It is given that $$\frac{1}{{{2^n}\sin \alpha }},1,{2^n}\sin \alpha$$    are in A.P. for some value of $$\alpha .$$ Let say for $$n = 1,$$  the $$\alpha $$ satisfying the above A.P. is $${\alpha _1},$$ for $$n = 2,$$  the value is $${\alpha _2}$$ and so on. If $$S = \sum\limits_{i = 1}^\infty {\sin {\alpha _i}} ,$$   then the value of $$S$$ is

Solution :
$$\eqalign{ & 2 = {2^n}\sin \alpha + \frac{1}{{{2^n}\sin \alpha }} \cr & 2 \cdot {2^n}\sin \alpha = {\left( {{2^n}\sin \alpha } \right)^2} + 1 \cr & {\left( {{2^n}\sin \alpha - 1} \right)^2} = 0 \cr & \sin \alpha = \frac{1}{{{2^n}}} \cr & {\text{for, }}n = 1,\sin {\alpha _1} = \frac{1}{2} \cr & {\text{for, }}n = 2,\sin{\alpha _2} = \frac{1}{4} \cr & {\text{for, }}n = 3,\sin {\alpha _3} = \frac{1}{8} \cr & S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .....\,{\text{upto }}\infty \cr & = \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = \frac{{\frac{1}{2}}}{{\frac{1}{2}}} = 1 \cr} $$