Let $$A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.$$ is purely imaginary $$\}$$. Then the sum of the elements in $$\mathrm{A}$$ is :

Let $$\mathrm{a}_{\mathrm{n}}$$ be the $$\mathrm{n}^{\text {th }}$$ term of the series $$5+8+14+23+35+50+\ldots$$ and $$\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$$. Then $$\mathrm{S}_{30}-a_{40}$$ is equal to :

$$25^{190}-19^{190}-8^{190}+2^{190}$$ is divisible by :

The integral $$ \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x $$ is equal to :