Find the 5th Term -810 , -270 , -90

$- 810$ , $- 270$ , $- 90$

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

Geometric Sequence: $r = \frac{1}{3}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = - 810$ and $r = \frac{1}{3}$ .

$a_{n} = - 810 {(\frac{1}{3})}^{n - 1}$

Step 4

Apply the product rule to $\frac{1}{3}$ .

$a_{n} = - 810 \frac{1^{n - 1}}{3^{n - 1}}$

Step 5

One to any power is one.

$a_{n} = - 810 \frac{1}{3^{n - 1}}$

Step 6

Combine $- 810$ and $\frac{1}{3^{n - 1}}$ .

$a_{n} = \frac{- 810}{3^{n - 1}}$

Step 7

Move the negative in front of the fraction.

$a_{n} = - \frac{810}{3^{n - 1}}$

Step 8

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

$a_{5} = - \frac{810}{3^{(5) - 1}}$

Step 9

Simplify the denominator.

Subtract $1$ from $5$ .

$a_{5} = - \frac{810}{3^{4}}$

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

$a_{5} = - \frac{810}{81}$

Step 10

Simplify the expression.

Divide $810$ by $81$ .

$a_{5} = - 1 \cdot 10$

Multiply $- 1$ by $10$ .

$a_{5} = - 10$

Find the 5th Term -810 , -270 , -90

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