Find the Fourth Term -128/27 , 32/9 , -8/3

$- \frac{128}{27}$ , $\frac{32}{9}$ , $- \frac{8}{3}$

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 $a_{1} = - \frac{128}{27}$ and $r = - \frac{3}{4}$ .

$a_{n} = - \frac{128}{27} {(- \frac{3}{4})}^{n - 1}$

Step 4

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

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

$a_{n} = - \frac{128}{27} ({(- 1)}^{n - 1} {(\frac{3}{4})}^{n - 1})$

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

$a_{n} = - \frac{128}{27} ({(- 1)}^{n - 1} \frac{3^{n - 1}}{4^{n - 1}})$

Step 5

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

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

$a_{n} = {(- 1)}^{n - 1} \cdot - 1 \frac{128}{27} \frac{3^{n - 1}}{4^{n - 1}}$

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

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

$a_{n} = {(- 1)}^{n - 1} \cdot {(- 1)}^{1} \frac{128}{27} \frac{3^{n - 1}}{4^{n - 1}}$

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

$a_{n} = {(- 1)}^{n - 1 + 1} \frac{128}{27} \frac{3^{n - 1}}{4^{n - 1}}$

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

Add $- 1$ and $1$ .

$a_{n} = {(- 1)}^{n + 0} \frac{128}{27} \frac{3^{n - 1}}{4^{n - 1}}$

Add $n$ and $0$ .

$a_{n} = {(- 1)}^{n} \frac{128}{27} \frac{3^{n - 1}}{4^{n - 1}}$

Step 6

Combine ${(- 1)}^{n}$ and $\frac{128}{27}$ .

$a_{n} = \frac{{(- 1)}^{n} \cdot 128}{27} \cdot \frac{3^{n - 1}}{4^{n - 1}}$

Step 7

Combine.

$a_{n} = \frac{{(- 1)}^{n} \cdot 128 \cdot 3^{n - 1}}{27 \cdot 4^{n - 1}}$

Step 8

Simplify the numerator.

Factor $- 1$ out of ${(- 1)}^{n}$ .

$a_{n} = \frac{- {(- 1)}^{n - 1} \cdot 128 \cdot 3^{n - 1}}{27 \cdot 4^{n - 1}}$

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

$a_{n} = \frac{- 1 {(- 1)}^{n - 1} \cdot 128 \cdot 3^{n - 1}}{27 \cdot 4^{n - 1}}$

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

$a_{n} = \frac{- 1 {(3 \cdot - 1)}^{n - 1} \cdot 128}{27 \cdot 4^{n - 1}}$

Multiply $128$ by $- 1$ .

$a_{n} = \frac{- 128 {(3 \cdot - 1)}^{n - 1}}{27 \cdot 4^{n - 1}}$

Step 9

Multiply $3$ by $- 1$ .

$a_{n} = \frac{- 128 {(- 3)}^{n - 1}}{27 \cdot 4^{n - 1}}$

Step 10

Move the negative in front of the fraction.

$a_{n} = - \frac{128 {(- 3)}^{n - 1}}{27 \cdot 4^{n - 1}}$

Step 11

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

$a_{4} = - \frac{128 {(- 3)}^{(4) - 1}}{27 \cdot 4^{(4) - 1}}$

Step 12

Simplify the numerator.

Subtract $1$ from $4$ .

$a_{4} = - \frac{128 {(- 3)}^{3}}{27 \cdot 4^{4 - 1}}$

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

$a_{4} = - \frac{128 \cdot - 27}{27 \cdot 4^{4 - 1}}$

Step 13

Simplify the denominator.

Subtract $1$ from $4$ .

$a_{4} = - \frac{128 \cdot - 27}{27 \cdot 4^{3}}$

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

$a_{4} = - \frac{128 \cdot - 27}{27 \cdot 64}$

Step 14

Simplify the expression.

Multiply $128$ by $- 27$ .

$a_{4} = - \frac{- 3456}{27 \cdot 64}$

Multiply $27$ by $64$ .

$a_{4} = - \frac{- 3456}{1728}$

Divide $- 3456$ by $1728$ .

$a_{4} = 2$

Multiply $- 1$ by $- 2$ .

$a_{4} = 2$

Find the Fourth Term -128/27 , 32/9 , -8/3

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