Let S_n denote the sum of the first n terms of an arithmetic progression. If S_10=390 and the ratio of the tenth and the fifth terms is 15 :7, then S_15-S_5 is equal to:

Question

Accepted Answer

Correct answer: C. (c) S_10=390 102 [2a+ (10-1 )d ]=390 ⇒2a+9 d=78 … (1 ) t_10t_5= 157 ⇒ a+9da+4d= 157 ⇒8a=3d … From (1) & (2), a=3 \&d=8 S_15-S_5= 152 (6+14 8 )- 52 (6+4 8 ) = 15 118-5 382=790

$Let S_{n} denote the sum of the first n terms of an arithmetic progression. If S_{10} = 390 and the ratio of the tenth and the fifth terms is 15 : 7, then S_{15} - S_{5} is equal to:$

Solution

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