Precalculus Examples Sequences and Series

Find the Next Term

$\frac{1}{2}$ , $\frac{1}{6}$ , $\frac{1}{18}$

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

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

Step 4

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

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

Step 5

Combine.

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

Step 6

Multiply $1$ by $1^{n - 1}$ by adding the exponents.

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Step 6.1

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

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Step 6.1.1

Raise $1$ to the power of $1$ .

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

Step 6.1.2

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

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

Step 6.2

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

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Step 6.2.1

Subtract $1$ from $1$ .

$a_{n} = \frac{1^{n + 0}}{2 \cdot 3^{n - 1}}$

Step 6.2.2

Add $n$ and $0$ .

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

Step 7

One to any power is one.

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

Step 8

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

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

Step 9

Simplify the denominator.

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Step 9.1

Subtract $1$ from $4$ .

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

Step 9.2

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

$a_{4} = \frac{1}{2 \cdot 27}$

Step 10

Multiply $2$ by $27$ .

$a_{4} = \frac{1}{54}$

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Find the Next Term

Sequences and Series

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