Question

The mean of the first “n” even numbers is?

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
Since we need to know about Arithmetic progression. An arithmetic progression can be given by a, a plus the d and a plus the two d, etc.., where $ a $ is the first term and $ d $ is a common difference.
Also, the mean which is the sum of the terms divides the number of the terms.
Formula used: $ a,(a + d),(a + 2d),(a + 3d),... $ ( Arithmetic progression) and $ mean = \dfrac{{\sum n }}{n} $

Complete step by step answer:
Let the set of all even numbers can be written as in the form of $ 2,4,6,8,...,n $ and it is countable because it is finite and countably infinite concerning the correspondence to the natural numbers.
Thus, first we are going to find the sum of the terms of the even numbers which are $ 2 + 4 + 6 + ... + n $ and by using the formula of the Arithmetic progression where a is the two and d is a common difference also two.
Hence applying that in the AP formula we get $ \dfrac{n}{2}(2.2 + (n - 1)2) $ , where two is the first and also common difference.
Hence further solving this we get $ \dfrac{n}{2}(2n + 2) = n(n + 1) $ which is the sum of the terms in the mean, also the number of the terms in even number is n (taking common terms out and cancel each other).
Thus, we get by using the mean formula as $ \dfrac{{n(n + 1)}}{n} = n + 1 $ (by division of common term n).
Hence $ n + 1 $ is the mean of the set of all even numbers.

Note: since the AP formula to consider for solving these questions $ {a_n} = a + (n - 1)d $ (another formula for AP), where $ d $ is the common difference, $ a $ is the first term.
Since we know that the difference between consecutive terms is constant in any A.P. Mean is also called the average.