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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the sum of $n$ terms of an $A.P.$ whose $k^{th}$ term is $5k+1$

$\begin{array}{1 1} =\large\frac{n(5n+1)}{2} \\=\large\frac{n(5n+6)}{2} \\=\large\frac{n(5n-7)}{2} \\ =\large\frac{n(5n+7)}{2} \end{array} $

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  • Sum of $n$ terms of an $A.P.=\large\frac{n}{2}$$[l+a]$ where $a=$ first term and $l=$last term=$t_n$
Given: $k^{th}$ term $t_k=5k+1$
$\therefore\:t_n=5n+1$
$\therefore\:t_1=a\:(first\:term)=5+1=6$
We know that $s_n=\large\frac{n}{2}$$(l+a)$
$\therefore\:S_n=\large\frac{n}{2}$$(t_n+6)=\large\frac{n}{2}$$(5n+1+6)$
$=\large\frac{n(5n+7)}{2}$
answered Feb 22, 2014 by rvidyagovindarajan_1
 

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