If the 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

#### 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 12th term (n = 12)

`a_12 = a + (12 - 1)d`

`-13 = a + 11d`

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

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

`S_n = n/2[2a +(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 = 4/2 [2(a) + (4 - 1)(d)]`

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

24 = 4a +6d

4a = 24 - 6d

` a= 6 - 6/4 d` ....(2)

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

`a - a = (6 - 6/4 d) - (-13 - 11)d`

`0 = 6 - 6/4 d + 13 + 11d`

`0 = 19 + 11d - 6/4 d`

`0 =19+ (44d - 6d)/4`

On further simplifying for *d, *we get,

`0 = 19 + (38d)/4`

`-19=19/2 d`

`d= (-19(2))/2`

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 = 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`