You can put this solution on YOUR website! S(n) = ....... a + (a+d) + (a+2d)+......+(a+(n-1)d)
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S(n) = (a+(n-1)d) + (a+(n-2)d) + (a+(n-3)d)+....+ a
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Both lines are the sum of n terms of the same AP.
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Add the 2 lines to get::
2*S(n) = (2a+(n-1)d) + (2a+(n-1)d)+....+(2a+(n-1)d)
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2*S(n) = n(2a+(n-1)d)
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Then S(n) = (n/2)(2a+(n-1)d) = (n/2)(a + [a+(n-1)d])
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Cheers,
Stan H.
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