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 $S =\displaystyle\sum_{i=1}^{\infty} \sin \alpha_{i}$, 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 $n=3, \sin \alpha_{3}=\frac{1}{8}$
$S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots$ upto $\infty$
$=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$