# If 12th 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?

Given:

The 12th term of an A.P. is $-13$ and the sum of the first four terms is 24.

To do:

We have to find the sum of the first 10 terms.

Solution:

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

We know that,

Sum of $n$ terms $S_n=\frac{n}{2}(2a+(n-1)d)$

$S_4=\frac{4}{2}[2a+(4-1)d]$

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

$2a+3d=12$

$2a=12-3d$

$a=\frac{12-3d}{2}$......(i)

$a_{12}=a+(12-1)d$

$-13=a+11d$

$-13=\frac{12-3d}{2}+11d$

$-13=\frac{12-3d+2(11d)}{2}$

$-13(2)=12-3d+22d$

$-26=12+19d$

$19d=-26-12$

$d=\frac{-38}{19}$

$d=-2$

This implies,

$a=\frac{12-3(-2)}{2}$

$=\frac{12+6}{2}$

$=\frac{18}{2}$

$=9$

Therefore,

$S_{10}=\frac{10}{2}[2(9)+(10-1)(-2)]$

$=5(18-18)$

$=0$

Hence, the sum of the first 10 terms is $0$.

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