geometric sequence

[xc630]

Hello, I need some help with this problem. I am given t1=2 and t2=2^3/2. I have to find t13, the 13th term. It is a geometric sequence. How would I go about doing this?

 

[pka]

Here is a big hint.
It is geometric with a common ratio: \(\displaystyle \L
2(\sqrt 2 ) = 2^{3/2}\).

 

[xc630]

I think the answer is 64. Can you please double check me.

 

[pka]

Your half right.
\(\displaystyle \L
g_1 = 2,\;g_2 = 2^{3/2} ,\;g_3 = 2^2 \;g_4 = 2^{5/2} \cdots\)

 

[soroban]

Hello, xc630!

Let's start over . . .

\(\displaystyle \text{Given the geometric sequence: }\,t_{_1}\,=\,2\,\text{ and }\, t_{_2}\,=\,2^{\frac{3}{2}}, \text{ find }t_{_{13}}\).

Do you know anything about geometric sequences?

For example, the n^th term is: \(\displaystyle \,t_n\;=\;t_{_1}\cdot r^{^{n-1}}\)
\(\displaystyle \;\;\)where \(\displaystyle t_1\) is the first term and \(\displaystyle r\) is the common ratio.


The common ratio is: \(\displaystyle \,r\;=\;\frac{2^{\frac{3}{2}}}{2}\;=\;2^{\frac{1}{2}}\)

For \(\displaystyle n\,=\,13\), we have: \(\displaystyle \,t_{_{13}}\;=\;2\cdot\left(2^{\frac{1}{2}}\right)^{12} \;= \;2\cdot2^6 \;=\;2^7\;=\;128\)