QJEEM20Ma167a: JEE Main 2020 Mathematics - Class 11 JEEM20Ma167a Solution: The common difference of the A.P. b 1 , b 2 , … , b m is 2 more …

Let common difference of series
a₁ , a₂ , a₃ ,..., a_n be d.

$\because$ a₄₀ = a₁ + 39d == –159 ...(i)

and a₁₀₀ = a₁ + 99d = –399 ...(ii)

From eqn. (ii) and (i)
d = –4 and a₁ = –3.

The common difference of the A.P.
b₁, b₂, … , b_m is 2 more than the common
difference of A.P. a₁, a₂, …, a_n.

$\therefore$ Common difference of b₁ , b₂ , b₃ , ..., be (–2).

$\because$ b₁₀₀ = a₇₀

$\therefore$ b₁ + 99(–2) = (–3) + 69(–4)

$\therefore$ b₁ = 198 – 279

$\therefore$ b₁ = – 81