Math Problem Solver

$$\frac{8}{27},\frac{4}{9},\frac{2}{3},1,...$$The next term in this geometric sequence is $a_{5}=\frac{3}{2}$.

The nth term in this sequence is given by the formula:

Explicit Formula: $$a_n=2^{4 - n} 3^{n - 4}$$

Recursive Formula: $$a_n=\frac{3 a_{n-1}}{2}, \text{where } a_{1}=\frac{8}{27}$$

$$\begin{matrix}\frac{8}{27}&&\frac{4}{9}&&\frac{2}{3}&&1\\[6pt]&\times\frac{3}{2}&&\times\frac{3}{2}&&\times\frac{3}{2}\\[6pt]\end{matrix}$$

Explicit Formula

The formula for a geometric sequence where $a_1$ is the 1^st term, $r$ is the common ratio, and $n$ is the term number is $$a_n=a_1 r^{n-1}$$

Find $a_1$ and $r$: $$\begin{aligned} a_1&=\frac{8}{27} \\ r&=\frac{3}{2} \end{aligned}$$

The n^th term rule is:$$\begin{aligned} a_n&=a_1 r^{n-1} \\ &=\frac{8}{27} \left(\frac{3}{2}\right)^{n-1} \\ &=2^{4 - n} 3^{n - 4} \end{aligned}$$

Recursive Formula

The formula for a geometric sequence where $a_{n-1}$ is the (n-1)^th term, $r$ is the common ratio, and $n>1$ is $$a_n=r a_{n-1}$$

Find $r$: $$\begin{aligned} r=\frac{3}{2} \end{aligned}$$

The n^th term rule is:$$\begin{aligned} a_n&=r a_{n-1} \\ &=\left(\frac{3}{2}\right) a_{n-1} \\ &=\frac{3 a_{n-1}}{2} \end{aligned}$$