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

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Answer 1 · 2016-06-06T18:05:37

There are two solutions:

$4$ and $32$

Or:

$-4$ and $-32$

If the general term of the sequence is $a_n$ ( $n = 1, 2, 3,...$ ) 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:

$a_n = a*r^(n-1)$

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

In our example, we find:

$r^2 = (a r^3)/(a r) = a_4/a_2 = 128/8 = 16 = 4^2$

So $r = +-4$ .

If $r = 4$ then:

$a_1 = a_2/r = 8/4 = 2$ and $a_3 = a_2*r = 8*4 = 32$

If $r = -4$ then:

$a_1 = a_2/r = 8/(-4) = -2$ and $a_3 = a_2*r = 8*(-4) = -32$

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