Solve this following


The common difference of the A.P. $b_{1}, b_{2}, \ldots$, $\mathrm{b}_{\mathrm{m}}$ 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. $-127$

  2. $-81$

  3. 81

  4. 127

Correct Option: , 2


$a_{1}, a_{2}, \ldots, a_{n} \rightarrow(C D=d)$

$\mathrm{b}_{1}, \mathrm{~b}_{2}, \ldots, \mathrm{b}_{\mathrm{m}} \rightarrow(\mathrm{CD}=\mathrm{d}+2)$

$a_{40}=a+39 d=-159$  ..........(1)

$a_{100}=a+99 d=-399$  ..........(2)

Subtract : $60 \mathrm{~d}=-240 \Rightarrow \mathrm{d}=-4$

using equation (1)



$a_{70}=a+69 d=-3+69(-4)=-279=b_{100}$



$\mathrm{~b}_{1}-198=-279 \Rightarrow \mathrm{b}_{1}=-81$


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