Question
Write the first five terms of the following sequences whose $n^{th}$ terms are:
$\text{a}_\text{n}=\frac{3\text{n}-2}{5}$
$\text{a}_\text{n}=\frac{3\text{n}-2}{5}$
✓
Answer
$\text{a}_\text{n}=\frac{3\text{n}-2}{5}$
Let $n = 1, 2, 3, 4, 5$, then
$\text{a}_1=\frac{3\times1-2}{5}=\frac{3-2}{5}=\frac{1}{5}$
$\text{a}_2=\frac{3\times2-2}{5}=\frac{6-2}{5}=\frac{4}{5}$
$\text{a}_3=\frac{3\times3-2}{5}=\frac{9-2}{5}=\frac{7}{5}$
$\text{a}_4=\frac{3\times4-2}{5}=\frac{12-2}{5}=\frac{10}{5}=2$
$\text{a}_5=\frac{3\times5-2}{5}=\frac{15-2}{5}=\frac{13}{5}$
Let $n = 1, 2, 3, 4, 5$, then
$\text{a}_1=\frac{3\times1-2}{5}=\frac{3-2}{5}=\frac{1}{5}$
$\text{a}_2=\frac{3\times2-2}{5}=\frac{6-2}{5}=\frac{4}{5}$
$\text{a}_3=\frac{3\times3-2}{5}=\frac{9-2}{5}=\frac{7}{5}$
$\text{a}_4=\frac{3\times4-2}{5}=\frac{12-2}{5}=\frac{10}{5}=2$
$\text{a}_5=\frac{3\times5-2}{5}=\frac{15-2}{5}=\frac{13}{5}$
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