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

Let the first term and the common difference of the A.P. Be a and d respectively. It is given thatfracS_7S_11=(6 / 11) => (2 / 11) (7 / 2) frac\2 a+6 d\\2 a+10 d\=(6 / 11) => (7(a+3 d) / (a+5 d))=6 => a=9 dIt is also given that => 130 => 130

Suppose that all the terms of an arithmetic progression... | Sequence & Series (Mathematics)

Let the first term and the common difference of the A.P. $\\$ Be $a$ and $d$ respectively. It is given that

$\frac{S_7}{S_{11}}=\frac{6}{11} \Rightarrow \frac{2}{11} \frac{7}{2} \frac{\{2 a+6 d\}}{\{2 a+10 d\}}=\frac{6}{11} \Rightarrow \frac{7(a+3 d)}{(a+5 d)}=6 \Rightarrow a=9 d$

It is also given that

$\Rightarrow \quad 130<a_7<140 \\$

$\Rightarrow \quad 130<9 d+6 d<140 \Rightarrow 130<15 d<140 \Rightarrow d=9$

Explanation

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