Question

If A is the A.M 's between a and b,prove that 4(a-A)(A-b)=(b-a)^2

Answer 1

A is the arithmatic mean of a and b. Then,
(a+b)/2=A
∴, 4(a-A)(A-b)
=4{a-(a+b)/2}{(a+b)/2-b}
=4{(2a-a-b)/2}{(a+b-2b)/2}
=4{(a-b)(a-b)/4}
={-(b-a)}{-(b-a)}
=(b-a)² (Proved)

Answer 2

Hence proved.

Step-by-step explanation:

Given:

$4(a-A)(A-b)=(b-a)^2$

Here A is the arithmetic mean between a and b

⇒ A = $\frac{a+b}{2}$

LHS = 4(a−A)(A−b)

= $4[a-(\frac{a+b}{2} )][(\frac{a+b}{2} )-b]$

= $4[\frac{2a-(a+b)}{2} ][\frac{a+b-2b}{2} ]$

= $4(\frac{2a-a-b}{2})(\frac{a-b}2} )$

= $4(\frac{a-b}{2} )(\frac{a-b}{2} )$

= $4(\frac{a-b}{2} )^2$

= $4\times\frac{(a-b)^2}{4}$

= $(a-b)^2$

= $(-b+a)^2$

= $[-(b-a)]^2$          $[(-(b-a))^2=(-b+a)^2]$

= $(b-a)^2$

= RHS.

Hence proved.