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

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

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 $a_{1} = \frac{1}{4}$ and $r = \frac{2}{3}$ .

$a_{n} = \frac{1}{4} {(\frac{2}{3})}^{n - 1}$

Step 4

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

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

Step 5

Combine.

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

Step 6

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

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

Step 7

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

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

Step 8

Simplify the numerator.

Subtract $1$ from $5$ .

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

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

$a_{5} = \frac{16}{4 \cdot 3^{5 - 1}}$

Step 9

Simplify the denominator.

Subtract $1$ from $5$ .

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

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

$a_{5} = \frac{16}{4 \cdot 81}$

Step 10

Reduce the expression by cancelling the common factors.

Multiply $4$ by $81$ .

$a_{5} = \frac{16}{324}$

Cancel the common factor of $16$ and $324$ .

Factor $4$ out of $16$ .

$a_{5} = \frac{4 (4)}{324}$

Cancel the common factors.

Factor $4$ out of $324$ .

$a_{5} = \frac{4 \cdot 4}{4 \cdot 81}$

Cancel the common factor.

$a_{5} = \frac{4 \cdot 4}{4 \cdot 81}$

Rewrite the expression.

$a_{5} = \frac{4}{81}$

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

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