JEE Main 2020 · 2 September · evening · Single correct · +4 / −1

Let S be the sum of the first 9 terms of the series :
x + ka + x² + (k + 2)a + x³ + (k + 4)a
+ x⁴ + (k + 6)a + .... where a ≠ 0 and x ≠ 1.
If S = (x¹⁰ - x + 45a( x - 1 ))/(x - 1), then k is equal to :

  1. (a) -3
  2. (b) 1
  3. (c) -5
  4. (d) 3
Answer and solution

Answer (a) -3

Solution

Maths sir · 4 steps

Split each bracket into its x part and its a part: the x part is a GP, the a part is an AP.

  1. Split the series. Each term is xⁿ + (k + 2(n-1))a for n = 1 to 9. So S is a GP sum in x plus an AP sum multiplied by a.
  2. GP part. x + x² + … + x⁹ = x(x⁹ - 1)/(x - 1) = (x¹⁰ - x)/(x - 1). This matches the first piece of the given S.
  3. Compare the rest. The given S splits as (x¹⁰ - x)/(x - 1) + 45a. So the a terms must add up to 45a.

    a[9k + 2(0 + 1 + … + 8)] = a(9k + 72) = 45a

  4. So. Since a ≠ 0, 9k + 72 = 45, which gives 9k = -27 and k = -3. That is option (a).

The idea to keep. When each term mixes a GP piece and an AP piece, sum the two families separately and match each part. Watch the AP's n-th coefficient: it is k + 2(n-1), not k + 2n.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Sequences & Series

Asked about 250 times in JEE Main since 2020. This question is from 2020.