JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

The common difference of the A.P.
b₁, b₂, … , bₘ is 2 more than the common
difference of A.P. a₁, a₂, …, aₙ. If
a₄₀ = –159, a₁₀₀ = –399 and b₁₀₀ = a₇₀, then b₁ is equal to :

  1. (a) 127
  2. (b) 81
  3. (c) –127
  4. (d) -81
Answer and solution

Answer (d) -81

Solution

Maths sir · 4 steps

Get the first A.P. fully from its two terms, then the second A.P. needs just one term and its difference.

  1. Difference of a. From a₄₀ to a₁₀₀ there are 60 steps of the common difference d. So 60d = -399 - (-159) = -240, which gives d = -4.
  2. First term of a. a₄₀ = a₁ + 39d, so a₁ = -159 - 39 × (-4) = -159 + 156 = -3. Then a₇₀ = -3 + 69 × (-4) = -3 - 276 = -279.
  3. Difference of b. The difference of the b A.P. is 2 more than -4, so it is -4 + 2 = -2. Now use b₁₀₀ = a₇₀ = -279.

    b₁₀₀ = b₁ + 99(-2) ⇒ -279 = b₁ - 198

  4. So. b₁ = -279 + 198 = -81, which is option (d).

The idea to keep. Two terms of an A.P. fix everything: the gap in positions times d equals the gap in values. Once you have a₁ and d, any term is a₁ + (n-1)d.

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.