Question

If the variance of the first n natural number is 10 and the variance of first n even natural number is 16, then m + n is equal to

Answer 1

Hint: Now we know the formula for variance of data ${{x}_{1}},{{x}_{2}},....{{x}_{n}}$ is given by $\dfrac{\sum\limits_{1}^{n}{{{x}_{i}}^{2}}}{n}-{{\mu }^{2}}$ where $\mu $ is the mean of given data. Hence in both the cases we will first find the mean of data. We know that the sum of first n natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$ . Using this we will find the mean of data. Now we will substitute the data and the mean in the formula of variance. Then we will simplify the equation with the help of formula ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ and solve the quadratic to find the values of m and n.

Complete step by step answer:
Now first let us consider first n natural numbers.
We know that the sum of n natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$ .
Hence using this we will get mean of first n natural numbers which is is $\dfrac{n\left( n+1 \right)}{2n}=\dfrac{\left( n+1 \right)}{2}$
Now let us check the formula for variance.
Variance of data ${{x}_{1}},{{x}_{2}},....{{x}_{n}}$ is given by $\dfrac{\sum\limits_{1}^{n}{{{x}_{i}}^{2}}}{n}-{{\mu }^{2}}$ where $\mu $ is the mean of given data.
Hence variance of the 1, 2, 3…n is given by
$\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.........{{n}^{2}}}{n}-{{\left( \dfrac{n+1}{2} \right)}^{2}}$
Now we know that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
Hence substituting this in the above equation we get
$\begin{align}
  & \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{{{\left( n+1 \right)}^{2}}}{4} \\
 & \Rightarrow \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{{{\left( n+1 \right)}^{2}}}{4}=\left( \dfrac{n+1}{2} \right)\left[ \dfrac{\left( 2n+1 \right)}{3}-\dfrac{n+1}{2} \right] \\
\end{align}$
Taking LCM we get,
$\begin{align}
  & \Rightarrow \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{{{\left( n+1 \right)}^{2}}}{4}=\dfrac{n+1}{2}\left[ \dfrac{4n+2-3n-3}{6} \right] \\
 & \Rightarrow \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{{{\left( n+1 \right)}^{2}}}{4}=\dfrac{n+1}{2}\left[ \dfrac{n-1}{6} \right] \\
 & \Rightarrow \dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{{{\left( n+1 \right)}^{2}}}{4}=\dfrac{{{n}^{2}}-1}{12} \\
\end{align}$
Now we know that the variance is equal to 10
Hence equating the obtained variance with 10 we get $\dfrac{\left( {{n}^{2}}-1 \right)}{12}=10$
Now let us multiply the whole equation with 12
${{n}^{2}}=120+1$
Hence we get n = 11.
Now let us consider m even natural numbers.
2, 4, 6, 8… 2m.
Now we will calculate mean of m even numbers
$\dfrac{2+4+6+....2m}{m}=\dfrac{2\left( 1+2+3+...m \right)}{m}$
We know that the sum of n natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$ . Hence we get
$\begin{align}
  & \Rightarrow \dfrac{2+4+6+....2m}{m}=\dfrac{2m\left( m+1 \right)}{2m} \\
 & \Rightarrow \dfrac{2+4+6+....2m}{m}=\left( m+1 \right) \\
\end{align}$
Now again we know that Variance of data ${{x}_{1}},{{x}_{2}},....{{x}_{n}}$ is given by $\dfrac{\sum\limits_{1}^{n}{{{x}_{i}}^{2}}}{n}-{{\mu }^{2}}$ where $\mu $ is the mean of given data.
Hence variance of 2, 4, 6, 8 … 2m is
\[\dfrac{{{2}^{2}}+{{4}^{2}}+{{6}^{2}}+...2{{m}^{2}}}{m}-{{\left( m+1 \right)}^{2}}=\dfrac{{{2}^{2}}\left( {{1}^{2}}+{{2}^{2}}+....2{{m}^{2}} \right)}{m}-{{\left( m+1 \right)}^{2}}\]
Now we are given that this variance is 16
Hence we have \[\dfrac{{{2}^{2}}\left( {{1}^{2}}+{{2}^{2}}+....2{{m}^{2}} \right)}{m}-{{\left( m+1 \right)}^{2}}=16\]
Dividing the equation by 4 we get
$\dfrac{\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{m}^{2}} \right)}{m}-\dfrac{{{\left( m+1 \right)}^{2}}}{4}=4$
Now again we know that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
Hence substituting this we get
$\dfrac{\left( m+1 \right)\left( 2m+1 \right)}{6}-\dfrac{{{\left( m+1 \right)}^{2}}}{4}=4$
$\Rightarrow \dfrac{\left( m+1 \right)}{2}\left[ \dfrac{\left( 2m+1 \right)}{3}-\dfrac{\left( m+1 \right)}{2} \right]=4$
Taking LCM we get,
$\begin{align}
  & \Rightarrow \dfrac{\left( m+1 \right)}{2}\left[ \dfrac{4m+2-3m-3}{6} \right]=4 \\
 & \Rightarrow \dfrac{\left( m+1 \right)\left( m-1 \right)}{12}=4 \\
 & \Rightarrow {{m}^{2}}-1=48 \\
 & \therefore {{m}^{2}}=49 \\
\end{align}$
Hence taking the square root on both sides we get the value of m is 7.
Hence m = 7 and n = 11.
This means m + n = 11.

Note:
Now if we have variance of 2, 4, …. 2m is 16 we can take 2 common and say that the variance of 2(1, 2, … m) is 16. Now when we have such a scenario we can directly write it as the variance of 1, 2, …. m is $\dfrac{16}{{{2}^{2}}}$ . Also note that here m and n are natural numbers hence we do not consider negative values while taking square root.