Question

The sum of first $8$ terms of the geometric series $2 + 6 + 18 + 54 + ........{\text{ is}}$ :
$\left( 1 \right)6506$
$\left( 2 \right)5650$
$\left( 3 \right)6650$
$\left( 4 \right)6560$

Answer 1

To solve this question, we should be familiar with the properties of Geometric progression (G.P.) . A geometric progression is a type of a sequence in which each term can be found out by multiplying the previous term with a constant or a fixed number. Generally, a geometric sequence is written like this; \left\\{ {a,{\text{ }}ar,{\text{ }}a{r^2},{\text{ }}a{r^3},{\text{ }}.........} \right\\} where, $\left( 1 \right)a$ is the first term of the sequence and $\left( 2 \right)r$ is the common factor between the terms and it is called “common ratio”. Example of a G.P. : $4,8,12,16,20$ where the common ratio is $2$ .

Complete step by step answer:
The given sequence is $2 + 6 + 18 + 54 + ........{\text{ }}$ and we are asked to calculate the sum of first $8$ terms of this geometric progression;
Here the first term is $a = 2$ , and the common ratio $r = \dfrac{{{\text{Second term}}}}{{{\text{First term}}}}$ ;
$\Rightarrow r = \dfrac{6}{2} = 3$
We know the formula for sum of $n$ terms for a geometric progression is given by;
$\Rightarrow {S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ (since $r \succ 1$ )
In this question we have to calculate the sum of first $8$ terms of the sequence $2 + 6 + 18 + 54 + ........{\text{ }}$
means the value of $n = 8$ ;
Put all the respective values in the formula for sum of $n$ terms, we get;
$\Rightarrow {S_8} = \dfrac{{2\left( {{3^8} - 1} \right)}}{{3 - 1}}$
The above expression can be further simplified like;
$\Rightarrow {S_8} = \left( {{3^8} - 1} \right)$
$\Rightarrow {S_8} = 6561 - 1$
The final value is, ${S_8} = 6560$.

So, the correct answer is “Option 4”.

Note: Here is the list of some important formulae for G.P. $\left( 1 \right)$The ${n^{th}}$ term of a G.P. is given by, ${a_n} = a{r^{n - 1}}$ (since the first term of the G.P. is $a$ which can also be written as $a{r^0}$ ) we can calculate any term of the G.P. using this formula . $\left( 2 \right)$ The sum of $n$ terms of a finite G.P. is given by, ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ if $r \ne 1{\text{ and r}} \succ {\text{1}}$ and ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}{\text{ if }}r \ne 1{\text{ and }}r \prec 1$ . $\left( 3 \right)$ 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 $0 \prec r \prec 1.$ $\left( 4 \right)$ If three quantities are in G.P. , then the middle quantity is called the geometric mean of the other two quantities, example: if $a,b,c$ are in G.P. then ${b^2} = ac$ or $b = \sqrt {ac}$ .