Let *a *and *d* be the first term and common difference of A.P. respectively.

Given, *S* _{10} = –150

Given, Sum of next 10 terms = *a* _{11} + *a* _{12} + *a* _{13} + ... + *a* _{20} = – 550

Solving (1) and (2), we get

When *d* = – 4, we get

2*a* + 9 � (– 4) = –30 (Using (1))

⇒ 2*a* = – 30 + 36 = 6

⇒ *a* = 3

Thus, the given A.P. is 3, –1, –5, –9, ...