Question

Let $a,b$ and $c$ be three real numbers satisfying
$\left[\begin{array}{ccc}a& b& c\end{array}\right]\left[\begin{array}{ccc}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$........ $\left(E\right)$

Let $b=6$, with $a$ and $c$ satisfying (E). lf $\alpha$ and $\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$, then $\sum _{n=0}^{\infty }{\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)}^n$ is:

• A. $6$
• B. $7$
• C. $\frac{6}{7}$
• D. $\infty$

Answer 1

The correct option is B $7$

Since, a ,b and c satisfies
$\left[\begin{array}{ccc}a& b& c\end{array}\right]\left[\begin{array}{ccc}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
So, we get the equations
$a+8b+7c=0$
$9a+2b+3c=0$
$7a+7b+7c=0$ or $a+b+c=0$
Since , $b=6$, so the equations become
$a+48+7c=0$ .....(1)
$9a+12+3c=0$ .....(2)
$a+6+c=0$ .....(3)
Subtracting (3) from (1), we get
$c=-7$
Using equation (3), we get $a=1$.
So, we have $a=1,b=6,c=-7$
So, the equation $a{x}^2+bx+c=0$ becomes $x^2+6x-7=0$
Since $\alpha$ and $\beta$ are the roots of this equation,
So, $\alpha +\beta =\frac{-b}{a}=\frac{-6}{1}=-6$
and $\alpha \beta =\frac{c}{a}=\frac{-7}{1}=-7$
Now, ${\sum }_{n=0}^{\infty }{\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)}^n={\sum }_{n=0}^{\infty }{\left(\frac{\alpha +\beta }{\alpha \beta }\right)}^n$
$={\sum }_{n=0}^{\infty }{\left(\frac{-6}{-7}\right)}^n={\sum }_{n=0}^{\infty }{\left(\frac{6}{7}\right)}^n$
$=1+\frac{6}{7}+{(\frac{6}{7})}^2+{(\frac{6}{7})}^3+...$
This is an infinite G.P.. $S_{\infty }=\frac{a}{1-r}$
${\sum }_{n=0}^{\infty }{\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)}^n=\frac{1}{1-\frac{6}{7}}=7$