Question

assignment 9.1 sequences and their notation
due wednesday by 11:59pm points 8 submitting an external tool a
assignment 9.1 sequences and their notation
score: 2/8 answered: 2/8
question 3
find the next term of the sequence:
\\( \frac { 1 } { 1 }, \frac { 1 } { 3 }, \frac { 1 } { 9 }, \frac { 1 } { 27 }, \frac { 1 } { 81 }, \\)
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Explanation:

Step1: Identify the pattern

This is a geometric sequence. The common ratio \(r\) is found by dividing a term by its previous term. For example, \(\frac{\frac{1}{3}}{\frac{1}{1}}=\frac{1}{3}\), \(\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}\), \(\frac{\frac{1}{27}}{\frac{1}{9}}=\frac{1}{3}\), \(\frac{\frac{1}{81}}{\frac{1}{27}}=\frac{1}{3}\). So the common ratio \(r = \frac{1}{3}\).

Step2: Find the next term

The formula for the \(n\)th term of a geometric sequence is \(a_{n}=a_{1}r^{n - 1}\). Here, \(a_{1}=\frac{1}{1}=1\), \(r=\frac{1}{3}\), and we want the 6th term (\(n = 6\)). Using the formula \(a_{6}=a_{5}\times r\) (since \(a_{n}=a_{n-1}\times r\) for a geometric sequence). Given \(a_{5}=\frac{1}{81}\), then \(a_{6}=\frac{1}{81}\times\frac{1}{3}\).

$$ \frac{1}{81}\times\frac{1}{3}=\frac{1\times1}{81\times3}=\frac{1}{243} $$

Answer:

\(\frac{1}{243}\)