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 $open parentheses 3 space plus space 2 square root of 2 close parentheses space : space open parentheses 3 minus 2 square root of 2 close parentheses$ .

9 Comment(s) on this Question

Write a Comment

Subhajit paul 2018-10-17 18:57:18

Thank you

Neetu 2018-08-19 12:29:13

Not clear cut

Neetu 2018-08-19 12:24:05

How did you get -4ab

Anjali 2018-06-19 10:16:42

Thanks a lot for this solution. This solution is very helpful for me.ð

Venkatesh Naidu 2017-09-23 13:29:06

Thank you sir/maâam. May Krishna shower you with his blessings.

Dark shadow 2017-07-14 23:02:41

Nice solition! Thanks a lot!

Sajad alam 2017-07-13 23:02:20

How to get (a-b) part

deepak 2017-06-21 13:32:09

Why we should not take two no. a and ar?

Navjot 2017-02-08 17:31:11

Thank you very much. Very simple and easy method.

NCERT Chapter

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

Popular Questions of Class 11 Mathematics

Recently Viewed Questions of Class 11 Mathematics

9 Comment(s) on this Question

Write a Comment:

NCERT Solutions

Quick Links

Resources

Support