The 12th term of an AP is −13 and the sum of its first four terms is 24.

Let a be the first term and d be the common difference of the AP. Then,

$a_{12}=-13$

$\Rightarrow a+11 d=-13 \quad \ldots \ldots(1) \quad\left[a_{n}=a+(n-1) d\right]$

Also,

$S_{4}=24$

$\Rightarrow \frac{4}{2}(2 a+3 d)=24 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

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

Solving (1) and (2), we get

$2(-13-11 d)+3 d=12$

$\Rightarrow-26-22 d+3 d=12$

$\Rightarrow-19 d=12+26=38$

$\Rightarrow d=-2$

Putting d = −2 in (1), we get

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

$\Rightarrow a=-13+22=9$

∴ Sum of its first 10 terms, S10

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

$=5 \times(18-18)$

$=5 \times 0$

$=0$

Hence, the required sum is 0.