Chapter 11.3, Problem 3OE

To determine

To calculate:

Formula for nth term of geometric sequence.

Answer to Problem 3OE

The formula for nth term of geometric sequence is $t_n=4\left(-\frac{1}{3}\right){\left(-3\right)}^n$ .

Explanation of Solution

Given information:

Sequence:

  $4,-12,36,-108,\mathrm{....}$

Formula used:

  $t_n=t_1{r}^{n-1}$

Where,

  $t_1$ = first number

  $r=$ ratio between any two terms.

Calculation:

Consider the sequence $4,-12,36,-108,\mathrm{....}$

Put the values in the following formula,

  $t_n=t_1{r}^{n-1}$ ,

Where,

  $\begin{array}{l}{t}_1=4\\ r=-3\end{array}$

So, the result is, $t_n=4{\left(-3\right)}^{n-1}$

Now simplify the $4{\left(-3\right)}^{n-1}$

So, raise the $-3$ to, each power separately and then multiply.

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

Then with the help of reciprocal of the number,

  ${\left(-3\right)}^{-1}=-\frac{1}{3}$

Now replacing ${\left(-3\right)}^{-1}$ with $-\frac{1}{3}$ , them simplify by multiplying $4$ by $-\frac{1}{3}$ to get final answer.

  $\begin{array}{l}{\left(-3\right)}^{-1}=-\frac{1}{3}\\ t_n=4\left(-\frac{1}{3}\right){\left(-3\right)}^n\end{array}$

So, the formula is $t_n=4\left(-\frac{1}{3}\right){\left(-3\right)}^n$ .