Question

Find the sum of the first 10 even numbers.

Answer 1

Hint:First of all write the first 10 even natural numbers. Now, assume the first term of the obtained A.P as ‘a’ and the common difference as ‘d’. Find the common difference by subtracting the first term from the second term. Now, use the formula for the sum of n terms of an A.P given as $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$ and substitute the value of n = 10 to get answer.

Complete step by step answer:
Here we have been asked to find the sum of the first 10 even natural numbers. First let us see the first 10 even natural numbers.
Now, natural numbers are counting numbers like 1, 2, 3, 4 and so on, and even numbers are the numbers that are divisible by 2 like 2, 4, 6 and so on. Therefore, the first 10 even natural numbers will be 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.
Clearly we can see that the given numbers form an A.P series with the first term as a = 2. Let us assume the common difference as d so we get,
$\Rightarrow$ d = 4 – 2
$\Rightarrow$ d = 2
We know that the sum of n terms of a A.P is given as $S_n={\displaystyle \frac{n}{2}}[2a+(n-1)d]$ and we have to find the sum of first 10 even natural numbers, so we have n = 10. Substituting the values of a, n and d we get,
$\begin{array}{rl}& \Rightarrow S_{10}={\displaystyle \frac{10}{2}}[2\left(2\right)+(10-1)2]\\ & \Rightarrow S_{10}=5[4+18]\\ & \therefore S_{10}=110\end{array}$

Hence, the required sum of the first 10 even natural numbers is 110.

Note: You can solve the question in less steps if you remember the formula of sum of first n even natural numbers which is given as $S_n=n(n+1)$. You must also remember the formula of the sum of first n odd natural numbers given as $S_n=n^2$. Remember the basic definitions of the terms like natural number, whole number, rational number etc. to avoid confusion.