# How do you use the geometric mean to find the 7th term in a geometric sequence if the 6th term is 12 and the 8th term?

Mar 23, 2016

If the common ratio is positive, then:

${a}_{7} = \sqrt{{a}_{6} \cdot {a}_{8}}$

#### Explanation:

If you are given the $6 t h$ and $8 t h$ terms of a geometric series then there are two possibilities for the $7 t h$ term, namely $\pm \sqrt{{a}_{6} \cdot {a}_{8}}$

If you are told that the common ratio is positive or that all of the terms of the sequence are positive then ${a}_{7} = \sqrt{{a}_{6} \cdot {a}_{8}}$ is the geometric mean of ${a}_{6}$ and ${a}_{8}$. Otherwise it could be $- \sqrt{{a}_{6} \cdot {a}_{8}}$.

Why does this work?

The general term of a geometric sequence can be written:

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

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

If $a , r > 0$ then:

$\sqrt{{a}_{6} \cdot {a}_{8}} = \sqrt{a {r}^{5} \cdot a {r}^{7}} = \sqrt{a {r}^{6} \cdot a {r}^{6}} = a {r}^{6} = {a}_{7}$

In fact, in general:

$\sqrt{{a}_{n} \cdot {a}_{n + 2}} = {a}_{n + 1}$

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In the given example ${a}_{6} = 12$ and let us suppose ${a}_{8} = 48$

Then ${a}_{7} = \sqrt{{a}_{6} \cdot {a}_{8}} = \sqrt{12 \cdot 48} = \sqrt{24 \cdot 24} = 24$