Write the first five terms of the sequences whose nth term is
Question:

Write the first five terms of the sequences whose nth term is  $\mathrm{a}_{\mathrm{n}}=\frac{2 \mathrm{n}-3}{6}$

Solution:

Substituting = 1, 2, 3, 4, 5, we obtain

$a_{1}=\frac{2 \times 1-3}{6}=\frac{-1}{6}$

$a_{2}=\frac{2 \times 2-3}{6}=\frac{1}{6}$

$a_{3}=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$

$a_{4}=\frac{2 \times 4-3}{6}=\frac{5}{6}$

$a_{5}=\frac{2 \times 5-3}{6}=\frac{7}{6}$

Therefore, the required terms are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}$, and $\frac{7}{6}$.