Question 5:Prove that a + (a + d) + (a + 2d) +...+ [a + (n-1)d]= (n/2)[2a + (n-1)d] for all natural numbers n.

S = a+ (a+d)+ (a+d) + ……[a+(n-1)d]

Writing in reverse order

S = [a+(n-1)d]+[a+(n-2)d]+…. (a+d) +a

Adding

2S = [2a+(n-1)d]+[2a+(n-1)d]+….[2a+(n-1)d]

2S = n[2a+(n-1)d]

S = n[2a+(n-1)d]/2

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