Question

Obtain the sum of the first $56$ terms of an A.P whose ${21}^{st}$ and ${36}^{th}$ terms are $52$ and $148$ respectively.

Answer 1

Given, ${21}^{st}$ term and ${36}^{th}$ terms of an A.P are $52$ and $148$ respectively.
Here, $t_n=a+(n-1)d$
$\therefore$ $t_{21}=a+(21-1)d$
$\therefore$ $t_{21}=a+(21-1)d$
$\therefore$ $52=a+20d.......\left(i\right)$
Also, $t_{36}=a(36-1)d$
$\therefore$ $148=a+35d.......\left(ii\right)$
Adding equations $\left(i\right)$ and $\left(ii\right)$ we get
$a+20d=52$
$a+35d=148$
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$2a+55d=\mathrm{200..........}\left(iii\right)$
Now $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{56}=\frac{56}{2}[2\times a+(56-1\left)d\right]$
$S_{56}=28(2a+55d)$
$\therefore$ $S_{56}=28\times 200$ (from $\left(iii\right)$
$\therefore$ $S_{56}=5600$
$\therefore$ The sum of the $56$ terms of an AP is $5600$.