$\frac{1}{x + y}, \frac{1}{2 y}, \frac{1}{y + z}$ are in A.P. Prove that x, y, z are the consecutive terms of G.P

Open in App

Solution

Given $\frac{1}{x + y}, \frac{1}{2 y}, \frac{1}{y + z}$ are in A.P.
To prove that x, y, z are the consecutive terms of G.P, then we have
Since, $\frac{1}{x + y}, \frac{1}{2 y}, \frac{1}{y + z}$ are in A.P.
then, $\frac{1}{2 y} = \frac{1 \frac{x + y}{+} \frac{1}{1} y + z}{2}$
Also we can calculate the common difference
$\frac{1}{2 y} - \frac{1}{x + y} = d \Rightarrow \frac{x - y}{x + y} = 2 y d - - - (1)$
$\frac{1}{y + z} - \frac{1}{2 y} =\Rightarrow \frac{x - y}{x + y} = 2 y d - - - (2)$
from (1) and (2) we get $\frac{x - y}{x + y} = \frac{y - z}{y + z}$
$\Rightarrow (x - y) (y + z) = (x + y) (y - z)$
$\Rightarrow x y + x z - y^{2} - y z - x y - x z + y^{2} - y z \Rightarrow 2 y^{2} - 2 x z = 0 \Rightarrow 2 (y^{2} - x z) = 0 \Rightarrow y^{2} - x z = 0 \Rightarrow y^{2} = x z$
Hence the terms x, y, z are the consecutive terms of a G.P

Suggest Corrections

Similar questions

\frac{1}{x + y}, \frac{1}{2 y}, \frac{1}{y + z}

\frac{1}{x + y}, \frac{1}{2 y}, \frac{1}{y + z} a r e i n A . P .

y^{2} = x z

x, y, z

\frac{1}{x + y} + \frac{1}{y + z} = \frac{1}{y}

\frac{1}{x + y}, \frac{1}{2 y}, \frac{1}{y + z}

x, y, z

Show that $x^{2} + x y + y^{2}, z^{2} + z x + x^{2}$ and $y^{2} + y z + z^{2}$ are consecutive terms of an A.P., if $x, y$ and $z$ are in A.P.

Join BYJU'S Learning Program