Let’s take a to be the first term and d to be the common difference.
And we know that, sum of first n terms
Sn = \(\frac{n}{2}\)(2a + (n − 1)d)
Given that sum of the first 10 terms of an A.P. is -150.
S10 = -150
And the sum of next 10 terms is -550.
So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms
S20 = -150 + -550 = -700
Now, having
S10 = \(\frac{10}{2}\)(2a + (10 − 1)d)
⟹ -150 = 5(2a + 9d)
⟹ -30 = 2a + 9d
⟹ 2a + 9d = -30 . . . . (1)
And,
S20 = \(\frac{20}{2}\)(2a + (20 − 1)d)
⟹ -700 = 10(2a + 19d)
⟹ -70 = 2a + 19d …. (2)
Now, subtracting (1) from (2), we get
⟹ 19d – 9d = -70 – (-30)
⟹ 10d = -40
⟹ d = -4
Using d in (1), we have
2a + 9(-4) = -30
2a = -30 + 36
a = \(\frac{6}{2}\) = 3
Hence, we have a = 3 and d = -4
So, the A.P is 3, -1, -5, -9, -13,…..