Question

1/x-1,1/x-2,1/x-5
is an ap so find value of x =?​

Answer 1

Answer:

x= -2,1

Step-by-step explanation:

Given:

a₁ = 1/x-1

a₂ = 1/x-2

a₃ = 1/x-5

As we know that, if a series is and an ap, then it must have a common difference.

∴a₂ - a₁ = a₃ - a₂

⇒ 1/x-2 - 1/x-1 = 1/x-5 - 1/x-2

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

⇒ 1/x²-x-2x+2 = 3/x²-2x-5x+10

By cross multiplication,

x²-2x-5x+10 = 3 (x²-x-2x+2)

⇒ x²-7x+10 = 3x²-9x+6

⇒ x²-3x²-7x+9x+10-6 = 0

⇒ -2x²-2x+4 = 0

⇒ -2(x²+x-2) = 0

⇒ x²+x-2 = 0

By middle term or you can use any factorizing method

x²-x+2x-2 = 0

⇒ x(x-1)+2(x-1) = 0

⇒ (x+2) (x-1) = 0

Now, x²+x-2 this term has two heroes

∴ x+2 = 0

⇒ x = -2

x-1 = 0

⇒ x = 1

Therefor both numbers can be there, or you can find it by substituting the numbers one by one in the ap.