Question

Write the first five terms of the sequences whose $n^{th}$ term is:

$a_n=\frac{2n-3}{6}$

Answer 1

Here $a_n=\frac{2n-3}{6}$

Putting n = 1, 2, 3, 4 and 5, we have

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

$a_2=\frac{2\times 2-3}{6}=\frac{4-3}{6}=\frac{1}{6}$

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

$a_4=\frac{2\times 4-3}{6}=\frac{8-3}{6}=\frac{5}{6}$

$a_5=\frac{2\times 5-3}{6}=\frac{10-3}{6}=\frac{7}{6}$

Thus, first five terms of sequence are :

$-\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6}and\frac{7}{6}.$