Question

How do you find the sum of a geometric series for which $a_1=48$, $a_n=3$, and $r=-\frac{1}{2}$?

Answer 1

Step-1: Model the given situation as a geometric progression:

Recall that the last term $a_n$ of a geometric sequence is given by $a_n=a_1{r}^{n-1}$, where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. Here, it is given that $a_1=48$, $r=-\frac{1}{2}$, and $a_n=3$.

Substitute $a_n=a_1{r}^{n-1}$ in the equation $a_n=3$:

$\begin{array}{rcl}{a}_n& =& 3\\ & \Rightarrow & a_1{r}^{n-1}=3\end{array}$

Thus, $\begin{array}{rcl}{a}_1{r}^{n-1}& =& 3\end{array}$.

Step-2: Find the number of terms $\begin{array}{l}n\end{array}$ in the given geometric progression.

Substitute $a_1=48$, and $r=-\frac{1}{2}$ in the equation $\begin{array}{rcl}{a}_1{r}^{n-1}& =& 3\end{array}$.

Then, solve for $n$:

$\begin{array}{rcl}& \Rightarrow & 48\times {\left(-\frac{1}{2}\right)}^{n-1}=3\\ & \Rightarrow & {\left(2\right)}^4\times 3\times {\left(-\frac{1}{2}\right)}^n÷\left(-\frac{1}{2}\right)=3\\ & \Rightarrow & {\left(2\right)}^4\times {\left(-\frac{1}{2}\right)}^n\times \left(-2\right)=1\\ & \Rightarrow & -{\left(2\right)}^5\times \frac{1}{{\left(-2\right)}^n}=1\\ & \Rightarrow & \frac{{\left(-2\right)}^5}{{\left(-2\right)}^n}=1\\ & \Rightarrow & {\left(-2\right)}^{5-n}={\left(-2\right)}^0\\ & \Rightarrow & 5-n=0\\ & \Rightarrow & 5=n\\ & & \end{array}$

Thus, the number of terms is and conditions $n=5$.

Step-3: Find the sum of the given geometric progression.

Recall that the sum of the first $n$ terms is $S_n=a_1\frac{\left(1-r^n\right)}{\left(1-r\right)}$.

Substitute $n=5$, $a_1=48$, and $r=-\frac{1}{2}$ in the equation $S_n=a_1\frac{\left(1-r^n\right)}{\left(1-r\right)}$ to find the sum of the present geometric progression.

$\begin{array}{rcl}{S}_n& =& a_1\frac{\left(1-r^n\right)}{\left(1-r\right)}\\ & \Rightarrow & S_5=\left(48\right)\frac{\left(1-{\left(-{\displaystyle \frac{1}{2}}\right)}^5\right)}{\left(1-\left(-{\displaystyle \frac{1}{2}}\right)\right)}\\ & \Rightarrow & S_5=\left(48\right)\times \frac{\left(1-\left({\displaystyle \frac{1}{{\left(-2\right)}^5}}\right)\right)}{\left(1+{\displaystyle \frac{1}{2}}\right)}\\ & \Rightarrow & S_5=48\times \frac{\left(1-{\displaystyle \frac{1}{\left(-32\right)}}\right)}{\left({\displaystyle \frac{3}{2}}\right)}\\ & \Rightarrow & S_5=48\times \left(1+\frac{1}{32}\right)÷\left(\frac{3}{2}\right)\\ & \Rightarrow & S_5=48\times \left(\frac{32+1}{32}\right)\times \frac{2}{3}\\ & \Rightarrow & S_5=48\times \frac{33}{32}\times \frac{2}{3}\\ & \Rightarrow & S_5=3\times \frac{11}{2}\times \frac{2}{1}\\ & \Rightarrow & S_5=33\end{array}$

Hence, the sum of the present geometric progression is $\begin{array}{rcl}{\mathit{S}}_{\mathbf{5}}& \mathbf{=}& \mathbf{33}\end{array}$.