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8. Sequences and Series
hard
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
$127$
(JEE MAIN-2020)
Solution
$a_{1}, a_{2}, \ldots, a_{n} \rightarrow(C D=d)$
$b _{1}, b _{2}, \ldots, b _{ m } \rightarrow( CD = d +2)$
$a_{40}=a+39 d=-159$
$a_{100}=a+99 d=-399$
Subtract : $60 d =-240 \Rightarrow d =-4$
using equation (1)
$a+39(-4)=-159$
$a=156-159=-3$
$a_{70}=a+69 d=-3+69(-4)=-279=b_{100}$
$b_{100}=-279$
$b_{1}+99(d+2)=-279$
$b_{1}-198=-279 \Rightarrow b_{1}=-81$
Standard 11
Mathematics
