# How do you write the first five terms of the geometric sequence a_1=81, a_(k+1)=1/3a_k and determine the common ratio and write the nth term of the sequence as a function of n?

Mar 27, 2017

First five terms are $\left\{81 , 27 , 9 , 3 , 1\right]$, common ratio is $\frac{1}{3}$ and ${n}^{t h}$ term ${a}_{n} = {3}^{5 - n}$

#### Explanation:

In a geometric series common ratio $r$ is the ratio of term to its immediately preceding term. Here we have a term ${a}_{k}$ and as next term is ${a}_{k + 1}$ and ${a}_{k + 1} / {a}_{k} = \frac{1}{3}$,

we have $r = \frac{1}{3}$

AS given first term as ${a}_{1}$ and common ratio as $r$, the ${n}^{t h}$ term ${a}_{n} = {a}_{1} \times {r}^{n - 1}$. Now given first term ${a}_{1} = 81$ and $r = \frac{1}{3}$

${n}^{t h}$ term ${a}_{n}$ is $81 \times \frac{1}{3} ^ \left(n - 1\right)$

= ${3}^{4} / {3}^{n - 1} = {3}^{4 - n + 1} = {3}^{5 - n}$

first five terms are $\left\{81 , 81 \times \frac{1}{3} = 27 , 27 \times \frac{1}{3} = 9 , 9 \times \frac{1}{3} = 3 , 3 \times \frac{1}{3} = 1\right\}$

i.e. $\left\{81 , 27 , 9 , 3 , 1\right\}$