Q25 / 60 · JEE Advanced 2015 (Paper 2) · Mathematics

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 terms lies in between $130$ and $140$ , then the common difference of this A.P. is

Correct answer: 9

Solution

\frac{\frac{7}{2} [2 a + 6 d]}{\frac{11}{2} [2 a + 10 d]} = \frac{6}{11}

\Rightarrow \frac{a + 3 d}{a + 5 d} = \frac{7}{6}

\Rightarrow a = 9 d

Also

130 < a + 6 d < 140

130 < 15 d < 140

\Rightarrow d = 9

Hence, the correct answer is 9.

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