Question

Given a1 = 3645 and a6 = 15, find a3

Answer 1

Answer:

$a_{3} = 405$

Step-by-step 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:

$a_{6} = a_{1}r^{6-1}$

$a_{6} = a_{1}r^{5}$

$3645r^{5} = 15$

$r^{5} = \frac{15}{3645}$

$r^{5} = \frac{1}{243}$

$r = \sqrt[5]{\frac{1}{243}}$

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

So

$a_{n} = 3645 \times (\frac{1}{3})^{n-1}$

$a_{3} = 3645 \times (\frac{1}{3})^{3-1} = 405$