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

**Given,** 2x+1, x^{2}+x+1 and 3x^{2}- 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,

(x^{2 }+ x + 1) - (2x + 1) = (3x^{2}- 3x+ 3) - (x^{2 }+ x + 1)

⇒ 2(x^{2 }+ x + 1) = (3x^{2}- 3x+ 3) + (2x + 1)

⇒ 2x^{2 }+ 2x + 2 = 3x^{2}- x + 4

⇒ 0 = x^{2 }- 3x + 2

⇒ x^{2 }- 3x + 2 = 0

⇒ x^{2 }- 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.

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