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(D) Let the first term be $a$ and the common difference be $d$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.Given $\f

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$9$

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(D) Let the first term be $a$ and the common difference be $d$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.Given $\frac{S_7}{S_{11}} = \frac{6}{11}$,we have $\frac{\frac{7}{2}(2a + 6d)}{\frac{11}{2}(2a + 10d)} = \frac{6}{11}$.Simplifying this,$\frac{7(a + 3d)}{11(a + 5d)} = \frac{6}{11}$ $\Rightarrow 7a + 21d = 6a + 30d$ $\Rightarrow a = 9d$.The seventh term $a_7 = a + 6d = 9d + 6d = 15d$.Given $130 Since all terms are natural numbers,$d$ must be an integer. Thus,$d = 9$.

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

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