Question

Given a1 = 3645 and a6 = 15, find a3

Answer 1

`a_(3) = 405`

Explanation:

A geometric sequence is based on the following equation:

`a_(n+1) = ra_(n)`

In which r is the common ratio.

This can be expanded for the nth term in the following way:

`a_(n) = a_(1)r^(n-1)`

In which
`a_(1)` is the first term.

In this question:

`a_(1) = 3645, a_(6) = 15`

Applying the equation:

`a_(6) = a_(1)r^(6-1)`

`a_(6) = a_(1)r^(5)`

`3645r^(5) = 15`

`r^(5) = (15)/(3645)`

`r^(5) = (1)/(243)`

`r = \sqrt[5]{(1)/(243)}`

`r = (1)/(3)`

So

`a_(n) = 3645 * ((1)/(3))^(n-1)`

`a_(3) = 3645 * ((1)/(3))^(3-1) = 405`