Question

The common difference of the $A.P$. $b _{1}, b _{2}, \ldots$ $b _{ m }$ is 2 more than the common difference of $A.P$. $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 :

• A. $-127$
• B. $-81$
• C. $81$
• D. $127$

Answer 1

Answer: (B)

$a_{1}, a_{2}, \ldots, a_{n} \rightarrow(C D=d)$
$b _{1}, b _{2}, \ldots, b _{ m } \rightarrow( CD = d +2)$
$a_{40}=a+39 d=-159\, \dots(1)$
$a_{100}=a+99 d=-399\, \dots(2)$
Subtract: $60 d =-240 \Rightarrow d =-4$
using equation (1)
$a+39(-4)=-159$
$a=156-159=-3$
$a_{70}=a+69 d=-3+69(-4)=-279=b_{100}$
$b_{100}=-279$
$b_{1}+99(d+2)=-279$
$b_{1}-198$
$=-279$
$ \Rightarrow b_{1}=-81$