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Question

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 first eleven terms is 6:11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

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Solution

Let a be the first term and d be the common difference of A.P.
Given, S7S11=611
72[2a+6d]112[2a+10d]=611
14a+42d2a+10d=6
14a+42d=12a+60d
2a=18d
a=9d
T7=a+6d=9d+6d=15d

Now, 130<T7<140
130<15d<140
13015<d<14015
8.6<d<9.3
d=9 (dN)



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