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

$\frac{S_7}{S_{11}}=\frac{6}{11}\Rightarrow \frac{7(2a+6d)}{11(2a+10d)}=\frac{6}{11}\Rightarrow 7(a+3d)=6(a+5d)\Rightarrow a=9d................\left(i\right)$
Now we know that,
$130<a_7<140\Rightarrow 130<a+6d<140$
Using the equation (i),
$\Rightarrow 130<15d<140\Rightarrow \frac{26}{3}<d<\frac{28}{3}$
As the is formed from natural numbers so $d=9$