Chapter 11.3, Problem 2OE

To determine

To calculate:

Formula for nth term of geometric sequence.

Answer to Problem 2OE

The formula for nth term of geometric sequence is $t_n=-3\left(\frac{1}{2}\right){\left(-2\right)}^n$ .

Explanation of Solution

Given information:

Sequence:

  $3,-6,12,-24,\mathrm{....}$

Formula used:

  $t_n=t_1{r}^{n-1}$

Where,

  $t_1$ = first number

  $r=$ ratio between any two terms.

Calculation:

Consider the sequence $3,-6,12,-24,\mathrm{....}$

Put the values in the following formula,

  $t_n=t_1{r}^{n-1}$ ,

Where,

  $\begin{array}{l}{t}_1=3\\ r=-2\end{array}$

So, the result is, $t_n=3{\left(-2\right)}^{n-1}$

Now simplify the $3{\left(-2\right)}^{n-1}$

So, raise the $-2$ to, each power separately and then multiply.

  $t_n=3{\left(-2\right)}^{-1}{\left(-2\right)}^n$

Then with the help of reciprocal of the number,

  ${\left(-2\right)}^{-1}=-\frac{1}{2}$

Now replacing ${\left(-2\right)}^{-1}$ with $-\frac{1}{2}$ , them simplify by multiplying $3$ by $-\frac{1}{2}$ to get final answer.

  $\begin{array}{l}{\left(-2\right)}^{-1}=-\frac{1}{2}\\ t_n=-3\left(\frac{1}{2}\right){\left(-2\right)}^n\end{array}$

So, the formula is $t_n=-3\left(\frac{1}{2}\right){\left(-2\right)}^n$ .