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