Sum of 1 to n natural numbers is 36, then find the value of n.

List of n natural number is

1, 2, 3, ……n

First term a = 1

Second term t₁ = 2

Third term t₃ = 3

Thus, common difference d = t₃ – t₂ = 3 – 2 = 1

Given S_n = 36

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find no. of terms n

⇒ n(1 + n) = 36 × 2 = 72

⇒ n² + n – 72 = 0

⇒ n² + 9n – 8n – 72 = 0

⇒ n(n + 9) – 8(n + 9) = 0

⇒ (n – 8)(n + 9) = 0

⇒ n – 8 = 0 or n + 9 = 0

⇒ n = 8 or n = – 9

Since, number of terms n can’t be negative

∴ n = 8