The common difference of the A.P. $b_{1}, b_{2}, \ldots, b_{m}$ is 2 more than the common difference of A.P. $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$. If $\mathrm{a}_{40}=-159$, $\mathrm{a}_{100}=-399$ and $\mathrm{b}_{100}=\mathrm{a}_{70}$, then $\mathrm{b}_{1}$ is equal to:
Correct Option: , 3
Let common difference of series
$a_{1}, a_{2}, a_{3}, \ldots \ldots, a_{n}$ be $d$
$\because a_{40}=a_{1}+39 d=-159 \quad \ldots$ (i)
and $a_{100}=a_{1}+99 d=-399 \quad$...(ii)
From equations (i) and (ii),
$d=-4$ and $a_{1}=-3$
Since, the common difference of $b_{1}, b_{2}, \ldots \ldots, b_{n}$ is 2 more
than common difference of $a_{1}, a_{2}, \ldots \ldots, a_{n}$
$\therefore$ Common difference of $b_{1}, b_{2}, b_{3}, \ldots \ldots$ is $(-2)$.
$\because b_{100}=a_{70}$
$\Rightarrow b_{1}+99(-2)=(-3)+69(-4)$
$\Rightarrow b_{1}=198-279 \Rightarrow b_{1}=-81$