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TMUA 2016 · Paper 2 · Question 12 of 20

TMUA 2016 Paper 2 Question 12

Sequences and series — Arithmetic series · deducing signs from an inequality on sums. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 2Sequences and seriesArithmetic series · deducing signs from an inequality on sums6 options
The first term of an arithmetic sequence is a and the common difference is d.

The sum of the first n terms is denoted by Sn.

If S8 > 3S6, what can be deduced about the sign of a and the sign of d?

  1. Aboth a and d are negative
  2. Ba is positive, d is negative
  3. Ca is negative, d is positive
  4. Da is negative, but the sign of d cannot be deduced
  5. Ed is negative, but the sign of a cannot be deduced
  6. Fneither the sign of a nor the sign of d can be deduced
Show the answer and worked solution
answer · F
  1. Aboth a and d are negative
  2. Ba is positive, d is negative
  3. Ca is negative, d is positive
  4. Da is negative, but the sign of d cannot be deduced
  5. Ed is negative, but the sign of a cannot be deduced
  6. Fneither the sign of a nor the sign of d can be deduced
Use Sn = n2(2a + (n−1)d). Then S8 = 8a + 28d and 3S6 = 3(6a + 15d) = 18a + 45d, so the condition becomes 8a + 28d > 18a + 45d, that is 10a + 17d < 0. That single inequality pins down neither sign on its own: a = 1, d = −1 satisfies it with a positive, and a = −10, d = 1 satisfies it with d positive. So neither sign can be deduced.