Find the Nth term in the geometric sequence where the first term is 3 and the fourth term is $6\sqrt{2}$

Emmy Swanson
2022-10-22
Answered

Find the Nth term in the geometric sequence where the first term is 3 and the fourth term is $6\sqrt{2}$

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Szulikto

Answered 2022-10-23
Author has **22** answers

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.

In our case we are given ${a}_{1}=3$ and $a}_{4}=6\sqrt{2$

So we find:

$r}^{3}=\frac{a{r}^{3}}{a{r}^{0}}=\frac{{a}_{4}}{{a}_{1}}=\frac{6\sqrt{2}}{3}=2\sqrt{2}={\left(\sqrt{2}\right)}^{3$

So (assuming the geometric sequence is Real):

$r=\sqrt[3]{{\sqrt{2}}^{3}}=\sqrt{2}$

and

$a}_{n}=3\cdot {\left(\sqrt{2}\right)}^{n-1}=3\cdot {2}^{\frac{n-1}{2}$

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

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

In our case we are given ${a}_{1}=3$ and $a}_{4}=6\sqrt{2$

So we find:

$r}^{3}=\frac{a{r}^{3}}{a{r}^{0}}=\frac{{a}_{4}}{{a}_{1}}=\frac{6\sqrt{2}}{3}=2\sqrt{2}={\left(\sqrt{2}\right)}^{3$

So (assuming the geometric sequence is Real):

$r=\sqrt[3]{{\sqrt{2}}^{3}}=\sqrt{2}$

and

$a}_{n}=3\cdot {\left(\sqrt{2}\right)}^{n-1}=3\cdot {2}^{\frac{n-1}{2}$

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We have the sequence $\{0,1,2,3,3,4,6,9,8,\dots \}$ for which we have to find the summation of terms to the nth term, $C}_{n$

However, the number of terms in the sub sequence,$N\ne n$ , the number of terms in the larger sequence. What is a way to represent N in terms of n in order to arrive at a single formula for $C}_{n$ ? I've determined that it can be $N=\frac{n}{3}$ but this is only true for cases where n is a multiple of 3. How do I find those two cases where n isn't a multiple of 3?

However, the number of terms in the sub sequence,