Question

Prove geometrically, the sum of first n natural numbers of an A. P. is n(n + 1)/2.

Answer 1

For an A.P., 

a = first term

d = common difference

The terms are:

a , a + d, a + 2d ............. a+ (n-1)d


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



Write in reverse order

Sn = [a + (n-1)d] + [a+ (n-2)d] + ............ + [(a+d) + a] -------- (ii)

Add (i) and (ii)


2Sn = 2a + (n-1)d + 2a + (n-1)d + 2a + (n-1)d + ...... up to the nth term


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

Divide both sides by 2


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

For natural numbers, a = 1 and d = 1


Sn = [n(2(1) + (n-1)(1)]/2

     = n(2 + n - 1)/2

     = n( n + 1)/2

Proved