Question

In an $A.P$ sum of first ten terms is $-150$ and the sum of its next ten terms is $-550$, Find the $A.P$.

Answer 1

The sum of the first n terms of an AP is
Sn = (n/2)[2a + (n-1)·d]
n = 10, S10 = -150
-150 = (10/2)[2a + (10 - 1)d]
-150 = 5[2a + 9d]
Dividing through by 5,
-30 = 2a + 9d ----------(1)

Given sum ofits next 10 terms is - 550.

n = 10, Sn = - 550
Here a = a + 10d
- 550 = (10/2)[2(a + 10d) + (10-1)d]
- 550 = 5[2a + 29d]
Dividing through by 10,
-110 = 2a + 29d ----------(2)
Solving (1) and (2) we get
d = -4 and a = 3.