Question

proof that Sn=n/2[2a(n-1)d]

Answer 1

Let Sn = Sum of 'n' terms, then,

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ a+(n-1)d

Let l = last term, then,

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ (l-3d) + (l -2d) + (l - d) + l

reversing the order of this gives:

Sn = l + (l-d) + (l-2d) + (l-3d) + ...+ (a+3d) + (a +2d) + (a + d) + a

Adding these last two equations gives,

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

2*Sn = (a+l) to n terms

2*Sn = (a+l)*n

Sn = n/2*(a+l)

but l = a+(n-1)d, so substituting this gives:

Sn = n/2*(2a+(n-1)d)

Answer 2

Let Sn = Sum of 'n' terms, then,

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ a+(n-1)d

Let l = last term, then,

Sn = a + (a+d) + (a+2d) + (a+3d) + ...+ (l-3d) + (l -2d) + (l - d) + l

reversing the order of this gives:

Sn = l + (l-d) + (l-2d) + (l-3d) + ...+ (a+3d) + (a +2d) + (a + d) + a

Adding these last two equations gives,

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

2*Sn = (a+l) to n terms

2*Sn = (a+l)*n

Sn = n/2*(a+l)

but l = a+(n-1)d, so substituting this gives:

Sn = n/2*(2a+(n-1)d)