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Find the first 5 terms of the series whose general term is given by $t_n=\large\frac{2n-3}{6}$

$\begin{array}{1 1}\large \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6} \frac{12}{6} \\\large \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6} \frac{9}{6} \\ \large\frac{-1}{6}, \frac{1}{6}, \frac{4}{6}, \frac{5}{6}, \frac{7}{6} \\\large\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6} \end{array} $

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Given: $t_n=\large\frac{2n-3}{6}$
By putting $n=1,2,3.......$ we can get the first 5 terms as
$t_1=\large\frac{2\times 1-3}{6}=-\frac{1}{6}$
$t_2=\large\frac{2\times 2-3}{6}=\frac{1}{6}$
$t_3=\large\frac{2\times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$
$t_4=\large\frac{2\times 4-3}{6}=\frac{5}{6}$
$t_5=\large\frac{2\times 5-3}{6}=\frac{7}{6}$
answered Feb 17, 2014 by rvidyagovindarajan_1

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