If 2(y – a) is the H.M. between y – x and y – z, then (x – a),

1.	If 2(y – a) is the H.M. between y – x and y – z, then (x – a), (y – a), (z – a) are in(a) A.P. (b) G.P. (c) H.P. (d) None of these
Answer» (b) G.P. Given, 2(y – a) is the H.M. between (y – x) and (y – z) ⇒ (y – x), 2 (y – a), (y – z) are in H.P ⇒ \(\frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}\) are in A.P. ⇒ \(\frac{1}{2(y-a)} - \frac{1}{(y-x)}\) = \(\frac{1}{(y-z)}-\frac{1}{2(y-a)}\) ⇒ \(\frac{y-x-2y+2a}{2(y-a)(y-x)}\) = \(\frac{2y-2a-y+z}{(y-z)\,2(y-a)}\) ⇒ \(\frac{-y-x-2y+2a}{(y-x)}\) = \(\frac{y+z-2a}{(y-z)}\) ⇒ \(\frac{x+y-2a}{x-y}\) = \(\frac{y+z-2a}{(y-z)}\) ⇒ \(\frac{(x-a)+(y-a)}{(x-a)-(y-a)}\) = \(\frac{(y-a)+(z-a)}{(y-a)-(z-a)}\) Now applying componendo and dividendo, we have \(\bigg(\text{By comp. and div,}\,\frac{x}{y}=\frac{a}{b}\implies\frac{x+y}{x-y}=\frac{a+b}{a-b}\bigg)\) ⇒\(\frac{2(x-a)}{2(y-a)}\) = \(\frac{2(y-a)}{2(z-a)}\) ⇒ (x – a) (z – a) = (y – a)² ⇒ (x – a), (y – a), (z – a) are in G.P.

If 2(y – a) is the H.M. between y – x and y – z, then (x – a), (y – a), (z – a) are in(a) A.P. (b) G.P. (c) H.P. (d) None of these

Answer»

(b) G.P.

Given, 2(y – a) is the H.M. between (y – x) and (y – z)

⇒ (y – x), 2 (y – a), (y – z) are in H.P

⇒ \(\frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}\) are in A.P.

⇒ \(\frac{1}{2(y-a)} - \frac{1}{(y-x)}\) = \(\frac{1}{(y-z)}-\frac{1}{2(y-a)}\)

⇒ \(\frac{y-x-2y+2a}{2(y-a)(y-x)}\) = \(\frac{2y-2a-y+z}{(y-z)\,2(y-a)}\)

⇒ \(\frac{-y-x-2y+2a}{(y-x)}\) = \(\frac{y+z-2a}{(y-z)}\)

⇒ \(\frac{x+y-2a}{x-y}\) = \(\frac{y+z-2a}{(y-z)}\)

⇒ \(\frac{(x-a)+(y-a)}{(x-a)-(y-a)}\) = \(\frac{(y-a)+(z-a)}{(y-a)-(z-a)}\)

Now applying componendo and dividendo, we have

\(\bigg(\text{By comp. and div,}\,\frac{x}{y}=\frac{a}{b}\implies\frac{x+y}{x-y}=\frac{a+b}{a-b}\bigg)\)

⇒\(\frac{2(x-a)}{2(y-a)}\) = \(\frac{2(y-a)}{2(z-a)}\)

⇒ (x – a) (z – a) = (y – a)²

⇒ (x – a), (y – a), (z – a) are in G.P.