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

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$-7$

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(D) Let the number to be added be $x$. The new numbers are $(13 + x), (15 + x), (19 + x)$.Since these numbers are in $H.P.$,their reciprocals are in $A.P.$Therefore,$\frac{1}{15 + x} - \frac{1}{13 + x} = \frac{1}{19 + x} - \frac{1}{15 + x}$.$\Rightarrow \frac{(13 + x) - (15 + x)}{(15 + x)(13 + x)} = \frac{(15 + x) - (19 + x)}{(19 + x)(15 + x)}$.$\Rightarrow \frac{-2}{(15 + x)(13 + x)} = \frac{-4}{(19 + x)(15 + x)}$.$\Rightarrow \frac{1}{13 + x} = \frac{2}{19 + x}$.$\Rightarrow 19 + x = 2(13 + x)$.$\Rightarrow 19 + x = 26 + 2x$.$\Rightarrow x = 19 - 26 = -7$.Thus,the number to be added is $-7$.

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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