Identify the Sequence 125 , -50 , 20 , -8

$125$ , $- 50$ , $20$ , $- 8$

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

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

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 125$ and $r = - \frac{2}{5}$ .

$a_{n} = 125 {(- \frac{2}{5})}^{n - 1}$

Step 4

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

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

$a_{n} = 125 ({(- 1)}^{n - 1} {(\frac{2}{5})}^{n - 1})$

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

$a_{n} = 125 ({(- 1)}^{n - 1} \frac{2^{n - 1}}{5^{n - 1}})$

Step 5

Multiply ${(- 1)}^{n - 1} \frac{2^{n - 1}}{5^{n - 1}}$ .

Combine ${(- 1)}^{n - 1}$ and $\frac{2^{n - 1}}{5^{n - 1}}$ .

$a_{n} = 125 \frac{{(- 1)}^{n - 1} \cdot 2^{n - 1}}{5^{n - 1}}$

Rewrite ${(- 1)}^{n - 1} \cdot 2^{n - 1}$ as ${(- 1 \cdot 2)}^{n - 1}$ .

$a_{n} = 125 \frac{{(- 1 \cdot 2)}^{n - 1}}{5^{n - 1}}$

Multiply $- 1$ by $2$ .

$a_{n} = 125 \frac{{(- 2)}^{n - 1}}{5^{n - 1}}$

Step 6

Combine $125$ and $\frac{{(- 2)}^{n - 1}}{5^{n - 1}}$ .

$a_{n} = \frac{125 {(- 2)}^{n - 1}}{5^{n - 1}}$

Identify the Sequence 125 , -50 , 20 , -8

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