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

If the sum of first 11 terms of an A.P.,
a₁, a₂, a₃, .... is 0 (a ≠ 0), then the sum of the A.P.,
a₁ , a₃ , a₅ ,....., a₂₃ is ka₁ , where k is equal to :

  1. (a) 121/10
  2. (b) -121/10
  3. (c) 72/5
  4. (d) -72/5
Answer and solution

Answer (d) -72/5

Solution

Maths sir · 3 steps

Use the zero sum to find d in terms of a, then the odd-numbered terms form a new A.P. with gap 2d.

  1. Use the zero sum. Let the first term be a and the common difference d. Then S₁₁ = 11/2(2a + 10d) = 0, so a + 5d = 0 and d = -a/5.
  2. The new A.P. The terms a₁, a₃, a₅, …, a₂₃ skip every other term, so their common difference is 2d = -2a/5. The subscripts 1, 3, …, 23 give (23 + 1)/2 = 12 terms.

    S = 12/2(2a + 11 × 2d) = 6(2a - 22a/5) = 6 × (-12a/5) = -72/5a

  3. So. The sum is k a₁ with k = -72/5, which is option (d).

The idea to keep. Picking every second term of an A.P. gives another A.P. with common difference 2d. Count its terms from the subscripts before you use the sum formula.

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.