Question

What should be added to each of the numbers 1, 21 and 121 so that the resulting numbers are in continued proportion?

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Answer 1

Let the required number be $x$.

$\therefore$ $1+x,21+x$ and $121+x$ are in continued proportion.

$\to$ $\frac{1+x}{21+x}=\frac{21+x}{121+x}$

$\to$ $(21+x{)}^2$ ​= $(1+x)(121+x)$

On solving we get $x=4$