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(D) Let the number to be added be $x$. Then the new numbers are $(13 + x), (15 + x), (19 + x)$. Since these numbers are in $H.P.$,their reciprocals mu

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$-7$

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(D) Let the number to be added be $x$. Then the new numbers are $(13 + x), (15 + x), (19 + x)$. Since these numbers are in $H.P.$,their reciprocals must be in $A.P.$ Therefore,$\frac{1}{15 + x} - \frac{1}{13 + x} = \frac{1}{19 + x} - \frac{1}{15 + x}$. This implies $\frac{(13 + x) - (15 + x)}{(15 + x)(13 + x)} = \frac{(15 + x) - (19 + x)}{(19 + x)(15 + x)}$. $\frac{-2}{(15 + x)(13 + x)} = \frac{-4}{(19 + x)(15 + x)}$. Canceling $(15 + x)$ from both sides (assuming $x 
eq -15$),we get $\frac{-2}{13 + x} = \frac{-4}{19 + x}$. $2(19 + x) = 4(13 + x)$. $38 + 2x = 52 + 4x$. $-14 = 2x$. $x = -7$.

Which number should be added to the numbers $13, 15, 19$ so that the resulting numbers are the consecutive terms of a $H.P.$?

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