The common difference of the A.P. b_{1}, b_{2 | vedclass

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(B) Let the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$ be $d.$ Then,the common difference of $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $d

Vedclass · Accepted Answer

$-81$

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(B) Let the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$ be $d.$ Then,the common difference of $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $d + 2.$For the first $A.P.:$$a_{40} = a_{1} + 39d = -159$  $(1)$$a_{100} = a_{1} + 99d = -399$  $(2)$Subtracting $(1)$ from $(2):$$60d = -240 \Rightarrow d = -4.$Substituting $d = -4$ in $(1):$$a_{1} + 39(-4) = -159 \Rightarrow a_{1} - 156 = -159 \Rightarrow a_{1} = -3.$Now,find $a_{70}:$$a_{70} = a_{1} + 69d = -3 + 69(-4) = -3 - 276 = -279.$Given $b_{100} = a_{70} = -279.$For the second $A.P.:$$b_{100} = b_{1} + 99(d + 2) = -279.$Since $d = -4,$ then $d + 2 = -2.$$b_{1} + 99(-2) = -279$$b_{1} - 198 = -279$$b_{1} = -279 + 198 = -81.$

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

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