# The common difference of the A.P

Question:

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:

1. (1) 81

2. (2) $-127$

3. (3) $-81$

4. (4) 127

Correct Option: , 3

Solution:

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 $(-2)$.

$\because b_{100}=a_{70}$

$\Rightarrow b_{1}+99(-2)=(-3)+69(-4)$

$\Rightarrow b_{1}=198-279 \Rightarrow b_{1}=-81$