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 + x2 = 676 + 26x + 26x + x2
640 + 56x + x2 = 676 + 52x + x2
56x + x2 – 52x – x2 = 676 – 640
So we get
4x = 36
x = 36/4 = 9
Hence, 9 is the number to be added to each of the numbers.