Question

In the following exercises, write the first five terms of the sequence whose general term is given.

$a_n=\frac{5n-3}{3^n}$

Short answer

The first five terms of the sequence $a_n=\frac{5n-3}{3^n}$ is $\frac{2}{3},\frac{7}{9}.\frac{4}{9},\frac{17}{81},\frac{22}{243}$

Step-by-step solution

Step 1. Given

The general term of the sequence is $a_n=\frac{5n-3}{3^n}$

To find the first five terms of the sequence.

Step 2. Substitute 1,2 in the general form

Substitute the values $1,2$in the general form of the sequence,

$$\begin{gathered} a_1=\frac{5\left(1\right)-3}{3^{\left(1\right)}} \\ =\frac{5-3}{3} \\ =\frac{2}{3} \end{gathered}$$

$$\begin{gathered} a_2=\frac{5\left(2\right)-3}{3^{\left(2\right)}} \\ =\frac{10-3}{9} \\ =\frac{7}{9} \end{gathered}$$

Step 3. Substitute 3,4,5 in the general from.

$$\begin{gathered} a_3=\frac{5\left(3\right)-3}{3^3} \\ =\frac{15-3}{27} \\ =\frac{4}{9} \end{gathered}$$

$$\begin{gathered} a_4=\frac{5\left(4\right)-3}{3^{\left(4\right)}} \\ =\frac{20-3}{81} \\ =\frac{17}{81} \end{gathered}$$

$$\begin{gathered} a_5=\frac{5\left(5\right)-3}{3^{\left(5\right)}} \\ =\frac{25-3}{243} \\ =\frac{22}{243} \end{gathered}$$

The five terms of the sequence are $\frac{2}{3},\frac{7}{9}.\frac{4}{9},\frac{17}{81},\frac{22}{243}$