Chapter 9, Problem 41RE

Solution

To identify: The common ratio, and an explicit rule for the nth term. If the geometric sequence that contains the two specified terms, $a_4=-192\text{and}{a}_9=196608$

The explicit rule is $a_n=3\cdot {\left(-4\right)}^{n-1}$ and common ratio is $r=-4$ .

Given information:

The given terms are $a_4=-192\text{and}{a}_9=196608$ .

Formula used:

The nth term of an 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 $a_4=-192\text{and}{a}_9=196608$

Determine the value of $r$ by using formula $r^{9-4}$ :

  $\begin{array}{l}{r}^{9-4}=\frac{196608}{-192}\\ r^5=-1024\\ r=-4\end{array}$

Put the value of $a_4=-192$ into $a_n=a_1\cdot r^{n-1}$ :

  $\begin{array}{c}{a}_4=a_1\cdot r^{4-1}\\ a_4=a_1\cdot r^3\\ -192=a_1\cdot r^3\left(\because a_4=-192\right)\\ -192=a_1\cdot {\left(-4\right)}^3\\ -192=a_1\left(-64\right)\\ a_1=3\end{array}$

Put the value of $a_1=3\text{and}r=-4$ in $a_n=a_1\cdot r^{n-1}$ .

  $a_n=3\cdot {\left(-4\right)}^{n-1}$

There are explicit rule.

  $a_n=3\cdot {\left(-4\right)}^{n-1}$