Find the 8th Term 4 , 8 , 16 , 32 , 64 , 128

$4$ , $8$ , $16$ , $32$ , $64$ , $128$

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

Geometric Sequence: $r = 2$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 4$ and $r = 2$ .

$a_{n} = 4 \cdot 2^{n - 1}$

Step 4

Multiply $4 \cdot 2^{n - 1}$ .

Rewrite $4$ as $2^{2}$ .

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

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

$a_{n} = 2^{2 + n - 1}$

Subtract $1$ from $2$ .

$a_{n} = 2^{n + 1}$

Step 5

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

$a_{8} = 2^{(8) + 1}$

Step 6

Add $8$ and $1$ .

$a_{8} = 2^{9}$

Step 7

Raise $2$ to the power of $9$ .

$a_{8} = 512$

Find the 8th Term 4 , 8 , 16 , 32 , 64 , 128

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