# What is the 7th term of the geometric sequence where a1 = 1,024 and a4 = -16?

May 14, 2016

${a}_{7} = 1024 {\left(\frac{1}{-} 4\right)}^{6} = \frac{1}{4}$

#### Explanation:

Each term in a sequence - whether geometric or arithmetic can be given as a formula or as a value.
In a GP, ${a}_{n} = a {r}^{n - 1}$

In this example, we have ${a}_{1} = - 16 \mathmr{and} {a}_{4} = 1024$, which can be written as $a {r}^{0} \mathmr{and} a {r}^{3}$

Let's divide them - formulae and values:

$\frac{a {r}^{3}}{a r} = \frac{- 16}{1024}$

$\frac{\cancel{a} {r}^{3}}{\cancel{a} r} = \frac{- 16}{1024} \text{ } \Rightarrow {r}^{3} = \frac{1}{- 64}$

Finding the cube root gives $r = \frac{1}{-} 4$.

Now that we have $a$ and $r$ we can find the value of any term in the sequence.

${a}_{7} = 1024 {\left(\frac{1}{-} 4\right)}^{6} = \frac{1}{4}$