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

Answer 1

Let $a$ be the first term and $d$ be the common difference of A.P.
Given, $\frac{S_7}{S_{11}}=\frac{6}{11}$
$\Rightarrow \frac{\frac{7}{2}[2a+6d]}{\frac{11}{2}[2a+10d]}=\frac{6}{11}$
$\Rightarrow \frac{14a+42d}{2a+10d}=6$
$\Rightarrow 14a+42d=12a+60d$
$\Rightarrow 2a=18d$
$\Rightarrow a=9d$
$\therefore T_7=a+6d=9d+6d=15d$

Now, $130<T_7<140$
$\Rightarrow 130<15d<140$
$\Rightarrow \frac{130}{15}<d<\frac{140}{15}$
$\Rightarrow 8.6<d<9.3$
$\Rightarrow d=9(\because d\in N)$