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If $1,2,3....$ are first terms $:\:1,3,5..........$ are common differences and $S_{1},S_{2},S_{3}.........$ are sums of $n$ terms of given $p\:\:AP's\:\:;$ then $S_{1}+S_{2}+S_{3}+.....+S_{p}$ is equal to
Option: 1
$\frac{np(np+1)}{2}$

Option: 2
$\frac{n(np+1)}{2}$

Option: 3
$\frac{np(p+1)}{2}$

Option: 4
$\frac{np(np-1)}{2}$

Answers (1)

best_answer

Sum of n terms of an AP -

$S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]$

and

Sum of n terms of an AP

$S_{n}= \frac{n}{2}\left [ a+l\right ]$

- wherein

$a\rightarrow$ first term

$d\rightarrow$ common difference

$n\rightarrow$ number of terms

$S_{1}+S_{2}+S_{3}+......S_{p}$

$\Rightarrow S_{1}=\frac{n}{2}\left [ 2.1+(n-1) 1\right ]$

$S_{2}=\frac{n}{2}\left [ 2.2+(n-1) 3\right ]$

$S_{3}=\frac{n}{2}\left [ 2.3+(n-1) 5\right ]$

$S_{p}=\frac{n}{2}\left [ 2.p+(n-1) (2p-1)\right ]$

so

$S_{1}+S_{2}+.....S_{p}=\frac{n}{2}\left [ 2(1+2+........p)+(n-1)(1+3+5+.......(2p-1)) \right ]$

$=\frac{n}{2}\left [ 2.\frac{p(p+1)}{2}+(n-1)p^{2} \right ]=\frac{np}{2}(n\:p+1)$

Posted by

Sanket Gandhi

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