What numbers must be added to each of the numbers 16, 26 and 40 so that the resulting numbers must be in continued proportion?

Let the number to be added to each number be x. So, the new numbers are 16 + x, 26 + x, 40 + x.

Since, new numbers are in continued proportion,

$\therefore \dfrac{16 + x}{26 + x} = \dfrac{26 + x}{40 + x} \\[0.5em] \Rightarrow (16 + x)(40 + x) = (26 + x)(26 + x) \\[0.5em] \Rightarrow 640 + 16x + 40x + x^2 = 676 + 26x + 26x + x^2 \\[0.5em] \Rightarrow x^2 - x^2 + 56x - 52x + 640 - 676 = 0 \\[0.5em] \Rightarrow 4x - 36 = 0 \\[0.5em] \Rightarrow 4x = 36 \Rightarrow x = 9.$

Hence, the number to be added to each number is 9.