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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the first 5 terms of the sequence whose general term is $t_n=n.\large\frac{n^2+5}{4}$

$\begin{array}{1 1} \large\frac{3}{4}, \frac{9}{4}, \frac{21}{4}, \frac{21}{2}, \frac{75}{2} \\ \large\frac{3}{4}, \frac{9}{4}, \frac{21}{4}, \frac{21}{2}, \frac{21}{1} \\\large\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, \frac{21}{4}, \frac{75}{2} \\\large\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, \frac{21}{1}, \frac{75}{2}\end{array} $

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Given: $t_n=n.\large\frac{n^2+5}{4}$
By putting $n=1,2,3...$ we get the first 5 terms as
$t_1=1.\large\frac{1^2+5}{4}=\frac{6}{4}=\frac{3}{2}$
$t_2=2.\large\frac{2^2+5}{4}=\frac{18}{4}=\frac{9}{2}$
$t_3=3.\large\frac{3^2+5}{4}=\frac{42}{4}=\frac{21}{2}$
$t_4=4.\large\frac{4^2+5}{4}=\frac{84}{4}$$=21$
$t_5=5.\large\frac{5^2+5}{4}=\frac{150}{4}=\frac{75}{2}$
answered Feb 17, 2014 by rvidyagovindarajan_1
 

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