Question

question

Answer 1

\(\large \color{black}{\begin{align}
&\normalsize \text{What is the arithmatic mean of first}\ n \ \text{natural numbers} \hspace{.33em}\\~\\
& a.)\ 2(n+1) \hspace{.33em}\\~\\
& b.)\ \dfrac{n(n+1)}{2} \hspace{.33em}\\~\\
& c.)\ \dfrac{(n+1)}{2} \hspace{.33em}\\~\\
& d.)\ \dfrac{n^{2}(n+1)}{2} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

What is your question I can not see it????!?

Answer 3

the sum of the first n numbers 1+2+3+..+n = n(n+1)/2
the average is found by dividing that by n

Answer 4

oh that is easy