Question

Obtain the sum of the first $56$ terms of an A.P whose ${18}^{th}$ and ${39}^{th}$ terms are $52$ and $148$ respectively.

Answer 1

Given $t_{18}=52$ and $t_{39}=148$
We know that, $t_n=a+(n-1)d$
Here, $t_{18}=52$
$\therefore$ $a+(18-1)d=52$
$\therefore$ $a+17d=\mathrm{52..........}\left(i\right)$
Also, $t_{39}=148$
$\therefore$ $a+(39-1)d=148$
$\therefore$ $a+38d=\mathrm{148.......}\left(ii\right)$
Adding equation $\left(i\right)$ and $\left(ii\right)$ we get
$a+17d=52$
$a+38d=148$
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$2a+55d=\mathrm{200.........}\left(iii\right)$
We know that
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{56}=\frac{56}{2}[2\times a+(56-1\left)d\right]$
$\therefore S_{56}==28(2a+55d)$
$\therefore S_{56}=28\times 200$
$\therefore S_{56}=5600$ [From $iii)$]
$\therefore$ The sum of the $56$ term of an A.P is $5600$