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 the 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:

Accepted Answer

$a_{40}=a_1+39d=-159$, $a_{100}=a_1+99d=-399$. Solving: $d=-4$, $a_1=-3$.

Common difference of $b$'s: $d+2=-2$.

$b_{100}=a_{70} \Rightarrow b_1+99(-2) = -3+69(-4) \Rightarrow b_1 = -279+198 = -81.$

Solutions · 1