Leave a Comment / Class 10th, Continued Proportion, ICSE / By

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

Ans: 

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

\frac{x}{y} = \frac{\times 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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