If 1/x + 2, 1/x + 3, 1/x + 5 are in A.P. Then, x =

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Here, we are given three terms, First term (a1) = 1/(x + 2) Second term (a2) = 1/(x + 3) Third term (a3) = 1/(x + 5) We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get, d = a2 - a1 `d = (1/(x + 3)) - (1/(x + 2 )) d = ((x + 2) - (x - 3))/((x + 2)(x + 3)) d =(x +2-x - 3)/((x + 2)(x + 3)) d =( -1) /((x + 2)(x +3 )) ...............(1) Also, d = a_3 - a_2 d = (1/(x +5 ) ) - (1/(x + 3)) d = (( x + 3) - ( x + 5) ) /((x + 5)(x +3 )) d = (x +3 - x - 5)/((x + 5)(x + 3)) d = (-2)/((x + 5)(x + 3))` .............(2) Now, on equating (1) and (2), we get, `(-2)/((x +5)(x + 3)) = (-1)/((x + 3)(x +2 ))` 2(x +3 )( x + 2) = 1 (x +5 ) ( x +3 ) 2x + 4 = x +5 2x - x = 5 - 4 x = 1 Therefore, for x = 1 , these three terms will form an A.P.

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