Question

In the arithmertic sequence series:
a+(a+d)+(a+2d)+...+(l-d)+l,a is the first term,d the common difference and l the last term. how do prove that Sn=n/2[a+1]

Answer 1

Typo:
you have Sn=n/2[a+1]
clearly should be : Sn=n/2[a+l]


let S(n)
= a + (a+d) + (a+2d) + ... +(l-2d) + (l-d) + l , for n terms
we could just write this turned around
S(n) = l + (l-d) + (l-2d) + ... + (a+2d) + (a+d) + a
add these two equations
2S(n) = a+l + (a+d+l-d) + ... + (l-d+a+d) + (l+a)
2S(n) = (a+l) + (a+l) + ...+(a+l) + (a+l) , n times
2S(n) = n(a+l)
S(n) = (n/2)(a+l)