Question

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

Answer 1

Since it is in AP, the common difference is constant

T2 - T1 = T3 - T2

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

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

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

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

=> x = - 4 + 5

=> x = 1

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Answer 2

since they are in ap , ( refer image )

now factorising the equation

(ab=21 , a+b =10 , we get a = 7 and b = 3 )
x squared + 7x + 3x + 21
x (x+7) + 3 (x+7)
(x+3)(x+7)

x ==> -7 or -3