Precalculus Examples

$- \frac{3}{2}$ , $3$ , $- 6$ , $12$ , $- 24$ , $48$

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

Geometric Sequence: $r = - 2$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = - \frac{3}{2}$ and $r = - 2$ .

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

Step 4

Combine ${(- 2)}^{n - 1}$ and $\frac{3}{2}$ .

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

Step 5

Cancel the common factor of ${(- 2)}^{n - 1}$ and $2$ .

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Step 5.1

Rewrite $2$ as $- 1 (- 2)$ .

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

Step 5.2

Factor $- 2$ out of ${(- 2)}^{n - 1} \cdot 3$ .

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

Step 5.3

Move the negative one from the denominator of $\frac{{(- 2)}^{n - 2} \cdot 3}{- 1}$ .

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

Step 6

Rewrite $- 1 \cdot ({(- 2)}^{n - 2} \cdot 3)$ as $- ({(- 2)}^{n - 2} \cdot 3)$ .

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

Step 7

Move $3$ to the left of ${(- 2)}^{n - 2}$ .

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

Step 8

Multiply $3$ by $- 1$ .

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

Step 9

Multiply $- 3$ by $- 1$ .

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

Step 10

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

$a_{7} = 3 {(- 2)}^{(7) - 2}$

Step 11

Subtract $2$ from $7$ .

$a_{7} = 3 {(- 2)}^{5}$

Step 12

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

$a_{7} = 3 \cdot - 32$

Step 13

Multiply $3$ by $- 32$ .

$a_{7} = - 96$