A-level Maths revision · Pure, year 2

Geometric series

The idea

A geometric sequence multiplies by the same ratio rr. The nnth term is arn−1ar^{n-1}, the sum of nn terms is Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r}, and if ∣r∣<1|r| < 1 the sum to infinity is S∞=a1−rS_\infty = \frac{a}{1 - r}.

  1. Find rr by dividing a term by the one before.
  2. Use arn−1ar^{n-1} for a term and a(1−rn)1−r\frac{a(1 - r^n)}{1 - r} for a sum.
  3. Use S∞S_\infty only when −1<r<1-1 < r < 1.

Watch out A sum to infinity needs ∣r∣<1|r| < 1, not just r<1r < 1.

See it worked

Worked example

Find the sum to infinity of 8+4+2+…8 + 4 + 2 + \dots

  1. a=8a = 8, r=12r = \frac{1}{2}.
  2. ∣r∣<1|r| < 1, so it converges.
  3. S∞=81−12=16S_\infty = \frac{8}{1 - \frac{1}{2}} = 16.

Answer 1616

Your turn

Question 1Non-calculator3 marks

Find the 6th term of the geometric sequence 5,15,45,…5, 15, 45, \dots

Question 2Non-calculator3 marks

Find the sum to infinity of 12+9+6.75+…12 + 9 + 6.75 + \dots

Question 3Non-calculator3 marks

Find the sum of the first 1010 terms of 2+4+8+…2 + 4 + 8 + \dots

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