Suppose that all the terms of an arithmetic progression (A.P.) are...

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

Step-by-step solution

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

⇒ \dfrac{a + 3d}{a + 5d} = \dfrac{7}{6}

⇒ a = 9d

Also

130 < a + 6d < 140

130 < 15d < 140

⇒ d = 9

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