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

Solution:

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 $S_{10}=0$.