Question

if 1/(x+2),1/(x+3)and 1/(x+5) are in A.P. find the value of x​

Answer 1

Solution :

1/(x+2), 1(x+3) and 1/(x+5) are in AP, therefore, common difference (d) must he same.

Here, a ( first term ) = 1/(x+2)

a2 ( second term ) = 1/(x+3)

a3 ( Third term ) = 1/(x+5)

=> a2 - a = a3 - a2

=> 1/(x+3) - 1/(x+2) = 1/(x+5) - 1/(x+3)

By taking LCM,

=> [(x+2)-(x+3)]/[(x+3)(x+2)] = [(x+3)-(x+5)]/[(x+3)(x+5)]

=> (x+2-x-3)/(x+2) = (x+3-x-5)/(x+5)

=> -1/(x+2) = -2/(x+5)

By cross multiplying,

=> -1(x+5) = -2(x+2)

=> -x-5 = -2x-4

=> -x+2x = -4+5

=> x = 1 ( required answer )

Answer 2

Solution

Given,

$\tt{\dfrac{1}{x + 2}, \: \dfrac{1}{x + 3} and \: \dfrac{1}{x + 5} are \: in \: AP}$

Since,the terms are in AP. Subtracting a term with its preceding term would give the common difference of the AP

Now,

$\tt{ {a}_{2} - {a}_{1}} \\ \\ \tt{= \dfrac{1}{x + 3} - \dfrac{1}{x + 2}} \\ \\ = \tt{ \frac{(x + 2) - (x + 3)}{(x + 2)(x + 3)} } \\ \\ = \tt{ \frac{ - 1}{(x + 2)(x + 3)} - - - - - - (1)}$

Also,

$\tt{ {a}_{3} - {a}_{2} } \\ \\ = \tt{ \frac{1}{x + 5} - \frac{1}{x + 3} } \\ \\ \tt{ = \frac{(x + 3) - (x + 5)}{(x + 3)(x + 5)} } \\ \\ = \tt{ \frac{ - 2}{(x + 3)(x + 5)} - - - - - (2) }$

Equating equations (1) and (2),we get :

$\longrightarrow \: \tt{ \frac{ - 1}{(x + 2) \cancel{(x + 3)}} = \frac{ - 2}{ \cancel{(x + 3)}(x + 5)} } \\ \\ \longrightarrow \: \tt{x + 5 = 2(x + 2)} \\ \\ \longrightarrow \: \tt{2x - x= 5 - 4} \\ \\ \longrightarrow \: \boxed{ \boxed{ \tt{x = 1}}}$