Find the Fourth Term -16/5 , 8 , -20

$- \frac{16}{5}$ , $8$ , $- 20$

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 $- \frac{5}{2}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = - \frac{5}{2}$

This is the form of a geometric sequence.

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

Substitute in the values of $a_{1} = - \frac{16}{5}$ and $r = - \frac{5}{2}$ .

$a_{n} = - \frac{16}{5} {(- \frac{5}{2})}^{n - 1}$

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

Apply the product rule to $- \frac{5}{2}$ .

$a_{n} = - \frac{16}{5} ({(- 1)}^{n - 1} {(\frac{5}{2})}^{n - 1})$

Apply the product rule to $\frac{5}{2}$ .

$a_{n} = - \frac{16}{5} ({(- 1)}^{n - 1} \frac{5^{n - 1}}{2^{n - 1}})$

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

Move ${(- 1)}^{n - 1}$ .

$a_{n} = {(- 1)}^{n - 1} \cdot - 1 \frac{16}{5} \frac{5^{n - 1}}{2^{n - 1}}$

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

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

$a_{n} = {(- 1)}^{n - 1} \cdot {(- 1)}^{1} \frac{16}{5} \frac{5^{n - 1}}{2^{n - 1}}$

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

$a_{n} = {(- 1)}^{n - 1 + 1} \frac{16}{5} \frac{5^{n - 1}}{2^{n - 1}}$

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

Add $- 1$ and $1$ .

$a_{n} = {(- 1)}^{n + 0} \frac{16}{5} \frac{5^{n - 1}}{2^{n - 1}}$

Add $n$ and $0$ .

$a_{n} = {(- 1)}^{n} \frac{16}{5} \frac{5^{n - 1}}{2^{n - 1}}$

Combine ${(- 1)}^{n}$ and $\frac{16}{5}$ .

$a_{n} = \frac{{(- 1)}^{n} \cdot 16}{5} \cdot \frac{5^{n - 1}}{2^{n - 1}}$

Combine.

$a_{n} = \frac{{(- 1)}^{n} \cdot 16 \cdot 5^{n - 1}}{5 \cdot 2^{n - 1}}$

Cancel the common factor of $5^{n - 1}$ and $5$ .

Factor $5$ out of ${(- 1)}^{n} \cdot 16 \cdot 5^{n - 1}$ .

$a_{n} = \frac{5 ({(- 1)}^{n} \cdot 16 \cdot 5^{n - 2})}{5 \cdot 2^{n - 1}}$

Cancel the common factors.

Factor $5$ out of $5 \cdot 2^{n - 1}$ .

$a_{n} = \frac{5 ({(- 1)}^{n} \cdot 16 \cdot 5^{n - 2})}{5 (2^{n - 1})}$

Cancel the common factor.

$a_{n} = \frac{5 ({(- 1)}^{n} \cdot 16 \cdot 5^{n - 2})}{5 \cdot 2^{n - 1}}$

Rewrite the expression.

$a_{n} = \frac{{(- 1)}^{n} \cdot 16 \cdot 5^{n - 2}}{2^{n - 1}}$

Move $16$ to the left of ${(- 1)}^{n}$ .

$a_{n} = \frac{16 {(- 1)}^{n} \cdot 5^{n - 2}}{2^{n - 1}}$

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

$a_{4} = \frac{16 {(- 1)}^{4} \cdot 5^{(4) - 2}}{2^{(4) - 1}}$

Simplify the numerator.

Raise $- 1$ to the power of $4$ .

$a_{4} = \frac{16 \cdot 1 \cdot 5^{4 - 2}}{2^{4 - 1}}$

Subtract $2$ from $4$ .

$a_{4} = \frac{16 \cdot 1 \cdot 5^{2}}{2^{4 - 1}}$

Raise $5$ to the power of $2$ .

$a_{4} = \frac{16 \cdot 1 \cdot 25}{2^{4 - 1}}$

Combine exponents.

Multiply $16$ by $1$ .

$a_{4} = \frac{16 \cdot 25}{2^{4 - 1}}$

Multiply $16$ by $25$ .

$a_{4} = \frac{400}{2^{4 - 1}}$

Simplify the denominator.

Subtract $1$ from $4$ .

$a_{4} = \frac{400}{2^{3}}$

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

$a_{4} = \frac{400}{8}$

Divide $400$ by $8$ .

$a_{4} = 50$

Find the Fourth Term -16/5 , 8 , -20

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