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 between $130$ and $140$, then the common difference of this A.P is

Answer 1

By using $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ we have,

$\frac{\frac{7}{2}(2a+6d)}{\frac{11}{2}(2a+10d)}=\frac{6}{11}$ .....[Given]

$\Rightarrow \frac{7a+21d}{a+5d}=6$

$\Rightarrow 7a+21d=6a+30d$

$\Rightarrow a=9d$ ------(1)

But, $130<a+6d<140$

from (1)

$\Rightarrow 130<15d<140$

As all terms in AP are natural.

$\therefore d=9$