Question -
Answer -
Let AP be 1, 2, 3, 4,…, n
Here,
First term, a = a1 =1
Common difference, d =a2 – a1 = 2 – 1 = 1
l = n
So, the sum of n terms= S = n/2 [2a + (n-1) d]
= n/2 [2(1) + (n-1) 1]
= n/2 [2 + n – 1]
= n/2 [n – 1]
∴ The sum of the firstn natural numbers is n(n-1)/2