Find the 8th Term 1 , 1/2 , 1/4 , 1/8 , 1/16 , 1/32

$1$ , $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{8}$ , $\frac{1}{16}$ , $\frac{1}{32}$

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

Geometric Sequence: $r = \frac{1}{2}$

Step 2

This is the form of a geometric sequence.

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

Step 3

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

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

Step 4

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

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

Step 5

Apply the product rule to $\frac{1}{2}$ .

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

Step 6

One to any power is one.

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

Step 7

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

$a_{8} = \frac{1}{2^{(8) - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $8$ .

$a_{8} = \frac{1}{2^{7}}$

Raise $2$ to the power of $7$ .

$a_{8} = \frac{1}{128}$

Find the 8th Term 1 , 1/2 , 1/4 , 1/8 , 1/16 , 1/32

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