Maths / Arithmetic Progression / General Term Of an AP

QUESTION

Find the value of x if

$\dpi{120} \fn_jvn \large 2x+1, x^2+x+1 \;and\; 3x^2-3x+3 \;are\; in\; AP$

 OPTIONS A. 1 B. 2 C. Both A and B D. 3
Right Option : C

EXPLANATION
Explain TypeExplanation Content
Text

$\dpi{120} \fn_jvn \large 2x+1, x^2+x+1 \;and\; 3x^2-3x+3 \;are\; in\; AP$

$\dpi{120} \fn_jvn \large d_1 = x^2+x+1 - (2x+1)$

$\dpi{120} \fn_jvn \large d_1 = x^2+x+1 - 2x-1=x^2-x$

$\dpi{120} \fn_jvn \large d_2 = 3x^2-3x+3 - (x^2 +x+1)$

$\dpi{120} \fn_jvn \large d_2 = 3x^2-3x+3 - x^2 -x-1= 2x^2-4x+2$

As the three terms are in AP

$\dpi{120} \fn_jvn \large d_1 =d_2$

$\dpi{120} \fn_jvn \large x^2-x =2x^2-4x+2$

$\dpi{120} \fn_jvn \large x^2-x -2x^2+4x-2=0$

$\dpi{120} \fn_jvn \large -x^2+3x-2=0$

$\dpi{120} \fn_jvn \large x^2-3x+2=0$

$\dpi{120} \fn_jvn \large x^2-2x-x+2=0$

$\dpi{120} \fn_jvn \large x(x-2)-1(x-2)=0$

$\dpi{120} \fn_jvn \large (x-1)(x-2)=0$

So either x = 1or x =2

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