Find the 6th Term 40 , 10 , 5/2 , 5/8

$40$ , $10$ , $\frac{5}{2}$ , $\frac{5}{8}$

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 $\frac{1}{4}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{4}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 40$ and $r = \frac{1}{4}$ .

$a_{n} = 40 {(\frac{1}{4})}^{n - 1}$

Step 4

Apply the product rule to $\frac{1}{4}$ .

$a_{n} = 40 \frac{1^{n - 1}}{4^{n - 1}}$

Step 5

One to any power is one.

$a_{n} = 40 \frac{1}{4^{n - 1}}$

Step 6

Combine $40$ and $\frac{1}{4^{n - 1}}$ .

$a_{n} = \frac{40}{4^{n - 1}}$

Step 7

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

$a_{6} = \frac{40}{4^{(6) - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $6$ .

$a_{6} = \frac{40}{4^{5}}$

Raise $4$ to the power of $5$ .

$a_{6} = \frac{40}{1024}$

Step 9

Cancel the common factor of $40$ and $1024$ .

Factor $8$ out of $40$ .

$a_{6} = \frac{8 (5)}{1024}$

Cancel the common factors.

Factor $8$ out of $1024$ .

$a_{6} = \frac{8 \cdot 5}{8 \cdot 128}$

Cancel the common factor.

$a_{6} = \frac{8 \cdot 5}{8 \cdot 128}$

Rewrite the expression.

$a_{6} = \frac{5}{128}$

Find the 6th Term 40 , 10 , 5/2 , 5/8

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