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27, 9, 3, 1, 1/3,1/9....What is the value of the 10th term in the sequence?

Sagot :

[tex]27,9,3,1,\frac{1}{3},\frac{1}{9}[/tex]

1. Identify if the sequence has a common difference or a common ratio.

Common difference: subtract each term from the next term:

[tex]\begin{gathered} 9-27=-18 \\ 3-9=-6 \\ 1-3=-2 \end{gathered}[/tex]

There is not a common difference.

Common ratio: Divide each term into the previous term:

[tex]\begin{gathered} \frac{9}{27}=\frac{1}{3} \\ \\ \frac{3}{9}=\frac{1}{3} \\ \\ \frac{1}{3}=\frac{1}{3} \\ \\ \frac{\frac{1}{3}}{1}=\frac{1}{3} \\ \\ \frac{\frac{1}{9}}{\frac{1}{3}}=\frac{3}{9}=\frac{1}{3} \end{gathered}[/tex]

The common ratio is 1/3; it is a geometric sequence.

2. Use the next fromula to write the formula to find the nth term in the sequence:

[tex]\begin{gathered} a_n=a_1*r^{n-1} \\ \\ r:common\text{ }ratio \end{gathered}[/tex][tex]a_n=27*(\frac{1}{3})^{n-1}[/tex]

Evaluare the formula above for n=10 to find the 10th term:

[tex]\begin{gathered} a_{10}=27*(\frac{1}{3})^{10-1} \\ \\ a_{10}=27*(\frac{1}{3})^9 \\ \\ a_{10}=27*\frac{1}{3^9} \\ \\ a_{10}=27*\frac{1}{19683} \\ \\ a_{10}=\frac{27}{19683} \\ \\ a_{10}=\frac{1}{729} \end{gathered}[/tex]

Then, the 10th term is 1/729