Ccan you figure it out?

Algebra Level 3

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 (assume it to be an integer)

This is a problem from JEE ADVANCED 2015 paper 2.


The answer is 9.

Harvindersingh Chavan by Harvindersingh Chavan , 24, India

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2 solutions

Let the first term of the A P AP be a a and difference d d . Then, we have:

7 ( 2 a + 6 d ) 2 11 ( 2 a + 10 d ) 2 = 6 11 7 a + 21 d 11 a + 55 d = 6 11 77 a + 231 d = 66 a + 330 d 11 a = 99 d ⇒ a = 9 d \begin{aligned} \dfrac {\dfrac{7(2a+6d)}{2}}{\dfrac{11(2a+10d)}{2}} & = \frac{6}{11} \\ \frac {7a+21d}{11a+55d} & = \frac{6}{11} \\ 77a + 231d & = 66a + 330d \\ 11a & = 99d \\ \Rightarrow a & = 9d \end{aligned}

The seventh term,

130 < a + 6 d < 140 130 < 15 d < 140 \begin{aligned} 130 < a +& 6d < 140 \\ 130 < 15& d < 140 \end{aligned}

For an integer d ⇒ 15 d = 135 ⇒ d = 9 d\quad \Rightarrow 15d = 135 \quad \Rightarrow d = \boxed{9}

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Good solution sir! Exactly what I did.

Harvindersingh Chavan - 6 years ago
Jaimin Pandya
Jun 3, 2015

same problem : https://brilliant.org/problems/jee-advanced-2015-problem/

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