Question

Find the thirteenth term of a sequence where the first term is $81$and the common ratio is $r=\frac{1}{3}$.

Short answer

The thirteenth term of a sequence is equal to$\frac{1}{19683}$.

Step-by-step solution

Step 1. Given information

The first term is $81$and the common ratio is$r=\frac{1}{3}$.

Step 2. Find the thirteenth term of a sequence. 

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

role="math" $\begin{array}{rcl}a& =& 81\\ r& =& \frac{1}{3}\\ a_n& =& a{r}^{n-1}\\ a_{13}& =& 81{\left(\frac{1}{3}\right)}^{13}\end{array}$

$$\begin{gathered} =3^4{\left(\frac{1}{3}\right)}^{13} \\ =\frac{1}{3^9} \\ =\frac{1}{19683} \end{gathered}$$