Question

Find the ninth term of the sequence $6,18,54,162,486,1,458,\dots$.Then find the general term for the sequence.

Short answer

The ninth term is and the general term are equal to$39366,6{\left(3\right)}^{n-1}$respectively.

Step-by-step solution

Step 1. Given information

Sequence is$6,18,54,162,486,1,458,...$

Step 2. Find the ninth term of the sequence.

Consider the sequence:

$6,18,54,162,486,1,458,...$

Let $a,r$denote the first term and the common ratio respectively.

$\begin{array}{rcl}a& =& 6\\ r& =& \frac{18}{6}\\ & =& 3\end{array}$

$\begin{array}{rcl}{a}_n& =& a{r}^{n-1}\\ a_9& =& 6{\left(3\right)}^{9-1}\\ & =& 6\left(3^8\right)\\ & =& 39366\end{array}$

Step 3, Find the general term of the sequence.

Put $a=6,r=3$in the formula$a_n=a{r}^{n-1}$.

So,

$a_n=6{\left(3\right)}^{n-1}$