given the AP a,a +d ,a+2d,a+3d,... show that S


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Question Number 45410 by Rio Michael last updated on 12/Oct/18
$given the AP a,a +d ,a+2d,a+3d,... show that S_n = (n/2){(2a+(n−1)d}$
$g i v e n t h e A P$ $a, a + d, a + 2 d, a + 3 d, . . .$ $s h o w t h a t$ $S_{n} = \frac{n}{2} {(2 a + (n - 1) d}$
Commented by maxmathsup by imad last updated on 13/Oct/18
$we have u_n =a +nd let S_n =u_0 +u_1 +....+u_(n−1) ⇒ S_n =(n/2){ u_0 +u_(n−1) }=(n/2){a +a +(n−1)d}=(n/2){2a+(n−1)d}.$
$w e h a v e u_{n} = a + n d l e t S_{n} = u_{0} + u_{1} + . . . . + u_{n - 1} \Rightarrow$ $S_{n} = \frac{n}{2} {u_{0} + u_{n - 1}} = \frac{n}{2} {a + a + (n - 1) d} = \frac{n}{2} {2 a + (n - 1) d} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
	$S_n =a+(a+d)+(a+2d)+..+{a+(n−1)d} S_n ={a+(n−1)d}+{a+(n−2)d}+..+a add them 2S_n =[a+a+(n−1)d]+[a+d+a+(n−2)d]+.. 2S_n =[2a+(n−1)d]+[2a+(n−1)d]+..n terms 2S_n =n[2a+(n−1)d](.nterms) 2S_n =n[2a+(n−1)d] S_n =(n/2)[2a+(n−1)d]$
	$S_{n} = a + (a + d) + (a + 2 d) + . . + {a + (n - 1) d}$ $S_{n} = {a + (n - 1) d} + {a + (n - 2) d} + . . + a$ $a d d t h e m$ $2 S_{n} = [a + a + (n - 1) d] + [a + d + a + (n - 2) d] + . .$ $2 S_{n} = [2 a + (n - 1) d] + [2 a + (n - 1) d] + . . n t e r m s$ $2 S_{n} = n [2 a + (n - 1) d] (. n t e r m s)$ $2 S_{n} = n [2 a + (n - 1) d]$ $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
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