Find the 5th Term 2/3 , 4/9 , 8/27 , 16/81

$\frac{2}{3}$ , $\frac{4}{9}$ , $\frac{8}{27}$ , $\frac{16}{81}$

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 $a_{1} = \frac{2}{3}$ and $r = \frac{2}{3}$ .

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

Step 4

Multiply $\frac{2}{3}$ by ${(\frac{2}{3})}^{n - 1}$ by adding the exponents.

Multiply $\frac{2}{3}$ by ${(\frac{2}{3})}^{n - 1}$ .

Raise $\frac{2}{3}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Combine the opposite terms in $1 + n - 1$ .

Subtract $1$ from $1$ .

$a_{n} = {(\frac{2}{3})}^{n + 0}$

Add $n$ and $0$ .

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

Step 5

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

$a_{n} = \frac{2^{n}}{3^{n}}$

Step 6

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

$a_{5} = \frac{2^{5}}{3^{5}}$

Step 7

Raise $2$ to the power of $5$ .

$a_{5} = \frac{32}{3^{5}}$

Step 8

Raise $3$ to the power of $5$ .

$a_{5} = \frac{32}{243}$

Find the 5th Term 2/3 , 4/9 , 8/27 , 16/81

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