Question

It is given that \[\dfrac{1}{{{2^n}\sin \alpha }},1,{2^n}\sin \alpha \]are in A.P. for some value of α. Let say for, the α satisfying the above A.P. is \[{\alpha _1}\], for \[n{\rm{ }} = {\rm{ }}2\], the value is \[{\alpha _2}\], and so on. If \[S = \sum\limits_{i = 1}^\infty {\sin {\alpha _i}} \]then the value of S is
A) \[1\]
B) \[\dfrac{1}{2}\]
C) \[2\]
D) None of these

Answer 1

Hint: in this question we have to find sum of infinite series. In order to find this first apply the property of AP which state that twice the middle term equal to sum of first term and last term. Then find some terms including first and second of series. Then apply the formula of sum of infinite terms of GP.

Formula Used: Sum of infinite term of GP calculated by using
\[{S_\infty } = \dfrac{a}{{1 - r}}\]
Where
a is first term of GP
r is common ratio of GP

Complete step by step solution: \[\dfrac{1}{{{2^n}\sin \alpha }},1,{2^n}\sin \alpha \] are in A.P
Now we know that twice the middle term of AP equal to sum of first term and last term of AP
\[2b = a + c\]
\[2 \times 1 = \dfrac{1}{{{2^n}\sin \alpha }} + {2^n}\sin \alpha \]
\[{2.2^n}\sin \alpha = {({2^n}\sin \alpha )^2} + 1\]
\[{({2^n}\sin \alpha - 1)^2} = 0\]
\[\sin \alpha = \dfrac{1}{{{2^n}}}\]
If \[n = 1\]then \[\sin \alpha = \dfrac{1}{{{2^1}}}\]
If \[n = 2\]then \[\sin \alpha = \dfrac{1}{{{2^2}}}\]
If \[n = 3\]then \[\sin \alpha = \dfrac{1}{{{2^3}}}\]
We know that \[S = \sum\limits_{i = 1}^\infty {\sin {\alpha _i}} \]
\[S = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + .......\infty \]
Sum of infinite term of GP calculated by using
\[{S_\infty } = \dfrac{a}{{1 - r}}\]
\[a = \dfrac{1}{2}\]and \[r = \dfrac{1}{2}\]
\[S = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}}\]
\[S = 1\]

Option ‘A’ is correct
Note: Here we must remember that if twice the middle term of AP equal to sum of first term and last term of AP. Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.