Question

Obtain the sum of the first $56$ terms of an A.P whose ${25}^{th}$ and ${32}^{nd}$ terms are $52$ and $148$ respectively.

Answer 1

Given 25th and 32nd terms of an A.P are $52$ and $148$ respectively.
We have to find $S_{56}$.
We know that $t_n=a+(n-1)d$
$t_{25}=52$ and $t_{32}=148$
Here, $t_{25}=a+(25-1)d$
$\therefore$ $t_{25}=a+(25-1)d$
$\therefore$ $52=a+24d..........\left(i\right)$
Also, $t_{32}=a+(32-1)d$
$\therefore$ $148=a+31d.........\left(ii\right)$
Adding equations $\left(i\right)$ and $\left(ii\right)$ we get
$a+24d=52$
$a+31d=148$
-----------------------
$2a+55d=\mathrm{200...............}\left(iii\right)$
We know that $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{56}=\frac{56}{2}[2\times a+(56-1\left)d\right]$
$S_{56}=28(2a+55d)$
$S_{56}=28\times 200$ (from $\left(iii\right)$)
$S_{56}=5600$
$\therefore$ The sum of the $56$ terms of an AP is $5600$.