# The sum of the first and third terms of a geometric sequence is 40 while the sum of its second and fourth terms is 96. How do you find the sixth term of the sequence?

Jan 15, 2016

Solve to find the initial term $a$, common ratio $r$ and hence:

${a}_{6} = a {r}^{5} = \frac{1990656}{4225}$

#### Explanation:

The general term of a geometric sequence is:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ is the common ratio.

We are given:

$40 = {a}_{1} + {a}_{3} = a + a {r}^{2} = a \left(1 + {r}^{2}\right)$

$96 = {a}_{2} + {a}_{4} = a r + a {r}^{3} = a r \left(1 + {r}^{2}\right)$

So:

$r = \frac{a r \left(1 + {r}^{2}\right)}{a \left(1 + {r}^{2}\right)} = \frac{96}{40} = \frac{12}{5}$

$a = \frac{40}{1 + {r}^{2}}$

$= \frac{40}{1 + {\left(\frac{12}{5}\right)}^{2}}$

$= \frac{40}{1 + \frac{144}{25}}$

$= \frac{40}{\frac{169}{25}}$

$= \frac{40 \cdot 25}{169}$

$= \frac{1000}{169}$

Then:

${a}_{6} = a {r}^{5}$

$= \frac{1000}{169} \cdot {\left(\frac{12}{5}\right)}^{5}$

$= \frac{{2}^{3} \cdot {5}^{3} \cdot {2}^{10} \cdot {3}^{5}}{{13}^{2} \cdot {5}^{5}}$

$= \frac{{2}^{13} \cdot {3}^{5}}{{5}^{2} \cdot {13}^{2}}$

$= \frac{1990656}{4225}$