Let $a, b, c$ be in arithmetic progression. Let the centroid of the triangle with vertices $(a, c),(2, b)$ and $(a, b)$ be $\left(\frac{10}{3}, \frac{7}{3}\right) .$ If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+1=0$, then the value of $\alpha^{2}+\beta^{2}-\alpha \beta$ is:
Correct Option: , 4
$2 b=a+c$
$\frac{2 a+2}{3}=\frac{10}{3}$ and $\frac{2 b+c}{3}=\frac{7}{3}$
$\left.a=4, \begin{array}{r}2 b+c=7 \\ 2 b-c=4\end{array}\right\}$, solving
$b=\frac{11}{4}$
$c=\frac{3}{2}$
$\therefore$ Quadratic Equation is $4 x^{2}+\frac{11}{4} x+1=0$
$\therefore$ The value of $(\alpha+\beta)^{2}-3 \alpha \beta=\frac{121}{256}-\frac{3}{4}=-\frac{71}{256}$
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