Identify the Sequence 128 , -16 , 2 , -1/4

$128$ , $- 16$ , $2$ , $- \frac{1}{4}$

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 $- \frac{1}{8}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = - \frac{1}{8}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 128$ and $r = - \frac{1}{8}$ .

$a_{n} = 128 {(- \frac{1}{8})}^{n - 1}$

Step 4

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

Apply the product rule to $- \frac{1}{8}$ .

$a_{n} = 128 ({(- 1)}^{n - 1} {(\frac{1}{8})}^{n - 1})$

Apply the product rule to $\frac{1}{8}$ .

$a_{n} = 128 ({(- 1)}^{n - 1} \frac{1^{n - 1}}{8^{n - 1}})$

Step 5

One to any power is one.

$a_{n} = 128 ({(- 1)}^{n - 1} \frac{1}{8^{n - 1}})$

Step 6

Combine ${(- 1)}^{n - 1}$ and $\frac{1}{8^{n - 1}}$ .

$a_{n} = 128 \frac{{(- 1)}^{n - 1}}{8^{n - 1}}$

Step 7

Combine $128$ and $\frac{{(- 1)}^{n - 1}}{8^{n - 1}}$ .

$a_{n} = \frac{128 {(- 1)}^{n - 1}}{8^{n - 1}}$

Identify the Sequence 128 , -16 , 2 , -1/4

Related Questions

Support LCMGCF.com by disabling your adblocker.