Question

the sum of first 1 to n natural number is 36 then find the value of n​

Answer 1

Answer:

8

Step-by-step explanation:

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

Answer 2

Answer:

n=+8

Step-by-step explanation:

Given,

Sn=36

Also, we know,

Sn= n(n+1)/2

Therefore,

36=(n^2 + n )/2

72=n^2+n

Now,

n^2+n-72=0

Further equating the equation, it comes out to be,

(n+9)(n-8)

Therefore,

n=-9 or n=+8

Now, as n can't be negative so, n=+8

Hope you understand!!

Thank you.