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AS & A Level · Practice questions · Geometric progressions and sum to infinity

Multiply and sum.

Six original Cambridge-style questions on geometric progressions, sums to infinity and problems involving both an arithmetic and a geometric progression.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Find \(a\) and \(r\), check \(|r|\), then the right formula.

01
Recall
[3 marks]

For the GP \(5, 15, 45, \ldots\) find (a) the 8th term, (b) the sum of the first 8 terms.

  • \(a = 5\), \(r = 3\). ✓
  • (a) \(u_8 = 5 \times 3^7 = 10\,935\) ✓
  • (b) \(S_8 = \dfrac{5(3^8 - 1)}{3 - 1} = \dfrac{5 \times 6560}{2} = 16\,400\) ✓
02
Calculation
[4 marks]

The 3rd term of a GP is 12 and the 6th term is \(-96\). Find the first term, the common ratio and the sum of the first 10 terms.

  • \(ar^2 = 12\), \(ar^5 = -96\), so \(r^3 = -8\) and \(r = -2\). ✓
  • \(a = \dfrac{12}{4} = 3\). ✓
  • \(S_{10} = \dfrac{3\left(1 - (-2)^{10}\right)}{1 - (-2)} = \dfrac{3(1 - 1024)}{3} = -1023\) ✓
03
Proof
[4 marks]

(a) Prove that the sum of the first \(n\) terms of a GP with first term \(a\) and common ratio \(r\) (\(r \ne 1\)) is \(\dfrac{a(1 - r^n)}{1 - r}\).

(b) Deduce the sum to infinity when \(|r| < 1\), explaining why the condition is needed.

  • (a) \(S_n = a + ar + ar^2 + \cdots + ar^{n-1}\) and \(rS_n = ar + ar^2 + \cdots + ar^{n}\). ✓
  • Subtract: \(S_n - rS_n = a - ar^n\), so \((1 - r)S_n = a(1 - r^n)\) and \(S_n = \dfrac{a(1 - r^n)}{1 - r}\). ✓
  • (b) If \(|r| < 1\), \(r^n \to 0\) as \(n \to \infty\), so \(S_n \to \dfrac{a}{1 - r}\). ✓
  • If \(|r| \ge 1\), \(r^n\) does not tend to 0, so \(S_n\) has no finite limit. ✓
04
Calculation
[4 marks]

(a) Use a geometric progression to express the recurring decimal \(0.363636\ldots\) as a fraction in its lowest terms.

(b) Find the sum to infinity of the GP \(18 - 12 + 8 - \cdots\)

  • (a) \(a = 0.36\), \(r = 0.01\): \(S_\infty = \dfrac{0.36}{0.99} = \dfrac{36}{99} = \dfrac{4}{11}\) ✓
  • (b) \(r = -\dfrac{12}{18} = -\dfrac{2}{3}\), \(|r| < 1\). ✓
  • \(S_\infty = \dfrac{18}{1 + \frac{2}{3}} = \dfrac{18}{5/3} = \dfrac{54}{5} = 10.8\) ✓
05
Analysis
[5 marks]

An arithmetic progression has first term 9 and common difference \(d\), where \(d \ne 0\). The 1st, 4th and 13th terms of the AP are the first three terms of a geometric progression.

(a) Find \(d\).

(b) Find the common ratio of the GP.

(c) Find the sum of the first 20 terms of the AP.

  • (a) The terms are \(9\), \(9 + 3d\), \(9 + 12d\). In GP: \((9 + 3d)^2 = 9(9 + 12d)\). ✓
  • \(81 + 54d + 9d^2 = 81 + 108d\), so \(9d^2 = 54d\); \(d \ne 0\), so \(d = 6\). ✓
  • (b) The terms are \(9, 27, 81\), so \(r = 3\). ✓
  • (c) \(S_{20} = \dfrac{20}{2}\left[18 + 19 \times 6\right] = 10 \times 132 = 1320\) ✓
06
Analysis
[6 marks]

A ball is dropped from a height of 10 m. After each bounce it rises to 60% of the height from which it last fell.

(a) Find the height the ball reaches after the 4th bounce.

(b) Find the total distance travelled by the ball up to the moment it hits the ground for the 5th time.

(c) Find the total distance the ball travels before it comes to rest.

  • (a) Heights after each bounce form a GP: \(6, 3.6, 2.16, \ldots\), so after the 4th bounce \(10 \times 0.6^4 = 1.296 = 1.30\) m (3 s.f.). ✓
  • (b) The first fall is 10 m; after each of the first 4 bounces the ball goes up and down the same height. ✓
  • \(10 + 2(6 + 3.6 + 2.16 + 1.296) = 10 + 2 \times 13.056 = 36.1\) m (3 s.f.) ✓
  • (c) The rises form a GP with \(a = 6\), \(r = 0.6\): \(S_\infty = \dfrac{6}{0.4} = 15\). ✓
  • Total \(= 10 + 2 \times 15 = 40\) m ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.