Find the Fourth Term 20 , -10 , 5

$20$ , $- 10$ , $5$

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 $- \frac{1}{2}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = - \frac{1}{2}$

This is the form of a geometric sequence.

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

Substitute in the values of $a_{1} = 20$ and $r = - \frac{1}{2}$ .

$a_{n} = 20 {(- \frac{1}{2})}^{n - 1}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{2}$ .

$a_{n} = 20 ({(- 1)}^{n - 1} {(\frac{1}{2})}^{n - 1})$

Apply the product rule to $\frac{1}{2}$ .

$a_{n} = 20 ({(- 1)}^{n - 1} \frac{1^{n - 1}}{2^{n - 1}})$

One to any power is one.

$a_{n} = 20 ({(- 1)}^{n - 1} \frac{1}{2^{n - 1}})$

Combine ${(- 1)}^{n - 1}$ and $\frac{1}{2^{n - 1}}$ .

$a_{n} = 20 \frac{{(- 1)}^{n - 1}}{2^{n - 1}}$

Combine $20$ and $\frac{{(- 1)}^{n - 1}}{2^{n - 1}}$ .

$a_{n} = \frac{20 {(- 1)}^{n - 1}}{2^{n - 1}}$

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

$a_{4} = \frac{20 {(- 1)}^{(4) - 1}}{2^{(4) - 1}}$

Simplify the numerator.

Subtract $1$ from $4$ .

$a_{4} = \frac{20 {(- 1)}^{3}}{2^{4 - 1}}$

Raise $- 1$ to the power of $3$ .

$a_{4} = \frac{20 \cdot - 1}{2^{4 - 1}}$

Simplify the denominator.

Subtract $1$ from $4$ .

$a_{4} = \frac{20 \cdot - 1}{2^{3}}$

Raise $2$ to the power of $3$ .

$a_{4} = \frac{20 \cdot - 1}{8}$

Reduce the expression by cancelling the common factors.

Multiply $20$ by $- 1$ .

$a_{4} = \frac{- 20}{8}$

Cancel the common factor of $- 20$ and $8$ .

Factor $4$ out of $- 20$ .

$a_{4} = \frac{4 (- 5)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

$a_{4} = \frac{- 5}{2}$

Move the negative in front of the fraction.

$a_{4} = - \frac{5}{2}$

Find the Fourth Term 20 , -10 , 5

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