Question

If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then 

विकल्प

• (a) \[\frac{1}{x} + \frac{1}{y} = 2\] 

• (b) \[\frac{1}{x} + \frac{1}{y} = \frac{1}{2}\] 

• (c) \[\frac{1}{x} + \frac{1}{y} = \frac{2}{a}\]

• (d) \[\frac{1}{x} + \frac{1}{y} = \frac{2}{b}\]

Answer 1

(d) \[\frac{1}{x} + \frac{1}{y} = \frac{2}{b}\] 

\[\text{ a, b and c are in G . P } . \]
\[ \therefore b^2 = ac . . . . . . . . (i)\]
\[\text{ a, x and b are in A . P } . \]
\[ \therefore 2x = a + b . . . . . . . . (ii)\]
\[\text{ Also, b, y and c are in A . P } . \]
\[ \therefore 2y = b + c \]
\[ \Rightarrow 2y = b + \frac{b^2}{a} \left[ \text{ Using } (i) \right]\]
\[ \Rightarrow 2y = b + \frac{b^2}{\left( 2x - b \right)} \left[ \text{ Using } (ii) \right]\] 
\[ \Rightarrow 2y = \frac{b\left( 2x - b \right) + b^2}{\left( 2x - b \right)}\]
\[ \Rightarrow 2y = \frac{2bx - b^2 + b^2}{\left( 2x - b \right)}\]
\[ \Rightarrow 2y = \frac{2bx}{\left( 2x - b \right)}\]
\[ \Rightarrow y = \frac{bx}{\left( 2x - b \right)}\]
\[ \Rightarrow y\left( 2x - b \right) = bx\]
\[ \Rightarrow 2xy - by = bx\]
\[ \Rightarrow bx + by = 2xy\]
\[\text{ Dividing both the sides by xy }: \]
\[ \Rightarrow \frac{1}{y} + \frac{1}{x} = \frac{2}{b}\]
\[\]