Suppose that all the terms of an arithmetic progression are natural numbers. I

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Suppose that all the terms of an arithmetic progression 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 AP is___

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Answer: Option A
:

Given, $\frac{S_{7}}{S_{11}} = \frac{6}{11} a n d 130 < t_{7} < 140$
$\Rightarrow \frac{\frac{7}{2} [2 a + 6 d]}{\frac{11}{2} [2 a + 10 d]} = \frac{6}{11}$
$\Rightarrow \frac{7 (2 a + 6 d)}{(2 a + 10 d)} = 6$
$\Rightarrow a = 9 d \dots \dots (i)$
Also, $130 < t_{7} < 140$
$\Rightarrow 130 < a + 6 d < 140$
$\Rightarrow 130 < 9 d + 6 d < 140$ [from Eq. (i)]
$\Rightarrow 130 < 15 d < 140$
$\Rightarrow \frac{26}{3} < d < \frac{28}{3}$
$∴$ d=9 [since, d is a natural number]

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