What number must be added to each of the numbers 7, 11 and 19 so that the resulting numbers may be in continued proportion?

1. −4

2. −3

3. 3

4. 4

Let the number to be added be x.

Thus, numbers 7 + x, 11 + x and 19 + x will be in continued proportion.

$\therefore \dfrac{7 + x}{11 + x} = \dfrac{11 + x}{19 + x} \\[1em] \Rightarrow (7 + x)(19 + x) = (11 + x)(11 + x) \\[1em] \Rightarrow 133 + 7x + 19x + x^2 = 121 + 11x + 11x + x^2 \\[1em] \Rightarrow x^2 + 26x + 133 = x^2 + 22x + 121 \\[1em] \Rightarrow x^2 - x^2 + 26x - 22x = 121 - 133 \\[1em] \Rightarrow 4x = -12 \\[1em] \Rightarrow x = \dfrac{-12}{4} \\[1em] \Rightarrow x = -3.$

Hence, option 2 is the correct option.