Lemmara logoLemmaraWorked solution

0606/21 May/Jun 2026, Question 3: Arithmetic Progressions in Problems

Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 8 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Solve an arithmetic progression problem: given the 6th term is 1.5 times the 3rd and the first ten terms sum to 255, find a and d, then the least term count for a sum above 40 000.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 4 marks

Find the first term a and common difference d from the ratio of the 6th to 3rd term and the sum of the first ten terms equal to 255.

  1. 1a + 5d = 1.5(a + 2d), so a = 4d
  2. 2255 = (10/2)(2a + 9d), so 2a + 9d = 51
  3. 38d + 9d = 51, d = 3
  4. 4a = 12

Answera = 12, d = 3

Part (b) · 4 marks

Using your a and d, find the smallest number of terms whose sum exceeds 40 000.

  1. 1(n/2)(24 + 3(n − 1)) > 40 000
  2. 23n² + 21n − 80 000 > 0
  3. 3n = (−21 + √(441 + 960 000))/6 ≈ 159.8
  4. 4Least n = 160 (S₁₅₉ = 39 591, S₁₆₀ = 40 080)

Answern = 160

Stuck on a step? Read the free lesson: Arithmetic Progressions in Problems

Practise more arithmetic progressions in problems questions

Real 0606 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on arithmetic progressions in problems

Also searched as: 0606/21/M/J/26 · 0606_s26_qp_21 · 0606 s26 21 · question 3.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.