If x, y and z are in continued proportion, Prove that:

Question

Q) If x, y and z are in continued proportion, Prove that:
$\frac{x}{y^2 . z^2} + \frac{y}{z^2 . x^2} + \frac{z}{x^2 . y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$

ICSE Specimen Question Paper (SQP)2025

Sapling Academy · Accepted Answer

Step 1: we know that if x, y and z are in continued proportion, then,

$\frac{x}{y} = \frac{	imes y}{z}$

y 2  = x z ........... (i)

Step 2: Let's solve LHS:

$\frac{x}{y^2 . z^2} + \frac{y}{z^2 . x^2} + \frac{z}{x^2 . y^2}$

= $\frac{x (x^2) + y (y^2) + z (z ^2)}{x^2 . y^2 . z^2} $

= $\frac{x ^3 + y^3 + z^3}{x^2 . y^2 . z^2}$

Step 3: By substituting y 2 = x z from equation (i), we get:

$\frac{x ^3 + y^3 + z^3}{x^2 . y^2 . z^2}$

= $\frac{x ^3 + y^3 + z^3}{x^2 . (x z). z^2}$

= $\frac{x ^3 + y^3 + z^3}{x^3 . z^3}$

= $\frac{x ^3}{x^3 . z^3} + \frac{y ^3}{x^3 . z^3} + \frac{z ^3}{x^3 . z^3}$

= $\frac{1}{z^3} + \frac{y ^3}{(x z)^3} + \frac{1}{x^3 }$

= $\frac{1}{z^3} + \frac{y ^3}{(y ^2)^3} + \frac{1}{x^3 }$

= $\frac{1}{z^3} + \frac{1}{y^3} + \frac{1}{x^3 }$

= RHS... Hence proved !

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