Find the 6th Term 81 , 27 , 9 , 3 , 1

$81$ , $27$ , $9$ , $3$ , $1$

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 $a_{1} = 81$ and $r = \frac{1}{3}$ .

$a_{n} = 81 {(\frac{1}{3})}^{n - 1}$

Step 4

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

$a_{n} = 81 \frac{1^{n - 1}}{3^{n - 1}}$

Step 5

One to any power is one.

$a_{n} = 81 \frac{1}{3^{n - 1}}$

Step 6

Combine $81$ and $\frac{1}{3^{n - 1}}$ .

$a_{n} = \frac{81}{3^{n - 1}}$

Step 7

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

$a_{6} = \frac{81}{3^{(6) - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $6$ .

$a_{6} = \frac{81}{3^{5}}$

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

$a_{6} = \frac{81}{243}$

Step 9

Cancel the common factor of $81$ and $243$ .

Factor $81$ out of $81$ .

$a_{6} = \frac{81 (1)}{243}$

Cancel the common factors.

Factor $81$ out of $243$ .

$a_{6} = \frac{81 \cdot 1}{81 \cdot 3}$

Cancel the common factor.

$a_{6} = \frac{81 \cdot 1}{81 \cdot 3}$

Rewrite the expression.

$a_{6} = \frac{1}{3}$

Find the 6th Term 81 , 27 , 9 , 3 , 1

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