JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

1 + \cos \alpha + {\cos ^2}\alpha + ...\infty = 2 - \sqrt 2 ,

then

\alpha ,\;(0 < \alpha < \pi )

View answer and explanation

Correct option: D

\frac{{3\pi }}{4}

Option D (

\frac{{3\pi }}{{4}}

) is correct because it is the value of

\alpha

that satisfies the given infinite geometric series equation.

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A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

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a,b,c

a,x,b

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{x^2},{b^2},{y^2}

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y

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y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

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G P

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\frac{3}{4}