Find the 5th Term 36 , 12 , 4 , 4/3

$36$ , $12$ , $4$ , $\frac{4}{3}$

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 $\frac{1}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{3}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 36$ and $r = \frac{1}{3}$ .

$a_{n} = 36 {(\frac{1}{3})}^{n - 1}$

Step 4

Apply the product rule to $\frac{1}{3}$ .

$a_{n} = 36 \frac{1^{n - 1}}{3^{n - 1}}$

Step 5

One to any power is one.

$a_{n} = 36 \frac{1}{3^{n - 1}}$

Step 6

Combine $36$ and $\frac{1}{3^{n - 1}}$ .

$a_{n} = \frac{36}{3^{n - 1}}$

Step 7

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

$a_{5} = \frac{36}{3^{(5) - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $5$ .

$a_{5} = \frac{36}{3^{4}}$

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

$a_{5} = \frac{36}{81}$

Step 9

Cancel the common factor of $36$ and $81$ .

Factor $9$ out of $36$ .

$a_{5} = \frac{9 (4)}{81}$

Cancel the common factors.

Factor $9$ out of $81$ .

$a_{5} = \frac{9 \cdot 4}{9 \cdot 9}$

Cancel the common factor.

$a_{5} = \frac{9 \cdot 4}{9 \cdot 9}$

Rewrite the expression.

$a_{5} = \frac{4}{9}$

Find the 5th Term 36 , 12 , 4 , 4/3

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