An important question from arithmetic progression chapter as it was already asked in previous year paper of 2010 in which we have been asked to find the AP if the sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550.

RS Aggarwal, Class 10, chapter 5C, question no 31.

## Given,

S10=−150

S20−S10= Next 10 term

S20−S10=−550

S20=−550+S10

S20=−550−150

S20=−700

Sn=n/2[2a+(n−1)d]

S10=10/2[2a+(10−1)d]=−150

2a+9d=−30 (i)

S20=−700

20/2[2a+19d]=−700

2a+19d=−70 (ii)

eq(i)-eq(ii)

(2a+9d=−30)−(2a+19d=−70)

−10d=40

d=−4

2a+9d=−30

2a=−30+36

⇒a=3

So the

A.P.will be as:-a, (a+d), (a+2d), …

3, (-1), (-5)& so on