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 differenc

Question

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 [2015]

Smart Achievers · Accepted Answer

(9)

$\frac{S_{7}}{S_{11}} = \frac{6}{11} \Rightarrow \frac{\frac{7}{2} [2 a + 6 d]}{\frac{11}{2} [2 a + 10 d]} = \frac{6}{11} \Rightarrow a = 9 d$

$a_{7} = a + 6 d = 15 d$

$Given 130 < a_{7} < 140$

$\Rightarrow 130 < 15 d < 140 \Rightarrow d = 9$

[Since $d$ is a natural number because all terms are natural numbers.]

Solution

Q.
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 [2015]

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Solution

Q. 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 [2015]

Ans. (9) S7S11=611⇒72[2a+6d]112[2a+10d]=611⇒a=9d a7=a+6d=15d Given 130<a7<140 ⇒130<15d<140⇒d=9 [Since d is a natural number because all terms are natural numbers.]

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Q.
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 [2015]