Question

In an A.P if the ${12}^{th}$ terms is $-13$ and the sum of its first four terms is $24$. Find sum of its first $10$ terms.

Answer 1

$a=$ first term, $d=$common difference

$t_n=a+(n-1)d$ $\therefore$ sum of $1^{st}$ ten term

As per given condition, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$-13=a+(12-1)d$ $S_n=\frac{10}{2}(2\times 9+(10-1\left)\right(-2\left)\right)$

$-13=a+11d$ _________ (1) $S_n=5(18+11(-2\left)\right)$

$a+a+d+a+2d+a+3d=24$ $S_n=5(18-22)$

$4a+6d=24$ _____ (2) $S_n=-20$

Solving (1) and (2) , we get,

$a=9$

$d=-2$