Question

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?

Answer 1

First five terms are `{81,27,9,3,1]`, common ratio is `1/3` and `n^(th)` 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=1/3`,

we have `r=1/3`

AS given first term as `a_1` and common ratio as `r`, the `n^(th)` term `a_n=a_1xxr^(n-1)`. Now given first term `a_1=81` and `r=1/3`

`n^(th)` term `a_n` is `81xx1/3^(n-1)`

= `3^4/3^(n-1)=3^(4-n+1)=3^(5-n)`

first five terms are `{81,81xx1/3=27,27xx1/3=9,9xx1/3=3,3xx1/3=1}`

i.e. `{81,27,9,3,1}`