Chapter 7.3, Problem 46E

Solution

To calculate: The geometric sequence that contains the two specified terms $a_3=\frac{7}{4}\text{and}{a}_5=\frac{7}{16}$

The required rule is $a_n=7\cdot {\left(\frac{1}{2}\right)}^{n-1}$

Given information:

The given terms are $a_3=\frac{7}{4}\text{and}{a}_5=\frac{7}{16}$ .

Formula used:

The nth term of a geometric sequence with first term $a_1$ and common ratio $r$ is given by: $a_n=a_1\cdot r^{n-1}$

calculation:

Consider the given sequence.

It is given that $a_3=\frac{7}{4}\text{and}{a}_5=\frac{7}{16}$ .

First write a system of equation using $a_n=a_1{r}^{n-1}$

Put the value of $a_3=\frac{7}{4}\text{and}{a}_5=\frac{7}{16}$ into $a_n=a_1{r}^{n-1}$

  $\begin{array}{l}{a}_n=a_1{r}^{n-1}\\ a_3=a_1{r}^{3-1}\\ \frac{7}{4}=a_1{r}^2\mathrm{...}\left(1\right)\end{array}$

  $\begin{array}{l}{a}_n=a_1{r}^{n-1}\\ a_5=a_1{r}^{5-1}\\ \frac{7}{16}=a_1{r}^4\mathrm{...}\left(2\right)\end{array}$

Solve the equation 1

  $\begin{array}{l}{a}_3=a_1{r}^{3-1}\\ \frac{7}{4}=a_1{r}^2\\ a_1=\frac{7}{4r^2}\end{array}$

Put the value of $a_1$ into equation $2$

  $\begin{array}{c}{a}_5=a_1{r}^{5-1}\\ \frac{7}{16}=a_1{r}^4\\ \frac{7}{16}=\frac{7}{4r^2}\cdot r^4\\ \frac{1}{4}=r^2\\ r=\frac{1}{2}\end{array}$

Put the value of $r$ into equation $1$

  $\begin{array}{c}\frac{7}{4}=a_1{r}^2\\ a_1=\frac{7}{4\cdot {\left(\frac{1}{2}\right)}^2}\\ a_1=7\end{array}$

Put the value of $a_1=7\text{and}r=\frac{1}{2}$ in $a_n=a_1\cdot r^{n-1}$

  $a_n=7\cdot {\left(\frac{1}{2}\right)}^{n-1}$

There is solution.

  $a_n=7\cdot {\left(\frac{1}{2}\right)}^{n-1}$