Question

If the mean of the squares of first $n$ natural numbers is 105, then the median of the first $n$ natural numbers is?
(a) 8
(b) 9
(c) 10
(d) 11

Answer 1

Hint: We will use the formula for calculating the mean which is given as
$mean=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{a}_{i}}}$
where $n$ is the count of the collection of numbers and ${{a}_{i}}$ are the elements in the collection of numbers. We will also use the formula for the sum of squares of first $n$ natural numbers. Using these two formulae, we will find the value of $n$. After finding the value of $n$, we will find the median.

Complete step-by-step answer:
We calculate the mean for a collection of numbers using the following formula,
 $mean=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{a}_{i}}}$
where $n$ is the count of the collection of numbers and ${{a}_{i}}$ are the elements in the collection of numbers. Now, the collection of numbers we have is the squares of first $n$ natural numbers. We know that the sum of first $n$ natural numbers is given by $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. We will now substitute the following values in the formula for calculating mean: $mean=105$ and $\sum\limits_{i=1}^{n}{{{a}_{i}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. We will get the following expression,
$105=\dfrac{1}{n}\times \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
Simplifying this expression, we get
$\begin{align}
  & 105\times 6=\left( n+1 \right)\left( 2n+1 \right) \\
 & \therefore 630=\left( n+1 \right)\left( 2n+1 \right) \\
\end{align}$
Now, we will form a quadratic equation by multiplying the factors in the RHS of the equation as follows,
$630=2{{n}^{2}}+n+2n+1$
Simplifying this equation we get,
$2{{n}^{2}}+3n-629=0$
Comparing this quadratic equation to the general quadratic equation $a{{x}^{2}}+bx+c=0$, we get $a=2$, $b=3$ and $c=-629$. We will use the quadratic formula to find the value of $n$. The quadratic formula to find the roots of the general quadratic equation is given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values $a=2$, $b=3$ and $c=-629$, we get
$n=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\left( 2 \right)\left( -629 \right)}}{2\times 2}$
Simplifying the above expression, we get
$\begin{align}
  & n=\dfrac{-3\pm \sqrt{9+5032}}{4} \\
 & =\dfrac{-3\pm \sqrt{5041}}{4} \\
 & =\dfrac{-3\pm 71}{4}
\end{align}$
So, we have two cases.
Case 1: $n=\dfrac{-3-71}{4}=\dfrac{-74}{4}$ .
We can discard this case since the value of $n$ is not a natural number.
Case 2: $n=\dfrac{-3+71}{4}=\dfrac{68}{4}=17$.
So, now we have to find the median of the first 17 natural numbers. We know that the median is the middle value of the set when its elements are ordered from least to greatest. The first 17 natural numbers are naturally ordered from least to greatest. Since there are 17 (an odd number) numbers in the set of first 17 natural numbers, the middle number will be the $\dfrac{n+1}{2}$th term. This means that the middle term is $\dfrac{17+1}{2}=\dfrac{18}{2}={{9}^{th}}$ term, which is 9. Hence, the median of the first 17 natural numbers is 9.

So, the correct answer is “Option b”.

Note: It is essential that we know the formula for the sum of squares of first $n$ natural numbers for this question. We should be able to use the formula we know to our advantage. In this type of question where we use multiple formulas to find the solution, it is useful to do the calculations explicitly to avoid making minor mistakes.