Question

I will mark you as a brainlist if you give me correct answer ,fastly .


In a AP ,if the 12th term is -13 and the sum of its four term is 24 , find the sum of its first 10 terms .

Answer 1

Given :-

$a_{12}$ = -13

$S_{4}$ = 24

To find :-

$S_{10}$

Now,

$a_{12}$ = -13

a + (12-1)d = -13

a + 11d = -13 ------------------------- 1

and,

$S_{4}$ = 24

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

2 (2a + 3d) = 24

2a + 3d = 12 -------------------------- 2

From 1 and 2,

a + 11d = -13

2a + 3d = 12

or,

2a + 22d = -26 ------------ 3

2a + 3d = 12     ------------- 4

By 4 - 3, we get,

19 d = - 38

    d = - 2

From 4,

2a + 3(-2) = 12

So,

2a - 6 = 12

2a = 18

a = 9

Now,

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

     = 5 (18 -18)

     = 0

So,

The sum of the first 10 terms will be 0