In an arithmetic progression, if S40=1030 and S12=57, then S30-S10 is equal to :
- 525
- 505
- 510
- 515
Answer
(4) 515
Solution
Concept: Sn= n2 [2a+(n-1)d ]
S40=1030 20(2a+39d)=1030 2a+39d=51.5
S30-S10=T11+T12+ +T30 (20 terms)
=10 (T11+T30)=10(2a+39d)
=10 51.5=515
Answer: (4)