Question

In an AP, given $a_n=4,d=2,S_n=-14,$ find $n$ and $a$.

• A. $n=11$ and $a=-3$
  
• B. $n=10$ and $a=-3$
• C. $n=7$ and $a=-8$
• D. $n=10$ and $a=3$

Answer 1

The correct option is C $n=7$ and $a=-8$

As we know nth term, $a_n=a+(n-1)d$

& Sum of first $n$ terms, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$, where $a$ & $d$ are the first term amd common difference of an AP.

Since, $a_n=4\Rightarrow a+(n-1)d=4$
$\Rightarrow a+(n-1)2=4$
$\Rightarrow a=6-2n$ ...(1)
Also, $S_n=-14$
$\Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]=-14$
$\Rightarrow \frac{n}{2}\left[2\right(6-2n)+(n-1\left)2\right]=-14$ ($\because from\left(1\right)$)
$\Rightarrow n[12-4n+2n-2]=-28$
$\Rightarrow n[n-5]=14$
$\Rightarrow n^2-5n-14=0$
$\Rightarrow (n-7)(n+2)=0$
$\Rightarrow n=7,-2$
But, $n=-2$ is not possible.
$\therefore n=7$
Putting $n=7$ in (1), we get
$\therefore a=6-2\times 7=-8$