Question

The $6^{th}$ and $7^{th}$ terms an $A.P.$ are $19$ and $41$ respectively, find the ${40}^{th}$ terms.

Answer 1

Given,

$a_7=41\Rightarrow a+6d=41$....(1)

$a_6=19\Rightarrow a+5d=19$.....(2)

solving (1) and (2), we get,

$d=22$

From (1)

$a=19-5\left(22\right)=-91$

$a_{40}=a+39d$

$-91+39\left(22\right)=767$