Sum to infinity" or "convergent"	Show $∣ r ∣ < 1$ , then $\frac{a}{1 - r}$
"Explain why there is no sum to infinity"	Show $r \geq 1$ or $r \leq - 1$
Terms of an AP are terms of a GP	AP terms in $a$ and $d$ , then equal ratios

Common mistake

Mixing up the two progressions. AP: $a + (n - 1) d$ . GP: $a r^{n - 1}$ .

Key idea

Name $a$ and $r$ first. Almost every GP question becomes easy once you know those two numbers.

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