This is a basic question from ML aggarwal book of class 10th, chapter 7, ratio and proportion., ICSE board

Here we have three different numbers and we have to find the value of the number that can be added in all of three to make them in continued proportion.

Question 10, exercise 7.2

Solution:Consider x be added to each number

16 + x , 26 + x and 40 + x are in continued proportion

It can be written as

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

By cross multiplication

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

On further calculation

640 + 16x + 40x + x

^{2}= 676 + 26x + 26x + x^{2}640 + 56x + x

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

^{2}– 52x – x^{2}= 676 – 640So we get

4x = 36

x = 36/4 = 9

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