The value of k that must be added to 7, 16, 4 | vedclass

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(B) For four numbers $a, b, c,$ and $d$ to be in proportion after adding a constant $k$,the condition is $\frac{a+k}{b+k} = \frac{c+k}{d+k}$.Given $a=

Vedclass · Accepted Answer

$5$

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(B) For four numbers $a, b, c,$ and $d$ to be in proportion after adding a constant $k$,the condition is $\frac{a+k}{b+k} = \frac{c+k}{d+k}$.Given $a=7, b=16, c=43, d=79$.Substituting these values: $\frac{7+k}{16+k} = \frac{43+k}{79+k}$.Cross-multiplying: $(7+k)(79+k) = (16+k)(43+k)$.Expanding both sides: $553 + 7k + 79k + k^2 = 688 + 16k + 43k + k^2$.Simplifying: $553 + 86k = 688 + 59k$.Rearranging terms: $86k - 59k = 688 - 553$.$27k = 135$.$k = \frac{135}{27} = 5$.Thus,the value of $k$ is $5$.

The value of $k$ that must be added to $7, 16, 43,$ and $79$ so that they are in proportion is

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