Carl Friedrich Gauss , a great German Mathematician , was in elementary
school. One day his teacher gave the class a problem of finding the sum of the first 100 natural numbers. While the rest of the class was struggling with the problem, Gauss gave the answer within no time.
He wrote the first 100 natural numbers as given below. S denotes the sum required to find .
S = 1 + 2 + 3 + 4 +...........+ 97 + 98 + 99 + 100
Reversing S = 100 + 99 + 98 + 97 +..........+ 4 + 3 + 2 + 1
Adding 2S = 101 + 101 +101 + 101 +.........+ 101 + 101 + 101 + 101
= 101 x 100
therefore S = 101 x 100 / 2 = 101 x 50 = 5050
We now give below a method of finding the sum of n terms of an A.P.
Let the first term of an A.P. be ' a ' and the common difference ' d ' . Let Sn denotes the sum of the first 'n' natural numbers . Then
Sn = a + ( a + d ) + ( a + 2d ) + ..+ ( a + n - 3d) + (a + n - 2 d ) + ( a + n - 1 d)----(1)
Rewriting Sn = (a + n -1 d) + ( a + n - 2 d) + (a + n - 3 d ) + .....+ (a + 2d) + ( a + d ) +a---(2)
Now we add (1) and (2) term by term . We observe that the sum of any term in (1) and the corresponding term in (2) is 2a + ( n - 1) . For instance how many times will get ( 2a = (n - 1)d) ? It is clear, that each Sn in (1) and (2) has n terms and therefore, we have
2 Sn = n {2a + (n - 1) d }
or S n = n/2 { 2a + ( n-1 ) d }, the formula for finding sum of n terms of an A. P .