Question

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 $S=\sum _{i=1}^{\infty }\sin {\alpha }_i$, then the value of $S$ is

• A. 1
• B. $\frac{1}{2}$
• C. 2
• D. None of these

Answer 1

Answer: (A)

$2=2^{n}sin\alpha +\frac{1}{2^n\sin \alpha }$
$\Rightarrow 2\cdot 2^n\sin \alpha ={\left(2^n\sin \alpha \right)}^2+1$
$\Rightarrow {\left(2^n\sin \alpha -1\right)}^2=0$
$\Rightarrow \sin \alpha =\frac{1}{2^n}$
for $n=1,\sin {\alpha }_1=\frac{1}{2};$ for $n=2,\sin {\alpha }_2=\frac{1}{4};$
for $n=3,\sin {\alpha }_3=\frac{1}{8}$
$S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$ upto $\infty$
$=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$