Chapter 8.3, Problem 40E

To determine

Write a rule for the nth term of the geometric sequence.

Answer to Problem 40E

Rule for the nth term of the geometric sequence is $\boxed{a_n=192\cdot {\left(-\frac{1}{4}\right)}^{n-1}}$

Explanation of Solution

Given information:

  $\begin{array}{l}{a}_2=-48\\ a_5=\frac{3}{4}\end{array}$

Formula used:

Nth term formula of a GP series $a_n=a_1{r}^{n-1}$

Write a system of equation using formula $a_n=a_1{r}^{n-1}$:

Equation 1:

  $\begin{array}{l}{a}_2=a_1{r}^{2-1}\\ -48=a_{1}r\end{array}$

Equation 2:

  $\begin{array}{l}{a}_5=a_1{r}^{5-1}\\ \frac{3}{4}=a_1{r}^4\end{array}$

Solve Equation 1 for $a_1$ :

  $a_1=\frac{-48}{r}$

Solve Equation 2 for $a_1$ :

  $a_1=\frac{3}{4r^4}$

Now equating $\frac{-48}{r}$ and $\frac{3}{4r^4}$ as they are equivalent to $a_1$ :

  $\begin{array}{l}\frac{-48}{r}=\frac{3}{4r^4}\\ \\ \frac{r^4}{r}=-\frac{3}{192}\\ \\ r^3=-\frac{1}{64}\\ \\ \boxed{r=-\frac{1}{4}}\end{array}$

Now putting value of $r$ in Equation 1 to find the value of $a_1$ :

  $\begin{array}{l}-48=a_1\left(-\frac{1}{4}\right)\\ \boxed{a_1=192}\end{array}$

Putting value of $a_1$ and $r$ in $a_n=a_1{r}^{n-1}$ to find rule for nth term:

  $\boxed{a_n=192\cdot {\left(-\frac{1}{4}\right)}^{n-1}}$