If 12th term of an A.P. is −13 and the sum of the first four terms is 24,

Solution:

Let a be the first term and d be the common difference.

$a_{12}=-13$

$\Rightarrow a+(12-1) d=-13$

$\Rightarrow a+11 d=-13 \quad \ldots(i)$

Also, $S_{4}=24$

$\Rightarrow \frac{4}{2}[2 a+(4-1) d]=24$

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

$\Rightarrow 2 a+3 d=12 \quad \ldots(i i)$

From (i) and (ii), we get:

$19 d=-38$

$\Rightarrow d=-2$

Putting the value of $d$ in (i), we get:

$a+11(-2)=-13$

$\Rightarrow a=9$

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

$\Rightarrow S_{10}=5[18+9(-2)]=0$