Chapter 9.3, Problem 32E

Solution

To identify: The first term, common ratio, and an explicit rule for the nth term .if the geometric sequence that contains the two specified terms, $a_2=3\text{and}{a}_8=192$

The explicit rule is $a_n=\frac{3}{2}\cdot 2^{n-1}$ first term and common ratio is $a_1=\frac{3}{2}\text{and}r=2$ respectively .

Given information:

The given terms are $a_2=3\text{and}{a}_8=192$ .

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 $a_2=3\text{and}{a}_8=192$

Determine the value of $r$ by using formula $r^{8-2}$

  $\begin{array}{l}{r}^{8-2}=\frac{192}{3}\\ r^6=64\\ r=2\end{array}$. Put the value of $a_2=3$ into $a_n=a_1\cdot r^{n-1}$

  $\begin{array}{l}{a}_2=a_1\cdot r^{2-1}\\ a_2=a_1\cdot r^1\end{array}$

  $\begin{array}{l}3=a_1\cdot r^1\\ 3=a_1\cdot {\left(2\right)}^1\\ 3=a_1\left(2\right)\\ a_1=\frac{3}{2}\end{array}$

Put the value of $a_1=\frac{3}{2}\text{and}r=2$ in $a_n=a_1\cdot r^{n-1}$

  $a_n=\frac{3}{2}\cdot 2^{n-1}$

There are solutions.

  $a_n=\frac{3}{2}\cdot 2^{n-1}$