Question

If the sum of a certain number of terms starting from first of an AP $25,22,19,..$is $116.$ Find the last term.

• A. $4$
• B. $9$
• C. $12$
• D. $31$

Answer 1

Answer: (A)

$4$

Let $a$ be the first term and $d$ be the common difference of the given AP. Then $a=25,d=-3$.

Let sum of first $n$ terms of the given AP is $116$. Therefore,

$S_n=116\Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]=116$

$\Rightarrow \frac{n}{2}[2\times 25+(n-1)\times -3]=116$

$\Rightarrow n[53-3n]=232$

$\Rightarrow 3n^2-53n+232=0$

$\Rightarrow 3n^2-24n-29n+232=0$

$\Rightarrow (3n-29)(n-8)=0$

$\Rightarrow n=\frac{29}{3},8$

But, $n=\frac{29}{3}$ is not possible.

$\therefore n=8$

Now, last term $=a_8=a+7d=25+7\times -3=4$

So, option A is correct.