Chapter 9, Problem 38RE

Solution

To identify: The sequences $\frac{1}{8},-\frac{1}{4},\frac{1}{2},1,\mathrm{....}$ are arithmetic or geometric. Find an explicit formula for the nth term. State the common difference or ratio.

The $r=-2$ and the rule is $a_n=\frac{1}{8}\cdot {(-2)}^{n-1}$ .

Given information:

The given sequence is $\frac{1}{8},-\frac{1}{4},\frac{1}{2},1,\mathrm{....}$ ..

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.

The common ratio is:

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ r=\frac{-\frac{1}{4}}{\frac{1}{8}}\\ r=-2\end{array}$

The given sequence is geometric with first term $a_1=\frac{1}{8}\text{ and }r=-2$ .

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

Put $a_1=\frac{1}{8}\text{ and }r=-2$ into $a_n=a_1\cdot r^{n-1}$ .

  $\begin{array}{}$ a_n=\frac{1}{8}\cdot {(-2)}^{n-1}$$ =\frac{1}{8}\cdot {(-2)}^{n-1}$\end{array}$

Therefore, the rule is $a_n=\frac{1}{8}\cdot {(-2)}^{n-1}$ .