Series problems: finding n and real-life series

Exam questions often ask how many terms are needed, or turn salaries and savings into arithmetic or geometric series.

Practise this lesson free (6 questions)

Finding n with logarithms

How many terms of 3 + 6 + 12 + ... are needed for the sum to exceed 1000?
Sn = 3(2n − 1) > 1000, so 2n > 334.33
n log 2 > log 334.33, so n > 8.39. The smallest whole number is n = 9.

Dividing by a negative log

When is the term 100(0.9)n − 1 first less than 1?
0.9n − 1 < 0.01, so (n − 1)log 0.9 < log 0.01.
log 0.9 is negative, so dividing by it reverses the inequality: n − 1 > 43.71.
So n = 45.

Series in context

A salary starts at £20 000 and rises by £1500 each year: arithmetic, a = 20 000, d = 1500.
Total over 10 years = 102(40 000 + 9 × 1500) = £267 500.
Savings of £1000 a year, each year 5% more than the last: geometric, a = 1000, r = 1.05.
Total over 10 years = 1000(1.0510 − 1)0.05 = £12577.89.
Worked example

2 + 5 + 8 + ... has sum 155. How many terms are there?

  1. n2(4 + 3(n − 1)) = 155
  2. n(3n + 1) = 310, so 3n2 + n − 310 = 0
  3. (3n + 31)(n − 10) = 0, so n = 10

Answer: n = 10

Key idea

Set up the inequality with Sn or un, take logs, and remember that dividing by the log of a number less than 1 reverses the sign. Salaries with a fixed rise are arithmetic; percentage rises are geometric.

Extra practice

2 more questions

Try each one, then tap to check.

  1. How many terms of 2 + 6 + 18 + ... are needed for the sum to exceed 5 000?

    Show the answer

    Answer: 8

    S7 = 2 186 and S8 = 6 560, so 8 terms are needed.

  2. A geometric sequence has un = 200(0.8)n − 1. Find the first term that is less than 1. Give n.

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    Answer: 25

    200(0.8)n − 1 < 1 gives n − 1 > log(1200)/log 0.8 = 23.74, so n = 25.

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Sources and references

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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