Q39 / 69 · JEE Advanced 2011 (Paper 1) · Mathematics

Paragraph: Let $a$, b and $c$ be three real numbers satisfying $\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$Question: Let $b=6$, with $a$ and $c$ satisfying Eq. (E). If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^2+b x+c=0$, then $\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^n$ is

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

Solution

$$ \begin{aligned} & \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right]_{3 \times 3}=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right] \\ & \Rightarrow \quad\left[\begin{array}{c} a+8 b+7 c \\ 9 a+2 b+3 c \\ 7 a+7 b+7 c \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow \quad a+8 b+7 c=0 \\ & \Rightarrow 9 a+2 b+3 c=0 \\ & \Rightarrow \quad a+b+c=0 \\ & \end{aligned} $$ On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get $$ 7 a+c=0 $$ Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get $$ \begin{array}{ll} & 6 a-b=0 \\ \therefore \quad & b=6 a \text { and } c=-7 a \end{array} $$ If $b=6, a=1$ and $c=-7$ $$ \begin{array}{ll} \therefore & a x^2+b x+c=0 \\ \Rightarrow & x^2+6 x-7=0 \\ \Rightarrow & (x+7)(x-1)=0 \\ \therefore & x=1,-7 \\ \Rightarrow & \quad \sum_{n=0}^{\infty}\left(\frac{1}{1}-\frac{1}{7}\right)^n \Rightarrow \sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^n \\ \Rightarrow & 1+\frac{6}{7}+\left(\frac{6}{7}\right)^n+\ldots \infty \\ & =\frac{1}{1-\frac{6}{7}}=\frac{1}{1 / 7}=7 \end{array} $$

Hence, the correct answer is option: 'B'.

Relevant Questions