find x so that 2x+1, x²+x+1&3x²- 3x+ 3 are the consecutive terms of an A.P.

Given,  2x+1, x²+x+1 and 3x²- 3x+ 3 are the consecutive terms of an A.P.

As the terms are in ap, therefore common difference between the given consecutive terms will be equal.

So,

(x² + x + 1) - (2x + 1) = (3x²- 3x+ 3) - (x² + x + 1)

⇒ 2(x² + x + 1) = (3x²- 3x+ 3) + (2x + 1)

⇒ 2x² + 2x + 2 = 3x²- x + 4 

⇒ 0 = x² - 3x + 2 

⇒ x² - 3x + 2 = 0

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) -1(x - 2) = 0

⇒ (x - 2)(x -1) = 0

⇒ x = 2 or x = 1

Here, x = 1 can't be possible because on putting x = 1 we are not getting the given terms in a.p.