geometric sequence

xc630

Junior Member
Joined
Sep 1, 2005
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164
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?
 
Here is a big hint.
It is geometric with a common ratio: \(\displaystyle \L
2(\sqrt 2 ) = 2^{3/2}\).
 
Your half right.
\(\displaystyle \L
g_1 = 2,\;g_2 = 2^{3/2} ,\;g_3 = 2^2 \;g_4 = 2^{5/2} \cdots\)
 
Hello, xc630!

Let's start over . . .

Given the geometric sequence: t1=2 and t2=232, find t13\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<sup>th</sup> term is: tn  =  t1rn1\displaystyle \,t_n\;=\;t_{_1}\cdot r^{^{n-1}}
    \displaystyle \;\;where t1\displaystyle t_1 is the first term and r\displaystyle r is the common ratio.


The common ratio is: r  =  2322  =  212\displaystyle \,r\;=\;\frac{2^{\frac{3}{2}}}{2}\;=\;2^{\frac{1}{2}}

For n=13\displaystyle n\,=\,13, we have: t13  =  2(212)12  =  226  =  27  =  128\displaystyle \,t_{_{13}}\;=\;2\cdot\left(2^{\frac{1}{2}}\right)^{12} \;= \;2\cdot2^6 \;=\;2^7\;=\;128
 
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