Let a, b, c be in arithmetic progression.
Question:

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:

1. (1) $\frac{71}{256}$

2. (2) $-\frac{69}{256}$

3. (3) $\frac{69}{256}$

4. (4) $-\frac{71}{256}$

Correct Option: , 4

Solution:

$2 b=a+c$

$\frac{2 a+2}{3}=\frac{10}{3}$ 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 $(\alpha+\beta)^{2}-3 \alpha \beta=\frac{121}{256}-\frac{3}{4}=-\frac{71}{256}$