Three numbers, a, b, and c are said to be in continued proportion if a : b :: b : c. What number must be added to each of the numbers 1, 7 and 25 so that the resulting sums, in that order, may be in continued proportion?

2

Let 'a' added to each number

⇒ \(\frac{1 + a}{7 + a}\) = \(\frac{7 + a}{25 + a}\)

⇒ (1 + a)(25 + a) = (7 + a)(7 + a)

⇒ 25 + 25a + a + \( {a }^{2 } \) = 49 + 14 + \( {a }^{2 } \)

⇒ 12 a = 24

⇒ a = 2.

Therefore, 2 must be added to each of the numbers 1, 7 and 25 so that the resulting sums, in that order, may be in continued proportion