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Prove that for any arithmetic sequence of which the first term is $a$ and the constant difference is $d$, the sum to n terms can be expressed as

$$S_{n} =\frac{n}{2}(2a + (n-1)d)$$
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let a1,a2,a3,... be first, second and third terms respectively and a(n),a(n-1) be last 2 terms.

let Sn= a1+a2+a3+...+ a(n-1)+a(n)

in reverse order Sn= a(n)+a(n-1)+a(n-2)+...+ a2+a1

2Sn(add 1st and reverse order terms)= [a1+a(n)]+[a2+a(n-1)]+[a3+a(n-2)] + ...+[a(n-1)+a2]+[a(n)+a1], NB-all bracket terms are equal

2Sn=[a1+a(n)].n              divide by 2 both sides

Sn= [(a1+a(n))/2].n

= [(a1+a1+ d(n-1))/2].n

= [(2a1 + a1 + d(n-1))/2].n

= [(a1 + d(n-1)/2].n

factoring out 2,                               = n/2 [2.a1 + (n-1)d]  ,proven
by Diamond (42,256 points)

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