Question

Obtain the sum of the 56 terms of an A.P. whose ${19}^{th}$ and ${38}^{th}$ terms are 52 and 148 respectively.

• A. 5500
• B. 6000
• C. 5600
• D. 5700

Answer 1

The correct option is C 5600

Let a and d be the first term and the common difference of the given AP respectively. Then,
$a_{19}=52\Rightarrow a+18d=52$ ...(1)
$a_{38}=148\Rightarrow a+37d=148$ ...(2)
On subtracting (1) from (2), we get
$19d=96\Rightarrow d=\frac{96}{19}$

Putting $d=\frac{96}{19}$ in (1), we get

$a=52-18\times \frac{96}{19}=\frac{-740}{19}$

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

$\therefore S_{56}=28[\frac{3800}{19}]=5600$