Find the 5th Term 1/6 , 1/9 , 2/27 , 4/81

$\frac{1}{6}$ , $\frac{1}{9}$ , $\frac{2}{27}$ , $\frac{4}{81}$

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by $\frac{2}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{2}{3}$

This is the form of a geometric sequence.

$a_{n} = a_{1} r^{n - 1}$

Substitute in the values of $a_{1} = \frac{1}{6}$ and $r = \frac{2}{3}$ .

$a_{n} = \frac{1}{6} {(\frac{2}{3})}^{n - 1}$

Apply the product rule to $\frac{2}{3}$ .

$a_{n} = \frac{1}{6} \cdot \frac{2^{n - 1}}{3^{n - 1}}$

Combine.

$a_{n} = \frac{1 \cdot 2^{n - 1}}{6 \cdot 3^{n - 1}}$

Multiply $2^{n - 1}$ by $1$ .

$a_{n} = \frac{2^{n - 1}}{6 \cdot 3^{n - 1}}$

Substitute in the value of $n$ to find the $n$ th term.

$a_{5} = \frac{2^{(5) - 1}}{6 \cdot 3^{(5) - 1}}$

Simplify the numerator.

Subtract $1$ from $5$ .

$a_{5} = \frac{2^{4}}{6 \cdot 3^{5 - 1}}$

Raise $2$ to the power of $4$ .

$a_{5} = \frac{16}{6 \cdot 3^{5 - 1}}$

Simplify the denominator.

Subtract $1$ from $5$ .

$a_{5} = \frac{16}{6 \cdot 3^{4}}$

Raise $3$ to the power of $4$ .

$a_{5} = \frac{16}{6 \cdot 81}$

Reduce the expression by cancelling the common factors.

Multiply $6$ by $81$ .

$a_{5} = \frac{16}{486}$

Cancel the common factor of $16$ and $486$ .

Factor $2$ out of $16$ .

$a_{5} = \frac{2 (8)}{486}$

Cancel the common factors.

Factor $2$ out of $486$ .

$a_{5} = \frac{2 \cdot 8}{2 \cdot 243}$

Cancel the common factor.

$a_{5} = \frac{2 \cdot 8}{2 \cdot 243}$

Rewrite the expression.

$a_{5} = \frac{8}{243}$

Find the 5th Term 1/6 , 1/9 , 2/27 , 4/81

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