Find the 5th Term 8 , -6 , 9/2

$8$ , $- 6$ , $\frac{9}{2}$

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 $- \frac{3}{4}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = - \frac{3}{4}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 8$ and $r = - \frac{3}{4}$ .

$a_{n} = 8 {(- \frac{3}{4})}^{n - 1}$

Step 4

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

Apply the product rule to $- \frac{3}{4}$ .

$a_{n} = 8 ({(- 1)}^{n - 1} {(\frac{3}{4})}^{n - 1})$

Apply the product rule to $\frac{3}{4}$ .

$a_{n} = 8 ({(- 1)}^{n - 1} \frac{3^{n - 1}}{4^{n - 1}})$

Step 5

Multiply ${(- 1)}^{n - 1} \frac{3^{n - 1}}{4^{n - 1}}$ .

Combine ${(- 1)}^{n - 1}$ and $\frac{3^{n - 1}}{4^{n - 1}}$ .

$a_{n} = 8 \frac{{(- 1)}^{n - 1} \cdot 3^{n - 1}}{4^{n - 1}}$

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

$a_{n} = 8 \frac{{(- 1 \cdot 3)}^{n - 1}}{4^{n - 1}}$

Multiply $- 1$ by $3$ .

$a_{n} = 8 \frac{{(- 3)}^{n - 1}}{4^{n - 1}}$

Step 6

Combine $8$ and $\frac{{(- 3)}^{n - 1}}{4^{n - 1}}$ .

$a_{n} = \frac{8 {(- 3)}^{n - 1}}{4^{n - 1}}$

Step 7

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

$a_{5} = \frac{8 {(- 3)}^{(5) - 1}}{4^{(5) - 1}}$

Step 8

Simplify the numerator.

Subtract $1$ from $5$ .

$a_{5} = \frac{8 {(- 3)}^{4}}{4^{5 - 1}}$

Raise $- 3$ to the power of $4$ .

$a_{5} = \frac{8 \cdot 81}{4^{5 - 1}}$

Step 9

Simplify the denominator.

Subtract $1$ from $5$ .

$a_{5} = \frac{8 \cdot 81}{4^{4}}$

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

$a_{5} = \frac{8 \cdot 81}{256}$

Step 10

Reduce the expression by cancelling the common factors.

Multiply $8$ by $81$ .

$a_{5} = \frac{648}{256}$

Cancel the common factor of $648$ and $256$ .

Factor $8$ out of $648$ .

$a_{5} = \frac{8 (81)}{256}$

Cancel the common factors.

Factor $8$ out of $256$ .

$a_{5} = \frac{8 \cdot 81}{8 \cdot 32}$

Cancel the common factor.

$a_{5} = \frac{8 \cdot 81}{8 \cdot 32}$

Rewrite the expression.

$a_{5} = \frac{81}{32}$

Find the 5th Term 8 , -6 , 9/2

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