Question 104572
prove that:
Sum of n numbers in a sequence is n/2[2a+(n-1)d].
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Write the sequence          : S(n)=a        ,a+d         ,a+2d    ,.....a+(n-1)d
Write the sequence backwards: S(n)=a+(n-1)d, a+(n-2d)    ,a+(n-3d)     ,a
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Add the two sequences to get: 2S(n)= n[a+a+(n-1)d]
2S(n) = n[2a+(n-1)d]
S(n) = (n/2)[2a+(n-1)d]
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Cheers,
Stan H.