Question

Write the first five terms of the sequence, whose $n^{th}$ term is $a_n=\frac{2n-3}{6}$.

Answer 1

Finding sequence
Given, $a_n=\frac{2n-3}{6},n\in N$
Substituting $n=1,2,3,4$ and $5$ in $a_n$, we get

$a_1=\frac{2\times 1-3}{6}\Rightarrow a_1=\frac{-1}{6}$

$a_2=\frac{2\times 2-3}{6}\Rightarrow a_2=\frac{1}{6}$

$a_3=\frac{2\times 3-3}{6}\Rightarrow a_3=\frac{1}{2}$

$a_4=\frac{2\times 4-3}{6}\Rightarrow a_4=\frac{5}{6}$

$a_5=\frac{2\times 5-3}{6}\Rightarrow a_5=\frac{7}{6}$ ;

Hence, the first five terms of sequence are $\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6}\text{and}\frac{7}{6}$.