Find the 7th Term 8 , 4 , 2 , 1 , 1/2 , 1/4

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

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 $a_{1} = 8$ and $r = \frac{1}{2}$ .

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

Step 4

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

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

Step 5

One to any power is one.

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

Step 6

Combine $8$ and $\frac{1}{2^{n - 1}}$ .

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

Step 7

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

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

Step 8

Simplify the denominator.

Subtract $1$ from $7$ .

$a_{7} = \frac{8}{2^{6}}$

Raise $2$ to the power of $6$ .

$a_{7} = \frac{8}{64}$

Step 9

Cancel the common factor of $8$ and $64$ .

Factor $8$ out of $8$ .

$a_{7} = \frac{8 (1)}{64}$

Cancel the common factors.

Factor $8$ out of $64$ .

$a_{7} = \frac{8 \cdot 1}{8 \cdot 8}$

Cancel the common factor.

$a_{7} = \frac{8 \cdot 1}{8 \cdot 8}$

Rewrite the expression.

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

Find the 7th Term 8 , 4 , 2 , 1 , 1/2 , 1/4

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