Write the first five terms of each of the following sequences whose nth terms are:
$ a_{n}=\frac{2 n-3}{6} $

Given:

\( a_{n}=\frac{2 n-3}{6} \)

To do:

We have to find the first five terms of the given sequence.

Solution:

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

Taking $n=1$, we get

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

Taking $n=2$, we get

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

Taking $n=3$, we get

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

Taking $n=4$, we get

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

Taking $n=5$, we get

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

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

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Updated on: 10-Oct-2022

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