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

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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

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  1. 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.

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