Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum…

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6:11 and the seventh terms lies in between 130 and 140, then the common difference of this A.P. is

Solution

72[2a+6d]112[2a+10d]=611
  a+3da+5d=76
a=9d
Also 130<a+6d<140
130<15d<140
d=9

Asked in: JEE Advanced 2015 (Paper 2)

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