Question

Obtain the sum of the $56$ term of an $A.P$. whose ${19}^{th}$ and ${38}^{th}$ terms are $52$ and $128$ respectively.

Answer 1

$T_n=a+(n-1)d$

Given,

$T_{38}=a+37d=128$...(1)

$T_{19}=a+18d=52$...(2)

(1)-(2) gives,

$d=4$

From (2)

$a=52-18\left(4\right)=-20$

$\therefore T_{56}=a+(56-1)d$

$=-20+55\times 4$

$=200$