Question

Obtain the sum of the first $56$ terms of an A.P. whose ${28}^{th}$ and ${29}^{th}$ terms are $52$ and $148$ respectively.

• A. $2100$
• B. $5600$
• C. $5200$
• D. $2600$

Answer 1

Answer: (B)

$5600$

We know that, $t_n=a+(n-1)d$

Here, $t_{28}=a+(28-1)d$

$\therefore$ $52=a+27d.......\left(i\right)$

Also, $t_{29}=a+(29-1)d$

$\therefore$ $148=a+28d........\left(ii\right)$

Adding equation $\left(i\right)$ and $\left(ii\right)$, we get:
$a+27d=52$
$a+28d=148$
----------------------
$2a+55d=\mathrm{200.............}\left(iii\right)$

Also, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{56}=\frac{56}{2}[2\times a+(56-1\left)d\right]$

$\Rightarrow$ $S_{56}=28(2a+55d)$

$\Rightarrow$ $S_{56}=28\times 200$

$\Rightarrow$ $S_{56}=5600$

$\therefore$ The sum of the $56$ terms of the given A.P. is $5600$.