Question

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

Answer 1

a=1,d=1

S  

n

​  

=36

2

n

​  

[2a+(n−1)d]=36

2

n

​  

[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



=9,n=8.