Question

sum of 1 to n natural numbers is 36 , then find the value of n​

Answer 1

Answer:

a=1, d=1

Sn=36

n/2[2a+(n-1) d]=36

n/2[2+(n-1) ]=36

n[2+(n-1) ]=72

n[n+1]=72

n^2+n-72=0

n^2+9n-8n-72=0

(n-8) (n-9) =0

n does not belong to 9 , n = 8