Question

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

Let b = 6, with a and c satisfying Eq. (i). 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 equal to

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

Answer 1

The correct option is B 7

On multiplying matrices, we have
$\left[\begin{array}{ccc}a+8b+7c& 9a+2b+3c& 7a+7b+7c\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
On solving the 3 equations, we get $b=6a\text{and}c=-7a$

If b = 6 $\Rightarrow$ a = 1 and c =-7
$\therefore a{x}^2+bx+c=0\Rightarrow x^2+6x-7=0$
$\Rightarrow$ (x+7) (x-1) = 0
$\therefore$ x = 1, - 7
$\Rightarrow {\sum }_{n=0}^{\infty }{(\frac{1}{1}-\frac{1}{7})}^n=1+\frac{6}{7}+{\left(\frac{6}{7}\right)}^2+\cdots +\infty =\frac{1}{1-\frac{6}{7}}$ . . . [Sum of infinite GP series]
$\Rightarrow \frac{1}{\frac{1}{7}}=7$