Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | The sum of two numbers is 6 times their

Welcome to the NCERT Solutions for Class 11th Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Exercise 3, Question 28: the sum of two numbers is 6 times their geometric....
Question 28

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio open parentheses 3 space plus space 2 square root of 2 close parentheses space : space open parentheses 3 minus 2 square root of 2 close parentheses.

Answer

Let the two numbers be a and b.

G.M. = square root of a b end root

According to the given condition,

a space plus space b space equals space 6 square root of a b end root space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

rightwards double arrow open parentheses a space plus space b close parentheses squared space equals space 36 open parentheses a b close parentheses

Also,

open parentheses a minus b close parentheses squared space equals space open parentheses a plus b close parentheses squared space minus space 4 a b space equals space 36 a b space minus 4 a b space equals space 32 a b

rightwards double arrow a minus b space equals space square root of 32 space square root of a b end root space
equals space 4 space square root of 2 space square root of a b end root space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Adding (1) and (2), we obtain

2 a space equals space open parentheses 6 space plus space 4 square root of 2 close parentheses square root of a b end root

rightwards double arrow a space equals space open parentheses 3 space plus space 2 square root of 2 close parentheses square root of a b end root

Substituting the value of a in (1), we obtain

b space equals space 6 square root of a b end root space minus space open parentheses 3 plus 2 square root of 2 close parentheses square root of a b end root

rightwards double arrow b space equals space open parentheses 3 space minus space 2 square root of 2 close parentheses space square root of a b end root
a over b space equals space fraction numerator open parentheses 3 space plus 2 square root of 2 close parentheses square root of a b end root over denominator open parentheses 3 minus 2 square root of 2 close parentheses square root of a b end root end fraction space equals fraction numerator 3 space plus space 2 square root of 2 over denominator 3 minus 2 square root of 2 end fraction space

Thus, the required ratio is open parentheses 3 space plus space 2 square root of 2 close parentheses space : space open parentheses 3 minus 2 square root of 2 close parentheses.

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