To prove: b2 is the AM between x2 and y2 .
Given: (i) a, b, c are in AP
(ii) x is the GM between a and b
(iii) y is the GM between b and c
Formula used:
(i) Arithmetic mean between a and b = \(\frac{a+b}{2}\)
(ii) Geometric mean between a and b = \(\sqrt{ab}\)
As a, b, c are in A.P.
⇒ 2b = a + c … (i)
As x is the GM between a and b
⇒ x = \((\sqrt{ab})\)
⇒ x2 = ab … (ii)
As y is the GM between b and c
y = \((\sqrt{bc})\)
⇒ y2 = bc … (iii)
Arithmetic mean of x2 and y2 is \((\frac{x^2 + y^2}{2})\)
Substituting the value from (ii) and (iii)
\((\frac{x^2 + y^2}{2})\) = \((\frac{ab + bc}{2})\)
Substituting the value from eqn. (i)
=\(\frac{b(a+c)}{2}\)
= b2
Hence Proved