Find the 7th Term 16/9 , 8/3 , 4 , 6 , 9

$\frac{16}{9}$ , $\frac{8}{3}$ , $4$ , $6$ , $9$

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 $\frac{3}{2}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{3}{2}$

This is the form of a geometric sequence.

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

Substitute in the values of $a_{1} = \frac{16}{9}$ and $r = \frac{3}{2}$ .

$a_{n} = \frac{16}{9} {(\frac{3}{2})}^{n - 1}$

Apply the product rule to $\frac{3}{2}$ .

$a_{n} = \frac{16}{9} \cdot \frac{3^{n - 1}}{2^{n - 1}}$

Combine.

$a_{n} = \frac{16 \cdot 3^{n - 1}}{9 \cdot 2^{n - 1}}$

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

$a_{7} = \frac{16 \cdot 3^{(7) - 1}}{9 \cdot 2^{(7) - 1}}$

Simplify the numerator.

Subtract $1$ from $7$ .

$a_{7} = \frac{16 \cdot 3^{6}}{9 \cdot 2^{7 - 1}}$

Raise $3$ to the power of $6$ .

$a_{7} = \frac{16 \cdot 729}{9 \cdot 2^{7 - 1}}$

Simplify the denominator.

Subtract $1$ from $7$ .

$a_{7} = \frac{16 \cdot 729}{9 \cdot 2^{6}}$

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

$a_{7} = \frac{16 \cdot 729}{9 \cdot 64}$

Reduce the expression by cancelling the common factors.

Multiply $16$ by $729$ .

$a_{7} = \frac{11664}{9 \cdot 64}$

Multiply $9$ by $64$ .

$a_{7} = \frac{11664}{576}$

Cancel the common factor of $11664$ and $576$ .

Factor $144$ out of $11664$ .

$a_{7} = \frac{144 (81)}{576}$

Cancel the common factors.

Factor $144$ out of $576$ .

$a_{7} = \frac{144 \cdot 81}{144 \cdot 4}$

Cancel the common factor.

$a_{7} = \frac{144 \cdot 81}{144 \cdot 4}$

Rewrite the expression.

$a_{7} = \frac{81}{4}$

Find the 7th Term 16/9 , 8/3 , 4 , 6 , 9

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