Question

if
$\frac{1}{x + 2} + \frac{1}{x + 3} + \frac{1}{x + 5}$
are in AP then x =???​

Answer 1

$\huge\underline\mathrm{Answer-}$

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$\large{\boxed{\red{\rm{x\:=\:1}}}}$

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$\huge\underline\mathrm{Explanation-}$

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$\sf{\dfrac{1}{x + 2} + \dfrac{1}{x + 3} + \dfrac{1}{x + 5}}$

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It is given that this is in AP, so the common difference must be equal.

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$\therefore$ $\sf{\dfrac{1}{x+3}\:-\:\dfrac{1}{x+2}\:=\:\dfrac{1}{x+5}\:-\:\dfrac{1}{x+3}}$

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$\implies$ $\sf{\dfrac{(x+2)-(x+3)}{(\cancel{x+3})(x+2)}\:=\:\dfrac{(x+3)-(x+5)}{(x+5)(\cancel{x+3})}}$

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$\implies$ $\sf{-1(x+5)\:=\:-2(x+2)}$

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$\implies$ $\sf{-x+5\:=\:-2x-4}$

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$\implies$ $\sf{-x+2x\:=\:-4+5}$

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$\large{\boxed{\red{\rm{x\:=\:1}}}}$

Answer 2

$\bf Given:\ \tt \dfrac{1}{x\ +\ 2}\ +\ \dfrac{1}{x\ +\ 3}\ +\ \dfrac{1}{x\ +\ 5}\ are\ in\ AP.\\\\\\\bf To\ find:\ \tt The\ value\ of\ x.\\\\\\\bf Answer:\\\\\tt Suppose\ a,\ a_2\ and\ a_3\ are\ in\ AP.\ Then,\\\\\\\bf a_2\ -\ a\ =\ a_3\ -\ a_2\\\\\\\bf From\ the\ data\ we\ have,\\\\\\\tt a\ =\ \dfrac{1}{x\ +\ 2}\\\\\\a_2\ =\ \dfrac{1}{x\ +\ 3}\\\\\\a_3\ =\ \dfrac{1}{x\ +\ 5}\\\\\\\implies\ \dfrac{1}{x\ +\ 3}\ -\ \dfrac{1}{x\ +\ 2}\ =\ \dfrac{1}{x\ +\ 5}\ -\ \dfrac{1}{x\ +\ 3}$

$\tt \dfrac{(x\ +\ 2)\ -\ (x\ -\ 3)}{(x\ +\ 3)(x\ +\ 2)}\ =\ \dfrac{(x\ +\ 3)\ -\ (x\ +\ 5)}{(x\ +\ 5)(x\ +\ 3}\\\\\\\\\dfrac{-1}{(x\ +\ 3)(x\ +\ 2)}\ =\ \dfrac{-2}{(x\ +\ 5)(x\ +\ 3)}\\\\\\\\-1(x\ +\ 5)(x\ +\ 3)\ =\ -2(x\ +\ 3)(x\ +\ 2)\\\\\\-1(x\ +\ 5)\ =\ -2(x\ +\ 2)\\\\\\-x\ -\ 5\ =\ -2x\ -\ 4\\\\\\-5\ +\ 4\ =\ -2x\ +\ x\\\\\\-1\ =\ -x\\\\\\\bf x\ =\ 1$