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

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Answer 1 · 2018-03-01T10:05:03

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

therefore, $x^2+x+1-(2x+1) = 3x^2-3x+3-(x^2+x+1)$
since difference between consecutive terms is the same
$=>$ common difference.

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

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

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

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

$x(x-1)-2(x-1)= 0$

$(x-2)(x-1)=0$

$x=1 or x=2$