Discussion Forum

1)

Let a, b and c be three real numbers satisfying [a b c] $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

Let b=6, with  a and c satisfying Eq. (E). If $\alpha$ and $\beta$ are the roots of the quadratic  equation  $ax^{2}+bx+c=0$ 

 then  $\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^{n}$ is 

Answer:

Option B

Explanation:

 Given  ,$[a  b c]_{1 \times 3}$  $\begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

 $\Rightarrow$ $\begin{bmatrix}a+8b+7c  \\9a+2b+3c \\7a+7b+7c\end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}$

$\Rightarrow$  a+8b+7c=0....(i)

$\Rightarrow$  9a+2b+3c=0 .....(ii)

$\Rightarrow$  a+b+c=0 .....(iii)

 On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get

 7a+c=0...(iv)

 Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get

 6a-b=0....(v)

 $\therefore$  b=6a and c=-7a

 if b=6 , a=1 and c=-7

 $\therefore$ $ax^{2}+bx+c=0$

 $\Rightarrow$    $x^{2}+6x-7=0$

$\Rightarrow$  $(x+7)(x-1)=0$

 $\therefore$ x=1,-7

 $\Rightarrow$  $\sum_{n=0}^{\infty}\left(\frac{1}{1}-\frac{1}{7}\right)^{n}\Rightarrow \sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^{n}$

$\Rightarrow$     $1+\frac{6}{7}+(\frac{6}{7})^{n}+....\infty$

 = $\frac{1}{1-\frac{6}{7}}= \frac{1}{1/7}=7$