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Find the product of xy if x, 2/3, 6/7, y are in GP

+1 vote
asked Mar 1, 2013 in BASIC MATH by Afeez Novice

2 Answers

+2 votes
 
Best answer

 Let the GP be a, a^r, ar^2, ar^3, . . . , ar^(n − 1). where r is the common ratio = ratio of sucessive terms.

given GP series is x, 2/3, 6/7, y.

First find the r = 3rd term/2nd term = (6/7) / (2/3) = 9/7.

So, r = 2nd term / 1st term ----------->(2/3)/ x = 9/7

3x /2 = 9/7 ----------> x = 14/27= a.

So, 4th terms is y = ar^3.

We know the a and r so, product of xy = a * ar^3 = (14/27)^2 * (9/7)^3 = 142884/250047

 

answered Mar 1, 2013 by John Lyn Pupil
selected Mar 1, 2013 by Afeez

The product of xy is 7/4.

0 votes

 Let the GP be a, ar^1, ar^2, ar^3, . . . , ar^(n − 1). where r is the common ratio = ratio of sucessive terms.

Given GP series is x, 2/3, 6/7, y.

Common ratio = 2 nd term/1 st term = 3 rd term/2 nd term = 4 rth term/3 rd term.

Consider that, 2 nd term/1 st term = 4 rth term/3 rd term

(2/3) / x = y / (6/7)

2 / (3x) = (7y) / 6

Cross multiplication.

2 * 6 = 7y * 3x

12 = 21xy

xy = 21/12

⇒ xy = 7/4.

Therefore, the product of xy is 7/4.

answered Aug 1, 2014 by lilly Expert

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