For a sequence $ ,{a_1} = 2$ and $\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}.$ Then, $\mathop \sum \limits_{r = 1}^{20} {a_r}$ is

$\frac{{20}}{2}\left[ {4 + 19 \times 3} \right]$

$2\left( {1 - {3^{20}}} \right)$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

For a sequence

< {a_n} > ,{a_1} = 2

and

\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}.

Then,

\sum \limits_{r = 1}^{20} {a_r}

View answer and explanation

Correct option: B

3\left( {1 - \frac{1}{{{3^{20}}}}} \right)

Option B (

3\left( {1 - \frac{1}{{{3^{20}}}}} \right)

) is correct because it is the sum of the first 20 terms of the geometric progression.

Practice All Chapter MCQs →

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

b,y,c

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

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

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

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

a, b

G P

a, b, c

\frac{3}{4}