Let $a, b$ and $c$ be three real numbers satisfying \[ \left[\begin{array}{lll} a & b & c \end{array}\right]\left[\begin{array}{lll} 1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \end{array}\right] \] Let $b=6$, with $a$ and $c$ satisfying (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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Sequences and Series

Let

a, b

and

c

be three real numbers satisfying

[a b c] 187923777 = [000]

Let

b = 6

, with

a

and

c

satisfying (E). If

α

and

β

are the roots of the quadratic equation

a x^{2} + b x + c = 0

, then

n = 0 \sum \infty (\frac{1}{α} + \frac{1}{β})^{n}

(A)6
(B)7✓ Correct
(C) $\frac{6}{7}$
(D) $\infty$

Answer: (B)

Solution

From

b = 6 a

, with

b = 6

, we get

a = 1

. Then

c = - 7 a = - 7

.

The quadratic equation is

x^{2} + 6 x - 7 = 0

, which factors as

(x + 7) (x - 1) = 0

.

The roots are

α = 1

and

β = - 7

.

Therefore

\frac{1}{α} + \frac{1}{β} = 1 - \frac{1}{7} = \frac{6}{7}

.

The sum is a geometric series:

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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option

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