420=n(n+1) solve bracket
420=n^{2 }+ n take 420 to the other side
n^{2}^{ }+ n - 420 = 0 apply quadritic form
n^{2 }- 20n+ 21n - 420 = 0 now factorize it
n(n-20) +21(n-20)=0
(n-20)(n+21)=0
n-20=0 , n+21=0
n=20 , n=-21
no of terms can never b negative, so the answer is 20
420 = n (n + 1 ) use distributive property
400 + 20 = n^{2 }+ n transpose
400 - n^{2 }= n - 20 get the square root of 400 - n^{2 }then it becomes
20 - n = n - 20 transpose
20 + 20 = n + n then add
40 = 2n then divide by 2 on each side to get the value of n
n = 20
Sum of first n even natural numbers is S=n(n-1)
Given Sum = 420
so, n(n-1) = 420
n^2-n=420
n^2-n-420=0
n^2+21n-20n-420=0
n(n+21)-20(n+21)=0
(n-20)(n+21)=0
n=20 or n=-21
But here n is the number of first n natural numbers so n cannot be negative. so n=20
given the sum of n even numbers n(n+1)=420
=n^2+n=420
=n^2+n-420---(1)
now we have to do factorization,so we have multiply the first term of and constant from equation(1) . then we will get, -420n^2
now we have divide the value into tow values .when we add or substarct the 420n^2 we have to get the middle term of the equation and also when we multiply it we have to get again -420n^2
so now we have to do, -420n^2= 21n*-20n
now we will rewrite the eqn(1),
=n^2+21n-20n-420=0
=n(n+21)-20(n+21)=0
=(n+21)(n-20)=0
=n+21=0,n-20=0
=n=-21,n=20
given problem is,the sum of even number so we will take n=20 only
so,n(n+1)=>20(20+1)=>20(21)=>420