# How do you find these missing terms [__,8,__,128,....]?

Jun 6, 2016

There are two solutions:

$4$ and $32$

Or:

$- 4$ and $- 32$

#### Explanation:

If the general term of the sequence is ${a}_{n}$ ($n = 1 , 2 , 3 , \ldots$) then we can rephrase the problem like this:

A geometric sequence has ${a}_{2} = 8$ and ${a}_{4} = 128$. What are ${a}_{1}$ and ${a}_{3}$ ?

The general term of a geometric sequence is described by the formula:

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

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

In our example, we find:

${r}^{2} = \frac{a {r}^{3}}{a r} = {a}_{4} / {a}_{2} = \frac{128}{8} = 16 = {4}^{2}$

So $r = \pm 4$.

If $r = 4$ then:

${a}_{1} = {a}_{2} / r = \frac{8}{4} = 2$ and ${a}_{3} = {a}_{2} \cdot r = 8 \cdot 4 = 32$

If $r = - 4$ then:

${a}_{1} = {a}_{2} / r = \frac{8}{- 4} = - 2$ and ${a}_{3} = {a}_{2} \cdot r = 8 \cdot \left(- 4\right) = - 32$