# 2x+1,x^2+x+1,3x^2-3x+3 are in A•P• then , find the value of X?

Mar 1, 2018

$2 x + 1 , {x}^{2} + x + 1 , 3 {x}^{2} - 3 x + 3$ are in AP

therefore, ${x}^{2} + x + 1 - \left(2 x + 1\right) = 3 {x}^{2} - 3 x + 3 - \left({x}^{2} + x + 1\right)$
since difference between consecutive terms is the same
$\implies$ common difference.

${x}^{2} + x + 1 - 2 x - 1 = 3 {x}^{2} - 3 x + 3 - {x}^{2} - x - 1$

${x}^{2} - x = 2 {x}^{2} - 4 x + 2$

${x}^{2} - 3 x + 2 = 0$

${x}^{2} - x - 2 x + 2 = 0$

$x \left(x - 1\right) - 2 \left(x - 1\right) = 0$

$\left(x - 2\right) \left(x - 1\right) = 0$

$x = 1 \mathmr{and} x = 2$