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JEE Main & Advanced Mathematics Sequence & Series Question Bank Self Evaluation Test - Sequences and Series

  • question_answer
    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

    A) 1

    B) \[\frac{1}{2}\]

    C) 2

    D) None of these

    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 \[n=3,\] \[\sin {{\alpha }_{3}}=\frac{1}{8}\] \[S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...upto\,\,\infty \] \[=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1\]


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