Find the 6th Term 8 , 12 , 18 , 27

$8$ , $12$ , $18$ , $27$

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

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

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 8$ and $r = \frac{3}{2}$ .

$a_{n} = 8 {(\frac{3}{2})}^{n - 1}$

Step 4

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

$a_{n} = 8 \frac{3^{n - 1}}{2^{n - 1}}$

Step 5

Combine $8$ and $\frac{3^{n - 1}}{2^{n - 1}}$ .

$a_{n} = \frac{8 \cdot 3^{n - 1}}{2^{n - 1}}$

Step 6

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

$a_{6} = \frac{8 \cdot 3^{(6) - 1}}{2^{(6) - 1}}$

Step 7

Simplify the numerator.

Subtract $1$ from $6$ .

$a_{6} = \frac{8 \cdot 3^{5}}{2^{6 - 1}}$

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

$a_{6} = \frac{8 \cdot 243}{2^{6 - 1}}$

Step 8

Simplify the denominator.

Subtract $1$ from $6$ .

$a_{6} = \frac{8 \cdot 243}{2^{5}}$

Raise $2$ to the power of $5$ .

$a_{6} = \frac{8 \cdot 243}{32}$

Step 9

Reduce the expression by cancelling the common factors.

Multiply $8$ by $243$ .

$a_{6} = \frac{1944}{32}$

Cancel the common factor of $1944$ and $32$ .

Factor $8$ out of $1944$ .

$a_{6} = \frac{8 (243)}{32}$

Cancel the common factors.

Factor $8$ out of $32$ .

$a_{6} = \frac{8 \cdot 243}{8 \cdot 4}$

Cancel the common factor.

$a_{6} = \frac{8 \cdot 243}{8 \cdot 4}$

Rewrite the expression.

$a_{6} = \frac{243}{4}$

Find the 6th Term 8 , 12 , 18 , 27

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