Find the 7th Term 4 , 6 , 9 , 13.5 , 20.25

$4$ , $6$ , $9$ , $13.5$ , $20.25$

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 $a_{1} = 4$ and $r = \frac{3}{2}$ .

$a_{n} = 4 {(\frac{3}{2})}^{n - 1}$

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

$a_{n} = 4 \frac{3^{n - 1}}{2^{n - 1}}$

Combine $4$ and $\frac{3^{n - 1}}{2^{n - 1}}$ .

$a_{n} = \frac{4 \cdot 3^{n - 1}}{2^{n - 1}}$

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

$a_{7} = \frac{4 \cdot 3^{(7) - 1}}{2^{(7) - 1}}$

Simplify the numerator.

Subtract $1$ from $7$ .

$a_{7} = \frac{4 \cdot 3^{6}}{2^{7 - 1}}$

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

$a_{7} = \frac{4 \cdot 729}{2^{7 - 1}}$

Simplify the denominator.

Subtract $1$ from $7$ .

$a_{7} = \frac{4 \cdot 729}{2^{6}}$

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

$a_{7} = \frac{4 \cdot 729}{64}$

Reduce the expression by cancelling the common factors.

Multiply $4$ by $729$ .

$a_{7} = \frac{2916}{64}$

Cancel the common factor of $2916$ and $64$ .

Factor $4$ out of $2916$ .

$a_{7} = \frac{4 (729)}{64}$

Cancel the common factors.

Factor $4$ out of $64$ .

$a_{7} = \frac{4 \cdot 729}{4 \cdot 16}$

Cancel the common factor.

$a_{7} = \frac{4 \cdot 729}{4 \cdot 16}$

Rewrite the expression.

$a_{7} = \frac{729}{16}$

Find the 7th Term 4 , 6 , 9 , 13.5 , 20.25

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