Question

If $12$th term of an A.P is $-13$ and the sum of the first four terms is $24$, what is the sum of first $10$ terms?

Answer 1

Let $a,d$ be the first term and the common difference respectively of the A.P.

nth term $t_n=a+(n-1)d$

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

Then the $12$ th term is $a+11d$.

According to the problem,

$a+11d=-13$.......(1).

Also the sum of first $4$ term is $\frac{4}{2}(2a+3d)$.

According to the problem,

$2(2a+3d)=24$.

or,$2a+3d=12$......(2).

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

$a=9,d=-2$.

Now, the sum of first $10$ terms =$\frac{10}{2}\{2\cdot 9+2\cdot (-9\left)\right\}=0$