JEE PYQ: Sequence And Series Question 34
Question 34 - 2020 (06 Sep Shift 2)
The common difference of the A.P. $b_1, b_2, \ldots, b_m$ is 2 more than the common difference of A.P. $a_1, a_2, \ldots, a_n$. If $a_{40} = -159$, $a_{100} = -399$ and $b_{100} = a_{70}$, then $b_1$ is equal to:
(1) 81
(2) $-127$
(3) $-81$
(4) 127
Show Answer
Answer: (3)
Solution
Let common difference of $a_1, a_2, \ldots, a_n$ be $d$.
$\therefore a_{40} = a_1 + 39d = -159$ …(i)
and $a_{100} = a_1 + 99d = -399$ …(ii)
From equations (i) and (ii),
$d = -4$ and $a_1 = -3$
Since, the common difference of $b_1, b_2, \ldots, b_n$ is 2 more than common difference of $a_1, a_2, \ldots, a_n$.
$\therefore$ Common difference of $b_1, b_2, \ldots$ is $(-2)$.
$\therefore b_{100} = a_{70}$
$\Rightarrow b_1 + 99(-2) = (-3) + 69(-4)$
$\Rightarrow b_1 = 198 - 279 \Rightarrow b_1 = -81$
BETA
