MCQ
If $\text{z}=\text{a}+\text{ib}$ lies in third quadrant, then $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant if:
- A$\text{a}>\text{b}>0$
- B$\text{a}<\text{b}<0$
- ✓$\text{b}<\text{a}<0$
- D$\text{b}>\text{a}>0$
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$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2.3}+\frac{1}{3^2}\right)+$
$\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2.3^2}-\frac{1}{3^3}\right)+$
$\left(\frac{1}{2^4}-\frac{1}{2^3 \cdot 3}+\frac{1}{2^2 \cdot 3^2}-\frac{1}{2 \cdot 3^3}+\frac{1}{3^4}\right)+\ldots$ is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to....