A.P.

Algebra Level 3

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 A.P. is:

The answer is 9.

1 solution

Sravanth C.
Jun 1, 2015

According to the question, the ratio of the sum of first $7$ terms to that of the sum of first $11$ terms is $6:11$ .

So, $\dfrac{\dfrac{7(2a+6d)}{2}}{\dfrac{11(2a+10d)}{2}}=\dfrac{6}{11} \\ \implies \dfrac{66}{2}×(2a+10d)=\dfrac{77}{2}×(2a+6d) \\ \implies 6×(2a+10d)=7×(2a+6d) \\ \implies 12a+60d=14+42d \\ \implies 2a=18d \\ \implies a=9d$

Now, we can say that, $t_n = 9d+6d \\ =15d \\ but, \quad 120<15d>130$

Hence, $15d$ is a multiple of 15 that lies between $120$ and $130$ .

$15d=135 \\ \boxed{d=9}$

Moderator note:

Simple standard approach.

why have you divided by 2 both the numerator and denominator in your first step? we can directly write it as [7a+21d]/[11a+55d]=6/11 then if we cross multiply ,we get 99d=11a so , a=9d

Srujan Deolasee - 5 years, 10 months ago

A very good solution.

You are worth following.

Sai Ram - 6 years ago

Thank you! $\huge\ddot\smile$

Sravanth C. - 6 years ago

15d lies between 130 and 140. Not 120 and 130.

Vishwak Srinivasan - 5 years, 10 months ago

You're right.

Sai Ram - 5 years, 10 months ago

A.P.

The answer is 9.

1 solution

Moderator note:

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