Question

If the sum of first n even natural numbers is 420, find the value of n.

Answer 1

Hint: Here we will use the general term used to denote an even number i.e. 2n and apply summation. We will get a resultant quadratic equation, solving which, we will get the value of n.

Complete step-by-step answer:
The general term denoting an even natural number is (2n) where n is a natural number as well (n = 1,2,3,4,.....), because when you multiply something by 2 it will result in an even number.
According to question,
$ \Rightarrow \sum\limits_{r = 1}^n {2n} = 420$
$ \Rightarrow 2\sum\limits_{r = 1}^n n = 420$
$ \Rightarrow \sum\limits_{r = 1}^n n = 210$
We know that, $\sum\limits_{r = 1}^n n = \dfrac{{n\left( {n + 1} \right)}}{2}$
$ \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2} = 210$
$ \Rightarrow {n^2} + n - 420 = 0$
$ \Rightarrow {n^2} + 21n - 20n - 420 = 0$ [$\because -420 = 21 \times (-20)$]
$ \Rightarrow n (n + 21) - 20 (n + 21) = 0$
$\Rightarrow$ $\left( {n - 20} \right)\left( {n + 21} \right) = 0$
$ \Rightarrow n = 20$ and $n = -21$
n cannot be negative.
$\therefore n=20$.
So, this is your answer.

Note: We can also solve these types of problems by using the concept of arithmetic progression as the sum is given, the first term and common difference can be determined and the number of terms n is found. In this case $S_n = 420$, $a = 2$ and $d =2$.