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 sum of 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

Answer 1

Let $a$ be the first term and $d$ be the common difference of A.P.
Given, $\frac{S_7}{S_{11}}=\frac{6}{11}$
$\Rightarrow \frac{7(2a+6d)}{11(2a+10d)}=\frac{6}{11}\Rightarrow 7(a+3d)=6(a+5d)\Rightarrow a=9d\cdots \left(1\right)$

Now, we know that,
$130<a_7<140\Rightarrow 130<a+6d<140$
Using equation $\left(1\right)$, we get
$130<15d<140\Rightarrow \frac{26}{3}<d<\frac{28}{3}$
As given A.P. consists of natural numbers, so $d=9$