prove that n/2 [2a+ (n-1)d]
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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)
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)
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