- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
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 $ax ^{2}+ bx +1=0$, then the value of $\alpha^{2}+\beta^{2}-\alpha \beta$ is ....... .
A
$\frac{71}{256}$
B
$\frac{69}{256}$
C
$-\frac{69}{256}$
D
$-\frac{71}{256}$
(JEE MAIN-2021)
Solution
$\frac{a+2+a}{3}=\frac{10}{3}$
$a=4$
and $\frac{c+b+b}{3}=\frac{7}{3}$
$c+2 b=7$
also $2 b=a+c$
$2 b-a+2 b=7$
$b=\frac{11}{4}$
now $4 x ^{2}+\frac{11}{4} x +1=0 (0=\alpha \,And \, \beta)$
$\alpha^{2}+\beta^{2}-\alpha \beta=(\alpha+\beta)^{2}-3 \alpha \beta$
$=\left(\frac{-11}{16}\right)^{2}-3\left(\frac{1}{4}\right)$
$=\frac{121}{256}-\frac{3}{4}=\frac{-71}{256}$
Standard 11
Mathematics