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Arithmetic sequences

For the derivation of the equation Sn = n/2[2a+(n-1)d], how does:

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ (l-3d) + (l -2d) + (l - d) + l
and
Sn = l + (l-d) + (l-2d) + (l-3d) + ...+ (a+3d) + (a +2d) + (a + d) + a <--reverse order

become:
2*Sn = a+l + (a+l) + (a+l) + (a+l) + ...+ (a+l) + (a+l) + (a+l) + a+l

Why do you multiply by two? I don't understand.

Also, what makes l = a+(n-1)d, l = a+(n-1)d ?
(edited 8 years ago)
Reply 1
Avatar for jamestg
jamestg
21
Where have you got l from?

Just watch this.
Reply 2
Avatar for CookieHero
CookieHero
OP
3
Original post by jamestg
Where have you got l from?

Just watch this.

I already seen the video but I am having issues about it.

-Why does she use a+(n-1)d in Sn = n/2(a+L) as L?
-Why does she add the two groups the normal order (Sn = a+a+d+a+2d...L-d+L) and the reverse order (Sn = L+L-d+L-2d...a+d+a) to create 2Sn?
(edited 8 years ago)

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