sum of 1 to n natural numbers is 36 then find the value of n
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Let S represent the sum of numbers.
S=n/2[2a+(n-1)d]=36
a=1 d=1
So,n/2[2+(n-1)]=36
n(n+1)=36
n square +n-72=0
n square +9n-8n-72=0
n(n+9)-8(n+9)=0
(n+9)(n-8)=0
So n is not equal to 9 because n would not be negative.
Therefore,n=8
Let S represent the sum of numbers.
S=n/2[2a+(n-1)d]=36
a=1 d=1
So,n/2[2+(n-1)]=36
n(n+1)=36
n square +n-72=0
n square +9n-8n-72=0
n(n+9)-8(n+9)=0
(n+9)(n-8)=0
So n is not equal to 9 because n would not be negative.
Therefore,n=8
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