Question

If the mean of n observations: ${1^2},{2^2},{3^2},{4^2}....{n^2}$ is $\dfrac{{46n}}{{11}}$, then n equals:
(A) $11$
(B) $12$
(C) $23$
(D) $22$

Answer 1

Hint: In the given question, we are required to find the value of mean of n observations that are the squares of first n natural numbers. So, we should know the formula for finding out the sum of n squares as $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$. We should also know the formula for finding the mean of n numbers as: $\dfrac{{{a_1} + {a_2} + {a_3}.... + {a_n}}}{n}$. Then, we will solve the equation obtained using the method of transposition.

Complete step-by-step answer:
So, we have n observations: ${1^2},{2^2},{3^2},{4^2}....{n^2}$.
Now, we have to calculate the mean of these n observations as: $\dfrac{{{\text{Sum of observations}}}}{{{\text{Number}}\,{\text{of}}\,{\text{observations}}}}$.
So, we get, Mean of the observations ${1^2},{2^2},{3^2},{4^2}....{n^2}$ $ = \dfrac{{{1^2} + {2^2} + {3^2} + {4^2}.... + {n^2}}}{n}$
Now, we know that the sum of squares of n natural numbers can be expressed as $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$. Putting this into the expression, we get,
\[ = \dfrac{{\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right)}}{n}\]
Simplifying the expression, we get,
\[ = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ = \dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
Now, we are also given the mean to be equal to $\dfrac{{46n}}{{11}}$. So, we get,
\[\dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = \dfrac{{46n}}{{11}}\]
Computing the product of terms, we get,
\[ \Rightarrow 11\left( {2{n^2} + 3n + 1} \right) = 276n\]
\[ \Rightarrow 22{n^2} + 33n + 11 = 276n\]
\[ \Rightarrow 22{n^2} - 243n + 11 = 0\]
Now, we use the method of splitting the middle term to find the roots of the above quadratic equation.
We split the middle term $ - 243x$ into two terms $ - 242x$ and $ - x$ since the product of these terms, $242{x^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $ - 243x$.
\[ \Rightarrow 22{n^2} - 242n - n + 11 = 0\]
\[ \Rightarrow 22n\left( {n - 11} \right) - \left( {n - 11} \right) = 0\]
\[ \Rightarrow \left( {22n - 1} \right)\left( {n - 11} \right) = 0\]
Now, either of the two terms have to be zero.
So, either \[\left( {n - 11} \right) = 0\] or \[\left( {22n - 1} \right) = 0\]
Either \[n = 11\] or \[n = \dfrac{1}{{22}}\].
Now, we know that n represents the number of observations. So, it should be a natural number. Hence, \[n = 11\].
Therefore, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease. : Splitting the middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Special care should be taken in such cases.