what number must be added to each of the numbers $16, 26$ and $40$ so that the real numbers may be in continued proportion?

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Solution

Let the number $x$ should be added then

$\frac{16 + x}{26 + x} = \frac{26 + x}{40 + x}$

$(16 + x) (40 + x) = (26 + x) (26 + x)$

$x^{2} + 56 x + 640 = x^{2} + 52 x + 676$

$4 x = 36$

$x = 9.$

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