Find the 6th Term 1/2 , 1/3 , 2/9 , 4/27

$\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{2}{9}$ , $\frac{4}{27}$

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

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

Step 4

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

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

Step 5

Combine.

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

Step 6

Cancel the common factor of $2^{n - 1}$ and $2$ .

Factor $2$ out of $1 \cdot 2^{n - 1}$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 3^{n - 1}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 7

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

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

Step 8

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

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

Step 9

Simplify the numerator.

Subtract $2$ from $6$ .

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

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

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

Step 10

Simplify the denominator.

Subtract $1$ from $6$ .

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

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

$a_{6} = \frac{16}{243}$

Find the 6th Term 1/2 , 1/3 , 2/9 , 4/27

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