Chapter 7.3, Problem 44E

Solution

To Calculate: The rule for geometric sequence that contains the two specified terms $a_2=-24\text{and}{a}_5=1536$

The required rule is $a_n=6\cdot {\left(-4\right)}^{n-1}$

Given information:

The given terms are $a_2=-24\text{and}{a}_5=1536$ .

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 $a_2=-24\text{and}{a}_5=1536$

Put the value of $a_2=-24,n=2$ into $a_n=a_1\cdot r^{n-1}$

  $\begin{array}{c}{a}_2=a_1{r}^{2-1}\\ a_2=a_{1}r\\ -24=a_{1}r\\ r=-\frac{24}{a_1}\end{array}$

Put the value of $a_5=1536,n=5$ into $a_n=a_1\cdot r^{n-1}$

  $\begin{array}{c}{a}_5=a_1{r}^{5-1}\\ a_5=a_1{r}^4\\ 1536=a_1{r}^4\end{array}$

Put the value of $r=-\frac{24}{a_1}$ into $1536=a_1\cdot r^4$

  $\begin{array}{c}1536=a_1\cdot {\left(-\frac{24}{a_1}\right)}^4\\ 1536=a_1\cdot \left(\frac{331776}{a^4{}_1}\right)\\ \frac{1536}{331776}=\frac{1}{a^3{}_1}\\ \frac{1}{216}=-\frac{1}{a^3{}_1}\\ a^3{}_1=\sqrt{216}\\ a_1=6\end{array}$

Put the value of $a_1$ into $r=-\frac{24}{a_1}$

  $\begin{array}{c}r=-\frac{24}{a_1}\\ r=-\frac{24}{6}\\ r=-4\end{array}$

Put the value of $a_1=6\text{and}r=-4$ in $a_n=a_1\cdot r^{n-1}$

  $a_n=6\cdot {\left(-4\right)}^{n-1}$

There is solution.

  $a_n=6\cdot {\left(-4\right)}^{n-1}$