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

In the given problem, we need to find the sum of first 10 terms of an A.P. Let us take the first term a and the common difference as d

Here, we are given that,

$a_{12}=-13$

$S_{4}=24$

Also, we know,

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

For the $12^{\text {th }}$ term $(n=12)$,

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

$-13=a+11 d$

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

So, as we know the formula for the sum of n terms of an A.P. is given by,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 4, we get,

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

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

$24=4 a+6 d$

$4 a=24-6 d$

$a=6-\frac{6}{4} d$ .........(2)

Subtracting (1) from (2), we get,

$a-a=\left(6-\frac{6}{4} d\right)-(-13-11 d)$

$0=6-\frac{6}{4} d+13+11 d$

$0=19+11 d-\frac{6}{4} d$

$0=19+\frac{44 d-6 d}{4}$

On further simplifying for d, we get,

$0=19+\frac{38 d}{4}$

$-19=\frac{19}{2} d$

$d=\frac{-19(2)}{19}$

$d=-2$

Now, to find a, we substitute the value of d in (1),

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

$a=-13+22$

$a=9$

Now, using the formula for the sum of n terms of an A.P. for n = 10, we get,

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

$=(5)[18+(9)(-2)]$

$=(5)(18-18)$

$=(5)(0)$

$=0$

Therefore, the sum of first 10 terms for the given A.P. is $S_{10}=0$.