How do you find the missing terms of the geometric sequence:2, __, __, __, 512, ...?

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6
Feb 8, 2017

The Missing Terms are, $8 , 32 , \mathmr{and} , 128.$

Explanation:

Let $r$ be the Common Ratio of the given Geo. Seq. denoted, by

${\left\{{a}_{n}\right\}}_{n \in \mathbb{N}} .$

Then, ${a}_{1} = 2 , \mathmr{and} , {a}_{5} = 512.$

But, we know that, ${a}_{n} = {a}_{1} \cdot {r}^{n - 1} , n \in \mathbb{N} .$

$\therefore {a}_{5} = 512 \therefore {a}_{1} \cdot {r}^{5 - 1} = 512 \therefore 2 \cdot {r}^{4} = 512 \therefore {r}^{4} = 256 = {4}^{4}$

$\therefore r = 4$.

Hence, the reqd. missing terms, known as, Intermediate

Geometric Means, are,

a_2=2*4^(2-1)=2*4=8, a_3=2*4^2=32, &, a_4=128.

Enjoy Maths.!

Then teach the underlying concepts
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2
Feb 8, 2017

There are four possibilities:

$8 , 32 , 128$

$- 8 , 32 , - 128$

$8 i , - 32 , - 128 i$

$- 8 i , - 32 , 128 i$

Explanation:

We are given:

$\left\{\begin{matrix}{a}_{1} = 2 \\ {a}_{5} = 512\end{matrix}\right.$

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

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

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

So we find:

${r}^{4} = \frac{a {r}^{4}}{a {r}^{0}} = {a}_{5} / {a}_{1} = \frac{512}{2} = 256 = {4}^{4}$

The possible values for $r$ are the fourth roots of ${4}^{4}$, namely:

$\pm 4$, $\pm 4 i$

For each of these possible common ratios, we can fill in ${a}_{2} , {a}_{3} , {a}_{4}$ as one of the following:

$8 , 32 , 128$

$- 8 , 32 , - 128$

$8 i , - 32 , - 128 i$

$- 8 i , - 32 , 128 i$

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