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TMUA 2017 · Paper 1 · Question 7 of 20

TMUA 2017 Paper 1 Question 7

Sequences and series — Arithmetic and geometric progressions sharing terms. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1Sequences and seriesArithmetic and geometric progressions sharing terms4 options
The first three terms of an arithmetic progression are p, q and p2 respectively, where p < 0.

The first three terms of a geometric progression are p, p2 and q respectively.

Find the sum of the first 10 terms of the arithmetic progression.

  1. A238
  2. B958
  3. C1158
  4. D1858
Show the answer and worked solution
answer · B
  1. A238
  2. B958
  3. C1158
  4. D1858
Use the geometric condition first: consecutive ratios are equal, so p2p = qp2, giving q = p3. The arithmetic condition is that consecutive differences are equal: q − p = p2 − q, so 2q = p + p2. Substituting gives 2p3 = p + p2, and dividing by p (non-zero since p<0) gives 2p2 − p − 1 = 0, so (2p+1)(p−1) = 0. As p < 0, p = −12, hence q = −18 and the common difference is d = q − p = 38. Then S10 = 102(2(−12) + 9⋅38) = 5⋅198 = 958.