Question

How to find the sum of the first 7 terms of the geometric sequence: 24, 12, 6, 3, ……?

Answer 1

We can apply the formula for finding the sum of G.P. which means geometric progression for the terms of the form $a,ar,a{r^2},a{r^3},...$. The formula is given by ${S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$ where $a$ is the first term of the series, $r = \dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}}$. Also ${S_n}$ is the sum of the first $n$ terms.

Complete step-by-step solution:
Now, we will consider the geometric series 24, 12, 6, 3, ……
Here, the first term $a = 24$ and we will apply the formula $r = \dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}}$ for finding $r$.
Basically, $r$ is the division of the second term by the first term. Therefore, we have that
$r = \dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}}$
$\Rightarrow r = \dfrac{{12}}{{24}}$
$\Rightarrow r = \dfrac{1}{2}$
Since, by dividing any second term with respect to its first consecutive term we will always get in this question $r$ as $\dfrac{1}{2}$. As we need to find the sum of first 7 terms as it is mentioned in the question. Therefore, we have $n = 7$. Now, we will apply the formula for finding the sum of G.P. which means geometric progression for the terms of the form $a,ar,a{r^2},a{r^3},...$. The formula is given by ${S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$ where $a$ is the first term of the series, $r = \dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}}$ and ${S_n}$ is the sum of the first $n$ terms.
Thus, by substituting $r = \dfrac{1}{2}$, $a = 24$ and $n = 7$ we have
${S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$
$\Rightarrow {S_7} = 24\left( {\dfrac{{1 - {{\left( {\dfrac{1}{2}} \right)}^7}}}{{1 - \dfrac{1}{2}}}} \right)$
As we know that the value of ${2^7} = 128$
therefore, we have
$\Rightarrow {S_7} = 24 \times 2 \times \dfrac {\left( {128 - 1}\right)} {128}$
$\Rightarrow {S_7} = 24 \times 2 \times \dfrac{127}{128}$
$\Rightarrow {S_7} = 47.625$
Hence, the value of the sum of the first 7 terms is 6096.

Note: The formula for finding the sum of G.P. which means geometric progression for the terms of the form $a,ar,a{r^2},a{r^3},...$. The formula is given by ${S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$ where $a$ is the first term of the series, $r = \dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}}$ and ${S_n}$ is the sum of the first $n$ terms should not be applied directly here. As to apply this formula we need to check whether the series given to us are in G.P. or not. This can be found out by the formula $r = \dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}}$ and using this formula to check $r$ at least first four terms. If the value of $r$ is coming to be equal for every division here then the formula for the sum is valid here.