JEE Main 2020 · 7 January · evening · Single correct · +4 / −1

Let a₁ , a₂ , a₃ ,....... be a G.P. such that
a₁ < 0, a₁ + a₂ = 4 and a₃ + a₄ = 16.
If Σᵢ₌₁⁹ aᵢ = 4λ, then λ is equal to:

  1. (a) 171
  2. (b) -171
  3. (c) -513
  4. (d) 511/3
Answer and solution

Answer (b) -171

Solution

Maths sir · 3 steps

Divide the two sums to get r², then let the condition a₁ < 0 choose the sign of r.

  1. Divide the equations. Write a₁ + a₂ = a₁(1 + r) = 4 and a₃ + a₄ = a₁ r²(1 + r) = 16. Dividing the second by the first gives r² = 4, so r = 2 or r = -2.
  2. Pick the sign. If r = 2, then 3a₁ = 4 and a₁ = 4/3, which is positive. That breaks a₁ < 0. If r = -2, then -a₁ = 4 and a₁ = -4, which works.

    Σᵢ₌₁⁹ aᵢ = a₁(r⁹ - 1)/(r - 1) = (-4)(-512 - 1)/(-2 - 1) = 2052/-3 = -684

  3. So. 4λ = -684, so λ = -171, which is option (b).

The idea to keep. Two sums of consecutive GP terms share the factor a₁(1 + r). Divide them to get a power of r, then use the sign condition to choose the root.

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

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