Answer:
The 10th term of the sequence is:
- [tex]a_{10}=\frac{1}{27}[/tex]
Step-by-step explanation:
Given the number
729, 243, 81, 27...
A geometric sequence has a constant ratio 'r' and is defined by
[tex]a_n=a_0\cdot r^{n-1}[/tex]
Computing the ratio of all the adjacent terms
[tex]\frac{243}{729}=\frac{1}{3},\:\quad \frac{81}{243}=\frac{1}{3},\:\quad \frac{27}{81}=\frac{1}{3}[/tex]
The ratio of all the adjacent terms is the same and equal to
[tex]r=\frac{1}{3}[/tex]
also
[tex]a_1=729[/tex]
Therefore, the nth term is computed by:
[tex]a_n=729\left(\frac{1}{3}\right)^{n-1}[/tex]
Putting n = 10 to find the 10th term
[tex]a_{10}=729\left(\frac{1}{3}\right)^{10-1}[/tex]
[tex]=729\cdot \frac{1}{3^9}[/tex]
[tex]=\frac{1\cdot \:729}{3^9}[/tex]
[tex]=\frac{729}{3^9}[/tex]
[tex]=\frac{3^6}{3^9}[/tex]
[tex]=\frac{1}{3^3}[/tex]
[tex]a_{10}=\frac{1}{27}[/tex]
Therefore, the 10th term of the sequence is:
- [tex]a_{10}=\frac{1}{27}[/tex]
