JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

The number of terms of the G.P.

3,\,\,\frac{3}{2},\,\,\frac{3}{4},\,...

needed to obtain a sum of

\frac{3069}{512}

is:

View answer and explanation

Correct option: B

Option B (10) is correct because by using the geometric progression sum formula, we can find the number of terms needed to obtain a sum of

\frac{3069}{512}

More Questions from SEQUENCE OF SERIES

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

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

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

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

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)

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a, b, c

\frac{3}{4}