The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.

According to ques,

 

\[S\text{ }=\text{ }n\left( n\text{ }+\text{ }1 \right)\]

Also, \[S\text{ }=\text{ }420\]

So, \[n\left( n\text{ }+\text{ }1 \right)\text{ }=\text{ }420\]

Or,

\[{{n}^{2}}~+\text{ }n\text{ }\text{ }420\text{ }=\text{ }0\]

or,

\[{{n}^{2}}~+\text{ }21n\text{ }\text{ }20n\text{ }\text{ }420\text{ }=\text{ }0\]

or,

\[n\left( n\text{ }+\text{ }21 \right)\text{ }\text{ }20\left( n\text{ }+\text{ }21 \right)\text{ }=\text{ }0\]

or,

\[\left( n\text{ }+\text{ }21 \right)\text{ }\left( n\text{ }\text{ }20 \right)\text{ }=\text{ }0\]

Or,

\[n\text{ }=\text{ }-21,\text{ }20\]

Since, n cannot be negative.

Hence, \[n\text{ }=\text{ }20.\]