Engineering
Mathematics
Arithmetic Progression and Its Properties
Question

Let Sn denote the sum of first n terms of an arithmetic progression. If S20 = 790 and S10 = 145, then S15 – S5 is :

395

410

405

390

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Solution
S20=202(2a+19d)=7902a+19d=79
S10=102(2a+9d)=1452a+9d=29
10 d = 50
d = 5
S15S5=152(2a+14d)52(2a+4d)
10a + 95d = 395