Question:
Question 6: If 1/x+2 , 1/x+3, 1/x+5 are in A.P. Then x =
Answer:
Given that 1/x+2 , 1/x+3, 1/x+5 are in A.P.
=> 2/(x+3) = 1/(x+2) + 1/(x+5)
=> 2/(x+3) = (x+5+x+2)/{(x+2)(x+5)}
=> 2/(x+3) = (2x+7)/{(x2 + 7x +10)}
=>2(x2 + 7x +10) = (x+3)*(2x+7)
=> 2x2 + 14x + 20 = 2x2 + 7x + 6x + 21
=> 2x2 + 14x + 20 = 2x2 + 13x + 21
=> 14x + 20 = 13x + 21
=> 14x - 13x = 21 - 20
=> x = 1
So value of x is 1
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