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in Geometric Progressions by (15.7k points)

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 .

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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

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