Question

find the sum of 32 terms of an A.P . whose 3rd term is 1 and 6th term is -11.

Answer 1

Here
Let 'a' be the first term and 'd' be the common difference.
3rd term = a + (n - 1)d
Here n = 3
= a + 2d
3rd term = 1. ............... (given)
Hence,
a + 2d = 1 ................ (1)
Now
6th term = a + (n - 1)d
Here n = 6
= a + 5d
6th term = -11. ............. (given)
Hence,
a + 5d = -11. ............... (2)
Subtracting equations (1) from (2)
a + 5d = -11
- a + 2d = 1
——————
3d = -12
d = -12/3 = -4
Hence
Putting d = -4 in equation (1)
a + 2d = 1
a + 2*(-4) = 1
a - 8 = 1
a = 9
Sum formula
S(n) = n/2 [2a+ (n - 1)d]
Sum of 32 terms =
S(32) = 32/2 [ 2*9 + (32 - 1)*(-4)]
= 16 [ 18 + 31*(-4)]
= 16 ( 18 - 124)
= 16 ( - 106)
= - 1696
So this is the final answer.
I hope so, you got it.