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0606/22 May/Jun 2025, Question 7: Arithmetic Progressions in Problems

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 9 marks

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The question, in short

Arithmetic progression with first three terms 2 ln(x³), 5 ln(x²), 2 ln(x⁷); for x > 1 find the least number of terms whose sum exceeds 43 ln(x²⁴), and given the 25th term equals 408 find x.

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Part (a) · 6 marks

Find the smallest number of terms needed for the sum of the progression to be greater than 43 ln(x²⁴), given x > 1.

  1. 1Terms are 6 ln x, 10 ln x, 14 ln x, so a = 6 ln x and d = 4 ln x
  2. 2Sₙ = (n/2)(12 ln x + 4(n − 1) ln x) = (2n² + 4n) ln x
  3. 343 ln(x²⁴) = 1032 ln x, and ln x > 0 since x > 1
  4. 42n² + 4n > 1032, so n² + 2n − 516 > 0
  5. 5n > −1 + √517 ≈ 21.7
  6. 6Least number of terms = 22

Answer22

Part (b) · 3 marks

Given the 25th term of the progression equals 408, find the exact value of x.

  1. 1u₂₅ = a + 24d = 6 ln x + 24 × 4 ln x
  2. 2u₂₅ = 102 ln x
  3. 3102 ln x = 408, so ln x = 4
  4. 4x = e⁴

Answerx = e⁴

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Also searched as: 0606/22/M/J/25 · 0606_s25_qp_22 · 0606 s25 22 · question 7.
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