If a, b, c are in G.P and x, y are the arithmetic means of a, b and b,

1.	If a, b, c are in G.P and x, y are the arithmetic means of a, b and b, c respectively, then \(\frac{1}{x}+\frac{1}{y}\) is equal to(a) \(\frac{b}{2}\) (b) \(\frac{b}{3}\)(c) \(\frac{2}{b}\) (d) \(\frac{3}{b}\)
Answer» (c) \(\frac{2}{b}\) a, b, c are in G.P ⇒ b² = ac ...(i) \(x\) is the A.M. of a, b ⇒ \(x\) = \(\frac{a+b}{2}\) ....(ii) y is the A.M. of b, c ⇒ y = \(\frac{b+c}{2}\) ....(iii) Now \(\frac{1}{x}+\frac{1}{y}\) = \(\frac{2}{a+b}\) + \(\frac{2}{b+c}\) = \(\frac{2(b+c)+2(a+b)}{(a+b)(b+c)}\) = \(\frac{2(a+2b+c)}{ab+b^2+ac+bc}\) = \(\frac{2(a+2b+c)}{ab+b^2+b^2+bc}\) = \(\frac{2(a+2b+c)}{b(a+2b+c)}\) = \(\frac{2}{b}\).

If a, b, c are in G.P and x, y are the arithmetic means of a, b and b, c respectively, then \(\frac{1}{x}+\frac{1}{y}\) is equal to(a) \(\frac{b}{2}\) (b) \(\frac{b}{3}\)(c) \(\frac{2}{b}\) (d) \(\frac{3}{b}\)

Answer»

(c) \(\frac{2}{b}\)

a, b, c are in G.P ⇒ b² = ac ...(i)

\(x\) is the A.M. of a, b ⇒ \(x\) = \(\frac{a+b}{2}\) ....(ii)

y is the A.M. of b, c ⇒ y = \(\frac{b+c}{2}\) ....(iii)

Now \(\frac{1}{x}+\frac{1}{y}\) = \(\frac{2}{a+b}\) + \(\frac{2}{b+c}\) = \(\frac{2(b+c)+2(a+b)}{(a+b)(b+c)}\)

= \(\frac{2(a+2b+c)}{ab+b^2+ac+bc}\) = \(\frac{2(a+2b+c)}{ab+b^2+b^2+bc}\) = \(\frac{2(a+2b+c)}{b(a+2b+c)}\) = \(\frac{2}{b}\).