What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion? Let x be added to each number then16 + x, 26 + x and 40 + xare in continued proportion.⇒ `(16 + x)/(26 + x) = (26 + x)/(40 + x)`Cross Multiplying(16 + x) (40 + x) = (26 + x) (26 + x) ⇒ 640 + 16x + 40x + x2 = 676 + 26x + 26x + x2 ⇒ 640 + 56x + x2 = 676 + 52x + x2 ⇒ 56x + x2 - 52x - x2 = 676 - 640⇒ 4x = 36⇒ x = `(36)/(4)` = 9 ∴ 9 is to be added. Concept: Concept of Proportion Is there an error in this question or solution?
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Answer:
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 \begin{array}{l} 640+16 x+40 x+x^{2}=676+26 x+26 x+x^{2} \\ 640+56 x+x^{2}=676+52 x+x^{2} \\ 56 x+x^{2}-52 x-x^{2}=676-640 \end{array} 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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