Let $A_r=\left(x+\frac{1}{x}\right)^3 \cdot\left(x^2+\frac{1}{x^2}\right)^3 \cdot\left(x^3+\frac{1}{x^3}\right)^3 \cdots\left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$, then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\ldots . \infty=$

If $z_1=x_1+i y_1, z_2=x_2+i y_2, z_3=x_1+\frac{i x_2}{2}, z_4=2 y_1+i y_2$ are complex numbers such that $\left|z_1\right|=1,\left|z_2\right|=2$ and $\operatorname{Re} \left(\begin{array}{ll}z_1 & z_2\end{array}\right)=0$, then

Assertion (A) If $z$ is a complex number such that $|z| \geq 3$, then the least value of $\left|z+\frac{3}{z}\right|$ is 1 .

Reason (R) $\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right|$, for any two complex numbers $z_1, z_2$

The correct option among the following is

$$ \text { If }\left(\frac{\cos \theta+i \sin \theta}{\sin \theta+i \cos \theta}\right)^{2020}+\left(\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}\right)^{2021}=x+i y, $$

then the value of $x+y$ at $\theta=\frac{\pi}{2}$ is