Find the Fourth Term 81/25 , 27/5 , 9

$\frac{81}{25}$ , $\frac{27}{5}$ , $9$

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 $\frac{5}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{5}{3}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = \frac{81}{25}$ and $r = \frac{5}{3}$ .

$a_{n} = \frac{81}{25} {(\frac{5}{3})}^{n - 1}$

Step 4

Apply the product rule to $\frac{5}{3}$ .

$a_{n} = \frac{81}{25} \cdot \frac{5^{n - 1}}{3^{n - 1}}$

Step 5

Combine.

$a_{n} = \frac{81 \cdot 5^{n - 1}}{25 \cdot 3^{n - 1}}$

Step 6

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

$a_{4} = \frac{81 \cdot 5^{(4) - 1}}{25 \cdot 3^{(4) - 1}}$

Step 7

Simplify the numerator.

Subtract $1$ from $4$ .

$a_{4} = \frac{81 \cdot 5^{3}}{25 \cdot 3^{4 - 1}}$

Raise $5$ to the power of $3$ .

$a_{4} = \frac{81 \cdot 125}{25 \cdot 3^{4 - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $4$ .

$a_{4} = \frac{81 \cdot 125}{25 \cdot 3^{3}}$

Raise $3$ to the power of $3$ .

$a_{4} = \frac{81 \cdot 125}{25 \cdot 27}$

Step 9

Simplify the expression.

Multiply $81$ by $125$ .

$a_{4} = \frac{10125}{25 \cdot 27}$

Multiply $25$ by $27$ .

$a_{4} = \frac{10125}{675}$

Divide $10125$ by $675$ .

$a_{4} = 15$

Find the Fourth Term 81/25 , 27/5 , 9

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