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JAMB 2012 Mathematics · Question 45

The sum to infinity of a geometric progression is −110-\frac{1}{10} and the first term is −18-\frac{1}{8}. Find the common ratio of the progression.

Options

  1. A.−15-\frac{1}{5}
  2. B.−14-\frac{1}{4}
  3. C.−13-\frac{1}{3}
  4. D.−12-\frac{1}{2}

Answer: B. −14-\frac{1}{4}

Explanation written with AI and checked against the answer key

Analyze the question: We are given the sum to infinity S∞ = -1/10 and the first term a = -1/8. We need the common ratio r. Concept: For a geometric progression with |r| < 1, the sum to infinity is S∞ = a / (1 − r) Solution: -1/10 = (-1/8) / (1 − r) Cross-multiply: (-1/10)(1 − r) = -1/8 Multiply both sides by -10: 1 − r = 10/8 = 5/4 r = 1 − 5/4 = -1/4 Check: a/(1 − r) = (-1/8) / (1 + 1/4) = (-1/8) / (5/4) = (-1/8)(4/5) = -4/40 = -1/10 ✓ Also |r| = 1/4 < 1, so the sum to infinity exists. Answer: The common ratio is -1/4, which is option B. Tip: Always start with S∞ = a/(1 − r). Rearranging gives r = 1 − a/S∞. Here, a/S∞ = (-1/8)/(-1/10) = 10/8 = 5/4, so r = 1 − 5/4 = -1/4. Always check that |r| < 1.

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