Constant average?

If a new number doesn't change the average, we must have added the average itself: $\begin{aligned} &&\frac{x_1+\cdots+x_n}{n} &= \frac{x_1+\cdots+x_n+x_{n+1}}{n+1} \\ &\Rightarrow& (n+1)(x_1+\cdots+x_n) &= n(x_1+\cdots+x_n+x_{n+1}) \\ &\Rightarrow& n(x_1+\cdots+x_n) + x_1+\cdots+x_n &= n(x_1+\cdots+x_n)+nx_{n+1} \\ &\Rightarrow& \frac{x_1+\cdots+x_n}{n} &= x_{n+1} \end{aligned}$

In this case, it means the new number we added was 2018. Adding 2018 again will not change the average.

When you go add same number the average will also remain constant and is equal to the number you are adding. In this case when we go on add the same number 2018 the average will also remain as 2018.

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$\dfrac{\overbrace{x + x + x + ...}^{\text{n times}}}{n} = \dfrac{nx}{n} = x \qquad where, x = 2018$

Let the total sum of T numbers is : H The new number is : x

Given, H/T=2018 =(H+x)/(T+1) Using addendo;

(H+x)/(T+1)=H/T=x/1=(H+2x)/(T+2)

Therefore the average =(H+2x)/(T+2)=2018=the same