Let Sn denote the sum of first n terms of an arithmetic progression. If S20=790 and S10=145, then S15-S5 is :
- 405
- 390
- 410
- 395
Answer
(4) 395
Solution
Sn= n2 [2a+(n-1)d ]
S20=10(2a+19d)=790 2a+19d=79
S10=5(2a+9d)=145 2a+9d=29
10d=50 d=5, a=-8
S15-S5= 152(-16+70)- 52(-16+20)
=405-10=395
Answer: (4)