If a, b, c are in AP, x is the GM between a and b; y is the GM between

1.	If a, b, c are in AP, x is the GM between a and b; y is the GM between b and c; then show that b2 is the AM between x2 and y2 .
Answer» 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

If a, b, c are in AP, x is the GM between a and b; y is the GM between b and c; then show that b2 is the AM between x2 and y2 .

Answer»

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