Q50 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics
Let $A_r=\left(x+\frac{1}{x}\right)^3 \cdot\left(x^2+\frac{1}{x^2}\right)^3 \cdot\left(x^3+\frac{1}{x^3}\right)^3 \ldots$. $\left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$, then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\ldots . . \infty=$
- A$\frac{1}{6}$
- B$\frac{2}{5}$
- C1
- Correct answerD$\frac{1}{7}$
Solution
It is given that, if $x^2+x+1=0$, then
$A_r=\left(x+\frac{1}{x}\right)^3\left(x^2+\frac{1}{x^2}\right)^3\left(x^3+\frac{1}{x^3}\right)^3 \ldots\left(x^r+\frac{1}{x^r}\right)^3$
As, we know that $w$ and $w^2$ are roots of quadratic equation $x^2+x+1=0$, so $w^3=1$ and
$1+w+w^2=0$
$\therefore \quad A_r=\left(w+\frac{1}{w}\right)^3\left(w^2+\frac{1}{w^2}\right)^3\left(w^3+\frac{1}{w^3}\right)^3$ $\ldots\left(w^r+\frac{1}{w^r}\right)^3$
$\therefore \quad A_3=(-1)(-1)(8)=8$
Similarly, $A_6=8^2, A_9=8^3, \ldots$ and so on.
$\therefore \quad \frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\ldots \infty$
$=\frac{1}{8}+\frac{1}{8^2}+\frac{1}{8^3}+\frac{1}{8^4}+\ldots \infty=\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac{1}{7}$
Hence, the correct answer is option: 'D'.