Question

If each n numbers \[{{x}_{i}}=i\] is replaced by \[\left( i+1 \right){{x}_{i}}\] then the new mean is
\[\left( a \right)\dfrac{\left( n+1 \right)\left( n+2 \right)}{n}\]
\[\left( b \right)n+1\]
\[\left( c \right)\dfrac{\left( n+1 \right)\left( n+2 \right)}{3}\]
(d) None of these

Answer 1

Hint: To solve this question, we will use the formula of mean stated as \[\text{Mean}=\overline{x},\] \[\text{Mean}=\overline{x}=\dfrac{\text{Sum of all observations}}{\text{number of observations}}.\] We will calculate the new sum of the observations by using the formula, \[\sum{{{n}^{2}}}=\dfrac{n\left( 2n+1 \right)\left( n+1 \right)}{6}\] and \[\sum{n}=\dfrac{n\left( n+1 \right)}{2}.\]

Complete step-by-step answer:
Let us first of all state the formula of the mean. The mean of n observations is given by \[\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{number of observations}}.\] Because each of the n numbers or observations is multiplied by ‘i' means no new term or observation is added. The number of observations is the same. The number of observations = n, as it was before. It is just the sum of all the observations which are going to change. Sum (new_ of all observation can be determined by multiplying \[{{x}_{i}}\] with (1 + i) and adding all the terms from the given terms.
Let the new sum be S.
\[\Rightarrow \text{New sum S}=\sum{\left( 1+i \right){{x}_{i}}}\]
Here, \[{{x}_{i}}=i\]
\[\Rightarrow {{x}_{1}}=1\]
\[\Rightarrow {{x}_{2}}=2\]
\[\Rightarrow {{x}_{3}}=3\]
\[\Rightarrow {{x}_{4}}=4\]
Similarly, all the terms are as above.
\[\Rightarrow \text{Sum (new) S}=\sum{\left( 1+i \right){{x}_{i}}}\]
\[\Rightarrow \text{Sum (new) S}=\sum{\left( 1+i \right)i}\]
\[\Rightarrow \text{S}=\sum{\left( 1+i \right)i}\]
\[\Rightarrow \text{S}=\sum{\left( i+{{i}^{2}} \right)}\]
\[\Rightarrow \text{S}=\sum{i}+\sum{{{i}^{2}}}\]
Now, finally, we will use the formula of the sum of n terms given as \[\sum{n}=\dfrac{n\left( n+1 \right)}{2}.\] Also, the sum of the square of n terms is given by \[\sum{{{n}^{2}}}=\dfrac{1}{6}n\left( n+1 \right)\left( 2n+1 \right).\]
So, finally, \[\sum{i}=\sum{n}=\dfrac{i\left( i+1 \right)}{2}\]
Using the formula, \[\sum{n}=\dfrac{n\left( n+1 \right)}{2},\]
And, \[\sum{{{i}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}=\dfrac{i\left( i+1 \right)\left( 2i+1 \right)}{6}\]
Where we have used the formula,
\[\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
\[\Rightarrow \text{Sum of observations S}=\sum{{{i}^{2}}+\sum{i}}\]
\[\Rightarrow \text{Sum of observations S}=\dfrac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)+\dfrac{n\left( n+1 \right)}{2}\]
\[\Rightarrow \text{S}=\dfrac{1}{6}\left( n\left( n+1 \right)\left( 2n+1 \right) \right)+\dfrac{n\left( n+1 \right)}{2}\]
Finally, we will use the formula of the mean as
\[\text{Mean}=\dfrac{S}{n}=\dfrac{\dfrac{1}{6}\left[ n\left( n+1 \right)\left( 2n+1 \right) \right]+\dfrac{n\left( n+1 \right)}{2}}{n}\]
\[\Rightarrow \text{Mean}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)+3n\left( n+1 \right)}{6n}\]
\[\Rightarrow \text{Mean}=\dfrac{\left( n+1 \right)\left( 2n+1 \right)+3\left( n+1 \right)}{6}\]
\[\Rightarrow \text{Mean}=\dfrac{n\left( 2n+1 \right)+2n+1+3n+3}{6}\]
\[\Rightarrow \text{Mean}=\dfrac{2{{n}^{2}}+3n+3n+1+3}{6}\]
\[\Rightarrow \text{Mean}=\dfrac{2{{n}^{2}}+6n+4}{6}\]
Using the splitting by the middle term, we have,
\[\Rightarrow \text{Mean}=\dfrac{{{n}^{2}}+3n+2}{3}\]
\[\Rightarrow \text{Mean}=\dfrac{{{n}^{2}}+2n+n+2}{3}\]
\[\Rightarrow \text{Mean}=\dfrac{n\left( n+2 \right)+1\left( n+2 \right)}{3}\]
\[\Rightarrow \text{Mean}=\dfrac{\left( n+2 \right)\left( n+1 \right)}{3}\]
Hence, the required mean is given by \[\text{Mean}=\dfrac{\left( n+2 \right)\left( n+1 \right)}{3}.\]

So, the correct answer is “Option c”.

Note: Always remember that in any case, if any term is multiplied by some factor or if all the terms are multiplied by some term, the number of observations is always the same, and the sum of observations will differ. This is the concept we have used here. Also, if all the observations are added by some term, then the sum and number of terms both may differ.