Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | If A and G be A.M. and G.M., respectivel

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 Excercise ".$ex_no." , Question 29: if a and g be a m and g m respectively between....
Question 29

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A plus-or-minus square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root
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Answer

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.

∴ A M space equals space A space equals space fraction numerator a space plus space b over denominator 2 end fraction space space space space space space space space space space space space... left parenthesis 1 right parenthesis
G M space equals space G space equals space square root of a b end root space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

From (1) and (2), we obtain

a + b = 2A … (3)

ab = G2 … (4)

Substituting the value of a and b from (3) and (4) in the identity (a – b)2 = (a + b)2 – 4ab, we obtain

(a – b)2 = 4A2 – 4G2 = 4 (A2G2)

(a – b)2 = 4 (A + G) (A – G)

open parentheses a minus b close parentheses space equals space 2 square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root space space space space space space space space space space... left parenthesis 5 right parenthesis

From (3) and (5), we obtain

2 a space equals space 2 A space plus space 2 square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root
rightwards double arrow a space equals space A space plus space square root of open parentheses A space plus space G close parentheses open parentheses A minus G close parentheses end root

Substituting the value of a in (3), we obtain

b space equals space 2 A space minus A space minus space square root of open parentheses A space plus space G close parentheses open parentheses A minus G close parentheses end root space equals space A space minus space square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root

Thus, the two numbers are A plus-or-minus square root of open parentheses A plus G close parentheses open parentheses A minus G close parentheses end root.

 

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