Question

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

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

Answer 1

The correct option is B 7

$\left[\begin{array}{ccc}a& b& c\end{array}\right]\left[\begin{array}{ccc}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
$\Rightarrow a+8b+7c=0....\left(i\right)$
and $9a+2b+3c=0....\left(ii\right)$
and $7a+7b+7c=0$ ....(iii)
$\frac{a}{1}=\frac{b}{6}=\frac{c}{7}=k\left(say\right)$
$\Rightarrow a=k,b=6k,c=-7k$
Given $b=6$
$\Rightarrow k=1$
$\Rightarrow a=1,c=-7$
Also, $\alpha$ and $\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$
$\Rightarrow \alpha +\beta =-6,\alpha \beta =-7$
Now, $\sum _{n=0}^{\infty }{(\frac{1}{\alpha }+\frac{1}{\beta })}^n$
$={\sum }_{n=0}^{\infty }{\left(\frac{\alpha +\beta }{\alpha \beta }\right)}^n$
$={\sum }_{n=0}^{\infty }{\left(\frac{6}{7}\right)}^n$
$=\frac{1}{1-\frac{6}{7}}=7$