Question

For a sequence $\left\{ {{a_n}} \right\}$, ${a_1} = 2$ and $\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{1}{3}$. Then what is the value of $\sum\limits_{r = 1}^{20} {{a_r}} $?
A. $\left( {\dfrac{{20}}{2}} \right)\left( {4 + 19 \times 3} \right)$
B. $3\left( {1 - \dfrac{1}{{{3^{20}}}}} \right)$
C. $2\left( {1 - {3^{20}}} \right)$
D. None of these

Answer 1

Hint: Use the given terms of a sequence $\left\{ {{a_n}} \right\}$ and find the values of some terms of the sequence. Check the difference between the two consecutive terms of the sequence. If the terms of the sequence are in geometric progression, then apply the formula of the sum of $n$ terms in GP. If the terms of the sequence are in arithmetic progression, then apply the formula of the sum of $n$ terms in AP.

Formula Used:
The sum of $n$ terms in geometric progression: ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$, $r < 1$, where $a$ is the first term and $r$ is the common ratio.

Complete step by step solution:
 The given term of the sequence $\left\{ {{a_n}} \right\}$ is ${a_1} = 2$, and $\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{1}{3}$.
From the equation $\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{1}{3}$, we get
${a_{n + 1}} = \dfrac{{{a_n}}}{3}$
Substitute $n = 1$ in the above equation.
${a_{1 + 1}} = \dfrac{{{a_1}}}{3}$
$ \Rightarrow {a_2} = \dfrac{2}{3}$
Again, substitute $n = 2$ in the equation ${a_{n + 1}} = \dfrac{{{a_n}}}{3}$.
${a_{2 + 1}} = \dfrac{{{a_2}}}{3}$
$ \Rightarrow {a_3} = \dfrac{{\dfrac{2}{3}}}{3}$
$ \Rightarrow {a_3} = \dfrac{2}{{{3^2}}}$

The sequence is, $\left\{ {{a_n}} \right\} = 2,\dfrac{2}{3},\dfrac{2}{{{3^2}}},....$
The ratio of the two consecutive terms of the sequence is: $r = \dfrac{{\dfrac{2}{3}}}{2} = \dfrac{1}{3}$.
Hence, the terms of the sequence are in geometric progression with first term ${a_1} = 2$.

Now calculate the sum of the first $20$ terms of the sequence.
Let ${S_{20}}$ be the sum of the first $20$ terms.
$\sum\limits_{r = 1}^{20} {{a_r}} = {S_{20}}$
Apply the formula of the sum of $n$ terms in geometric progression.
$\sum\limits_{r = 1}^{20} {{a_r}} = \dfrac{{2\left( {1 - {{\left( {\dfrac{1}{3}} \right)}^{20}}} \right)}}{{1 - \dfrac{1}{3}}}$
$ \Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = \dfrac{{2\left( {1 - \dfrac{1}{{{3^{20}}}}} \right)}}{{\dfrac{2}{3}}}$
$ \Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = \dfrac{{\left( {1 - \dfrac{1}{{{3^{20}}}}} \right)}}{{\dfrac{1}{3}}}$
$ \Rightarrow \sum\limits_{r = 1}^{20} {{a_r}} = 3\left( {1 - \dfrac{1}{{{3^{20}}}}} \right)$

Option ‘B’ is correct

Note: A geometric series is a sequence in which each consecutive element is obtained by multiplying the preceding element by a constant. The constant value is called a common ratio.
The general form of a geometric series is $a,ar,a{r^2},a{r^3},....$. Where $a$ is the first term and $r$ is a common ratio.