Suppose all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term is, lies in between 130 and 140, then the common difference of this A.P. is

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answer is D.

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Detailed Solution

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

$\begin{array}{l} \Rightarrow 130 < a_{7} < 140 \\ \Rightarrow 130 < 9 d + 6 d < 140 \Rightarrow 130 < 15 d < 140 \Rightarrow d = 9 . \end{array}$

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