Suppose that all terms of an arithmetic progression ( A.P. ) are natural numbers. If ratio of sum of

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8. Sequences and Series

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

A

$6$

B

$7$

C

$8$

D

$9$

(IIT-2015)

Solution

Let seventh term be ' $a$ ' and common difference be ' $d$ ' Given $\frac{ S _7}{ S _{11}}=\frac{6}{11} \Rightarrow a =15 d$

Hence, $130 < 15 d < 140$

$\Rightarrow d=9$

Standard 11

Mathematics

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