Find the 5th Term 42 , 6 , 6/7 , 6/49

$42$ , $6$ , $\frac{6}{7}$ , $\frac{6}{49}$

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 $\frac{1}{7}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{7}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 42$ and $r = \frac{1}{7}$ .

$a_{n} = 42 {(\frac{1}{7})}^{n - 1}$

Step 4

Apply the product rule to $\frac{1}{7}$ .

$a_{n} = 42 \frac{1^{n - 1}}{7^{n - 1}}$

Step 5

One to any power is one.

$a_{n} = 42 \frac{1}{7^{n - 1}}$

Step 6

Combine $42$ and $\frac{1}{7^{n - 1}}$ .

$a_{n} = \frac{42}{7^{n - 1}}$

Step 7

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

$a_{5} = \frac{42}{7^{(5) - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $5$ .

$a_{5} = \frac{42}{7^{4}}$

Raise $7$ to the power of $4$ .

$a_{5} = \frac{42}{2401}$

Step 9

Cancel the common factor of $42$ and $2401$ .

Factor $7$ out of $42$ .

$a_{5} = \frac{7 (6)}{2401}$

Cancel the common factors.

Factor $7$ out of $2401$ .

$a_{5} = \frac{7 \cdot 6}{7 \cdot 343}$

Cancel the common factor.

$a_{5} = \frac{7 \cdot 6}{7 \cdot 343}$

Rewrite the expression.

$a_{5} = \frac{6}{343}$

Find the 5th Term 42 , 6 , 6/7 , 6/49

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