Question

if one upon X + 2, 1 upon X + 3 and 1 upon X + 5 are in AP find the value of x​

Answer 1

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Answer 2

${\bold{\mathcal {\huge {\red {ANSWER :}}}}}$

$a.p. = \frac{1}{x + 2} , \: \frac{1}{x + 3} , \: \frac{1}{x + 5}$

Given:

$a = \frac{1}{x + 2} \\d = a_2 - a_1\\ d = \frac{1}{x + 3} - \frac{1}{x + 2} \\ d \frac{(x +2) - (x + 3)}{(x + 3)(x + 2)}$

$d = \frac{ - 1}{ {x}^{2} + 5x + 6}$

$d = \frac{ - 1}{ {x}^{2} + 2x + 3x + 6 }$

By the formula of A.p.

${an}^{th} = a + (n - 1)d$

${an}^{th} = \frac{1}{x + 2} + (3 - 1) \times \frac{ - 1}{ {x}^{2} + 5x + 6 }$

$= \frac{1}{x + 2} - \frac{ - 2}{ {x }^{2} - 5x + 6} \\ = \frac{ ({x}^{2} - 5x + 6) - 2(x + 2)}{(x + 2)( {x}^{2} - 5x + 6)}$

on solving this complex

we get

${x=1}$