Question

For a sequence \left\\{ {{a_n}} \right\\},{\text{ }}{a_1} = 2{\text{ and }}\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{1}{3}{\text{ , then }}\sum\limits_{r = 1}^{20} {{a_r}} {\text{ is}} ;
$\left( 1 \right)\left( {\dfrac{{20}}{2}} \right)\left( {4 + 19 \times 3} \right)$
$\left( 2 \right)3\left( {1 - \dfrac{1}{{{3^{20}}}}} \right)$
$\left( 3 \right)2\left( {1 - {3^{20}}} \right)$
$\left( 4 \right)$ None of these

Answer 1

To solve this question we must be familiar with the basic concept of geometric progression. A geometric progression is a sequence in which each next term is generated by multiplying the previous term with a constant value. General representation of a geometric sequence is \left\\{ {a,ar,a{r^2},a{r^3},.....} \right\\} where $a$ is the first term of the sequence and $r$ is called the common ratio between the terms $\left( {r = \dfrac{{{\text{First term}}}}{{{\text{Second term}}}}} \right)$ . Example of a geometric progression: $4,8,16,32,64$ is in G.P. having a common ratio of $2$ .

Complete step by step solution:
The first term of the sequence is ;
$\Rightarrow {a_1} = 2$
$\Rightarrow \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{1}{3}$ ( common ratio )
Which means on rearranging we can say; ${a_{n + 1}} = \dfrac{{{a_n}}}{3}{\text{ }}......\left( 1 \right)$
This series is in geometric progression with common ratio $r = \dfrac{1}{3}$ and ${a_1} = 2$ ;
Therefore, we can calculate successive or preceding terms as;
Put the value of $n = 1$ , in equation $\left( 1 \right)$ ;
$\Rightarrow {a_2} = \dfrac{{{a_1}}}{3} = \dfrac{2}{3}$
Put the value of $n = 2$ , in equation $\left( 1 \right)$ ;
$\Rightarrow {a_3} = \dfrac{{{a_2}}}{3} = \dfrac{2}{3}\left( {\dfrac{1}{3}} \right)$
Put the value of $n = 3$ , in equation $\left( 1 \right)$ ;
$\Rightarrow {a_4} = \dfrac{{{a_3}}}{3} = \dfrac{2}{3}{\left( {\dfrac{1}{3}} \right)^2}$
Put the value of $n = 4$ , in equation $\left( 1 \right)$ ;
$\Rightarrow {a_5} = \dfrac{{{a_4}}}{3} = \dfrac{2}{3}{\left( {\dfrac{1}{3}} \right)^3}$
We notice the pattern is ${a_r} = \dfrac{2}{3}{\left( {\dfrac{1}{3}} \right)^{r - 2}}$ ;
Therefore, the geometric progression can be written as ( Since we have to calculate the first $20$ terms) ;
$\Rightarrow {a_1} + {a_2} + {a_3} + {a_4} + ........... + {a_{20}} = 2 + \dfrac{2}{3} + \dfrac{2}{3}\left( {\dfrac{1}{3}} \right) + \dfrac{2}{3}\left( {\dfrac{1}{{{3^2}}}} \right) + ........ + \dfrac{2}{3}\left( {\dfrac{1}{{{3^{18}}}}} \right)$
The above geometric progression can also be as;
$\Rightarrow 2 + \dfrac{2}{3}\left[ {1 + \dfrac{1}{3} + {{\left( {\dfrac{1}{3}} \right)}^2} + {{\left( {\dfrac{1}{3}} \right)}^3} + .....{{\left( {\dfrac{1}{3}} \right)}^{18}}} \right]$
$\Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = {S_{20}}$
We know that for a geometric progression $a + ar + a{r^2} + a{r^3} + a{r^n}$ , the formula of sum of n-terms of G.P. is given by;

$\Rightarrow {S_n} = a\left[ {\dfrac{{\left( {{r^n} - 1} \right)}}{{r - 1}}} \right]{\text{ if }}r \ne 1{\text{ and }}r \succ 1{\text{ }}......\left( 2 \right)$
$\Rightarrow {S_n} = a\left[ {\dfrac{{\left( {1 - {r^n}} \right)}}{{1 - r}}} \right]{\text{ if }}r \ne 1{\text{ and }}r \prec 1{\text{ }}......\left( 3 \right)$
Since, $r = \dfrac{1}{3}$ here means $r \prec 1$ ; therefore we will use the formula specified in equation $\left( 3 \right)$ ;
$\Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = \dfrac{{{a_1}\left( {1 - {r^{20}}} \right)}}{{1 - r}}$
$\Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = \dfrac{{2\left( {1 - {{\left( {\dfrac{1}{3}} \right)}^{20}}} \right)}}{{1 - \dfrac{1}{3}}}$
Simplifying the above expression, we get;
$\Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = \dfrac{{2\left( {1 - \dfrac{1}{{{3^{20}}}}} \right)}}{{\dfrac{2}{3}}}$
On further simplification;
$\Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = 3\left[ {1 - \dfrac{1}{{{3^{20}}}}} \right]$
Therefore, the correct answer for this question is option $\left( 2 \right)$.

Note:
The ${n^{th}}$ term of a G.P. is given by ${a_n} = a{r^{n - 1}}$ ( the first term of a G.P. is $a$ which equals $a{r^0}$) . A geometric series can be finite or infinite . We have discussed the formula to calculate the sum of finite geometric series above. The sum of infinite geometric series is given by $\sum\limits_{n = 0}^\infty {\left( {a{r^n}} \right)} = a\left( {\dfrac{1}{{1 - r}}} \right)$ such that $0 \prec r \prec 1$ .