Sequence And Series Question 407
Question: It is given that $ \frac{1}{2^{n}\sin ,\alpha },1,,2^{n}\sin $ a 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
Options:
A) 1
B) $ \frac{1}{2} $
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ 2=2^{n}\sin \alpha +\frac{1}{2^{n}\sin \alpha } $ $ {{2.2}^{n}}\sin \alpha ={{(2^{n}\sin \alpha )}^{2}}+1 $ $ {{(2^{n}\sin \alpha -1)}^{2}}=0 $ $ \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}+…upto\infty $ $ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1 $
BETA
