Question

Find the mean of the following data: Range of first $n$ natural numbers range of negative integers from $-n$ to $-1$ (where $-n<-1$), range of first $n$ positive even integers and range of first $n$ positive odd integers

• A. $\frac{2n+1}{4}$
• B. $\frac{3}{2}(n-1)$
• C. $\frac{3n}{2}$
• D. $3n$

Answer 1

The correct option is A $\frac{2n+1}{4}$

Sum of n natural number $=\frac{n}{2}[2n+(n-1)d]=\frac{n}{2}[2+n-1]=\frac{n}{2}[n+1]$

Sum of $n$ negative integers $=\frac{n}{2}[-2n+(n-1)]=\frac{n}{2}[-n-1]=-\frac{n}{2}[n+1]$

Sum of $n$ positive even integers $=\frac{n}{2}[2\times 2+(n-1)2]=\frac{n}{2}[4+2n-2]=\frac{n}{2}[2+2n]=n(n+1)$

Sum of n positive odd Integers $=\frac{n}{2}[2\times 1+(n-1)2]=\frac{n}{2}[2+2n-2]=n^2$

Total sum $=\frac{n}{2}(n+1)-\frac{n}{2}(n+1)+n(n+1)+n^2=2n^2+n$

Mean $=\frac{2n^2+n}{n+n+n+n}=\frac{2n^2+n}{4n}=\frac{n(2n+1)}{4n}=\frac{2n+1}{4}Ans.$