Find the 5th Term 5/6 , 5/9 , 10/27 , 20/81

$\frac{5}{6}$ , $\frac{5}{9}$ , $\frac{10}{27}$ , $\frac{20}{81}$

Step 1

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

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = \frac{5}{6}$ and $r = \frac{2}{3}$ .

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

Step 4

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

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

Step 5

Combine.

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

Step 6

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

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

Step 7

Simplify the numerator.

Subtract $1$ from $5$ .

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

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

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

Step 8

Simplify the denominator.

Subtract $1$ from $5$ .

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

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

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

Step 9

Reduce the expression by cancelling the common factors.

Multiply $5$ by $16$ .

$a_{5} = \frac{80}{6 \cdot 81}$

Multiply $6$ by $81$ .

$a_{5} = \frac{80}{486}$

Cancel the common factor of $80$ and $486$ .

Factor $2$ out of $80$ .

$a_{5} = \frac{2 (40)}{486}$

Cancel the common factors.

Factor $2$ out of $486$ .

$a_{5} = \frac{2 \cdot 40}{2 \cdot 243}$

Cancel the common factor.

$a_{5} = \frac{2 \cdot 40}{2 \cdot 243}$

Rewrite the expression.

$a_{5} = \frac{40}{243}$

Find the 5th Term 5/6 , 5/9 , 10/27 , 20/81

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