A geometric progression multiplies by the same ratio each time. When the ratio lies between \(-1\) and \(1\) the terms shrink so fast that adding infinitely many of them gives a finite total, the sum to infinity.
A GP with first term \(a\) and common ratio \(r\) has \(u_n = ar^{n-1}\) and \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) (\(r \ne 1\)). It converges when \(|r| < 1\), and then \(S_\infty = \dfrac{a}{1 - r}\). Three numbers \(a, b, c\) are in GP when \(b^2 = ac\).
Fig. 1: the partial sums of the GP 6 + 3 + 1.5 + ... (a = 6, r = 0.5) creep up towards, but never pass, the sum to infinity 6 ÷ (1 − 0.5) = 12
Section 01
The common ratio.
In the lab, choose a first term and a multiplier and watch the terms. A ratio bigger than 1 makes them explode; a ratio between 0 and 1 makes them shrink towards zero; a negative ratio makes them alternate in sign.
GP: each term is the previous term multiplied by a constant \(r\), the common ratio \(= \dfrac{u_{n+1}}{u_n}\).
\(n\)th term: \(u_n = ar^{n-1}\) (the first term has had no multiplications yet).
Three terms in GP: \(\dfrac{b}{a} = \dfrac{c}{b}\), i.e. \(b^2 = ac\).
Worked example A term
Find the 10th term of the GP \(3, 6, 12, \ldots\)
\(a = 3\), \(r = \dfrac{6}{3} = 2\).
\(u_{10} = 3 \times 2^{9} = 3 \times 512\).
Answer: \(1536\)
Worked example Two terms give a and r
The 3rd term of a GP is 18 and the 6th term is 486. Find the first term and the common ratio.
\(ar^2 = 18\) and \(ar^5 = 486\).
Divide: \(r^3 = \dfrac{486}{18} = 27\), so \(r = 3\).
\(a = \dfrac{18}{9} = 2\).
Answer: \(a = 2\), \(r = 3\)
Worked example Using b squared = ac
The numbers \(x\), \(x + 6\) and \(4x\) are consecutive terms of a GP. Find the possible values of \(x\) and the common ratio in each case.
\((x + 6)^2 = x \times 4x\), so \(x^2 + 12x + 36 = 4x^2\).
Dividing \(ar^5\) by \(ar^2\) gives \(r^3\), not \(r^4\): subtract the powers, \(5 - 2 = 3\).
Section 02
The sum of a GP.
Write \(S_n = a + ar + \cdots + ar^{n-1}\), multiply every term by \(r\), and subtract. Everything cancels except the first and last terms: \(S_n - rS_n = a - ar^n\), so \(S_n = \dfrac{a(1 - r^n)}{1 - r}\). For \(r > 1\) the equivalent form \(\dfrac{a(r^n - 1)}{r - 1}\) avoids negative numbers.
Worked example A growing GP
Find the sum of the first 8 terms of the GP \(3, 6, 12, \ldots\)
With a negative ratio, bracket it: \(\left(-\tfrac{1}{3}\right)^6 = +\dfrac{1}{729}\), and \(1 - r = 1 - \left(-\tfrac{1}{3}\right) = \tfrac{4}{3}\), not \(\tfrac{2}{3}\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is the common ratio of the GP \(48, -12, 3, \ldots\)?
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Find the 6th term of the GP \(2, 6, 18, \ldots\)
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Which of these progressions has a sum to infinity?
Quick check+10 XP
Find the sum to infinity of \(10 + 5 + 2.5 + \cdots\)
Section 03
Sum to infinity.
If \(|r| < 1\), then \(r^n \to 0\) as \(n\) grows, so \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) gets as close as we like to \(\dfrac{a}{1 - r}\). That limit is the sum to infinity (Fig. 1). If \(|r| \ge 1\) the terms do not shrink and there is no sum to infinity.
Convergence condition: \(-1 < r < 1\). Always state it when the ratio contains an unknown.
Sum to infinity: \(S_\infty = \dfrac{a}{1 - r}\).
Recurring decimals are sums to infinity: \(0.\dot{2}\dot{7} = 0.27 + 0.0027 + \cdots\)
Worked example Sum to infinity and a recurring decimal
(a) Find the sum to infinity of \(24 + 12 + 6 + \cdots\) (b) Write \(0.272727\ldots\) as a fraction.
\((4r - 1)(4r - 3) = 0\), so \(r = \tfrac{1}{4}\) or \(r = \tfrac{3}{4}\); both satisfy \(|r| < 1\).
\(a = \dfrac{6}{r}\): \(a = 24\) or \(a = 8\).
Answer: \(r = \tfrac{1}{4}, a = 24\) or \(r = \tfrac{3}{4}, a = 8\)
Your turn
Find the set of values of \(x\) for which the GP \(1 + 2x + 4x^2 + \cdots\) has a sum to infinity, and find that sum when \(x = 0.2\).
\(r = 2x\), so \(-1 < 2x < 1\), i.e. \(-\tfrac{1}{2} < x < \tfrac{1}{2}\). When \(x = 0.2\), \(r = 0.4\) and \(S_\infty = \dfrac{1}{0.6} = \dfrac{5}{3}\).
Examiner trap
\(S_\infty\) only exists when \(|r| < 1\). A GP such as \(2 - 4 + 8 - \cdots\) has \(r = -2\), so \(\dfrac{a}{1 - r} = \dfrac{2}{3}\) is meaningless.
Section 04
More than one progression.
Paper 1 questions often link an AP and a GP: some terms of one are terms of the other. Write each term in terms of \(a\) and \(d\) (or \(a\) and \(r\)), use \(b^2 = ac\) or \(2b = a + c\), and solve.
Worked example AP terms that form a GP
An AP has first term 5 and common difference \(d\), where \(d \ne 0\). The 1st, 2nd and 4th terms of the AP are the first three terms of a GP. Find \(d\), the common ratio, and the sum of the first 10 terms of the GP.
The terms are \(5\), \(5 + d\), \(5 + 3d\). In GP: \((5 + d)^2 = 5(5 + 3d)\).
When an equation like \(d^2 = 5d\) appears, do not divide by \(d\) blindly: say \(d \ne 0\) (given), so \(d = 5\). If \(d = 0\) were allowed it would also be a solution.
Stage 2 · Exam
Exam-style questions
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The 2nd term of a GP is 12 and the 5th term is 324. Find the sum of the first 6 terms.
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A GP has second term 6 and sum to infinity 32. Find the possible values of the first term.
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Express the recurring decimal \(0.454545\ldots\) as a fraction in its lowest terms.
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The GP \(3 + 6x + 12x^2 + \cdots\) has a sum to infinity of 5. Find \(x\).
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