Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be in arithmetic progression. Let the centroid of the triangle with vertices $(\mathrm{a}, \mathrm{c}),(2, \mathrm{~b})$ 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:

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Sequences and Series

Let

a, b, c

be in arithmetic progression. Let the centroid of the triangle with vertices

(a, c), (2, b)

and

(a, b)

(\frac{10}{3}, \frac{7}{3})

. If

α, β

are the roots of the equation

a x^{2} + b x + 1 = 0

, then the value of

α^{2} + β^{2} - α β

is:

Answer: (D)

The centroid condition on the abscissae gives

\frac{a + 2 + a}{3} = \frac{10}{3}

, so

a = 4

, and on the ordinates

\frac{c + b + b}{3} = \frac{7}{3}

, so

c + 2 b = 7

. Since

a, b, c

are in arithmetic progression,

2 b = a + c = 4 + c

, and substituting this gives

c + (4 + c) = 7

, so

c = \frac{3}{2}

and

b = \frac{11}{4}

. The quadratic is therefore

4 x^{2} + \frac{11}{4} x + 1 = 0

, for which

α + β = - \frac{11}{16}

and

α β = \frac{1}{4}

. Then

α^{2} + β^{2} - α β = (α + β)^{2} - 3 α β = \frac{121}{256} - \frac{192}{256},

which equals

- \frac{71}{256}

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Single correct option