A sequence of odd positive integer written as 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 the mean of the \[{{n}^{th}}\] row is
(A) \[\dfrac{{{n}^{2}}\left( 2{{n}^{2}}+1 \right)}{3}\]
(B) \[\dfrac{{{n}^{2}}\left( 4{{n}^{2}}+1 \right)}{6}\]
(C) \[\dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{3}\]
(D) \[\dfrac{n\left( 2{{n}^{2}}+1 \right)}{3}\]
Answer
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Hint: The first row contains 1 term, the second row contains 4 terms, and the third row contains 9 terms. So, the total number of terms for the \[{{n}^{th}}\] row is \[{{n}^{2}}\] . In the \[{{1}^{st}}\] row, we have only one term that is 1. In the \[{{2}^{nd}}\] row, we have four terms that are 3, 5, 7, and 9. In the \[{{3}^{rd}}\] row, we have nine terms that are 11, 13, 15, 17, 19, 21, 23, 25, and 27. The first term of the first row, second row, and the third row is 1, 3, and 11 respectively. Similarly, the last term of the first row, second row, and the third row is 1, 9, and 27 respectively. The first term and last term for the \[{{n}^{th}}\] row is \[\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)\] and \[\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] respectively. We can observe that the terms of every row is in arithmetic progression. Now, use the formula of summation of all terms of an Arithmetic Progression, \[\dfrac{N}{2}\left( First\,term+Last\,term \right)\] where, N is the total number of terms. We also know the formula of mean, \[Mean=\dfrac{summation\,of\,all\,terms}{total\,number\,of\,terms}\] . The total number of terms for the \[{{n}^{th}}\] row is \[{{n}^{2}}\] . Now, solve it further and calculate the mean.
Complete step-by-step solution:
According to the question, we have the sequence of odd positive integers written as 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27.
In the \[{{1}^{st}}\] row, we have only one term that is 1 ……………………………………(1)
In the \[{{2}^{nd}}\] row, we have four terms that are 3, 5, 7, and 9 …………………………………………….(2)
In the \[{{3}^{rd}}\] row, we have nine terms that are 11, 13, 15, 17, 19, 21, 23, 25, and 27 …………………………………….(3)
It means we can say that for \[{{n}^{th}}\] row, we have a total of \[{{n}^{2}}\] terms ………………………………………….(4)
Now, from equation (1), equation (2), and equation (3), we have
The first row is starting with the number 1. So, the first number of the first row is 1.
Similarly, we can observe that the second and the third row are starting with the numbers 3 and 11 respectively. So, the first number of the second and the third row is 3 and 11 respectively.
Let us assume that \[{{T}_{n}}\] is the first number of the \[{{n}^{th}}\] row.
\[{{T}_{1}}=1\]
\[{{T}_{2}}=3\]
\[{{T}_{3}}=11\]
Since every \[{{n}^{th}}\] row has \[{{n}^{2}}\] terms so, the first term of the \[{{n}^{th}}\] row will be
\[{{\left[ \left\{ {{1}^{2}}+{{2}^{2}}+.......+{{\left( n-1 \right)}^{2}} \right\}+1 \right]}^{th}}\] term of the Arithmetic Progression, 1, 3, 5, 7, ………… ………………………………(4)
We know the formula for the summation of the squares of \[n\] natural numbers,
\[{{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\] …………………………………………(5)
Now, from equation (4) and equation (5), we get
\[\Rightarrow {{S}_{n}}=\dfrac{\left( n-1 \right)\left( n-1+1 \right)\left( 2n-2+1 \right)}{6}\]
\[\Rightarrow {{S}_{n}}=\dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}\] ……………………………………………(6)
So, the first term of \[{{n}^{th}}\] row is \[{{\left\{ \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right\}}^{th}}\] term of the A.P, 1, 3, 5, 7, ……………………………………..(7)
Now, from equation (5) and equation (6), we get
\[\begin{align}
& \Rightarrow {{T}_{n}}=1+\left[ \left\{ \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}+1 \right\}-1 \right]2 \\
& \Rightarrow {{T}_{n}}=1+\left\{ \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right\}2 \\
& \Rightarrow {{T}_{n}}=1+\left\{ \dfrac{2{{n}^{3}}-3{{n}^{2}}+n}{3} \right\} \\
\end{align}\]
\[{{T}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)\] ………………………………………..(8)
So, the first term of the \[{{n}^{th}}\] row is \[{{T}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)\] ……………………………………(9)
Now, again from equation (1), equation (2), and equation (3), we have
The first row is ending with the number 1. So, the last number of the first row is 1.
Similarly, we can observe that the second and the third row are ending with the numbers 9 and 27 respectively. So, the last number of the second and the third row is 9 and 27 respectively.
Similarly, let us assume that \[{{L}_{n}}\] is the last number of the \[{{n}^{th}}\] row.
\[{{L}_{1}}=1\]
\[{{L}_{2}}=9\]
\[{{L}_{3}}=27\]
Since every \[{{n}^{th}}\] row has \[{{n}^{2}}\] terms so, the last term of the \[{{n}^{th}}\] row will be
\[{{\left\{ {{1}^{2}}+{{2}^{2}}+.......+{{n}^{2}} \right\}}^{th}}\] term of the Arithmetic Progression, 1, 9, 27, ………… ………………………………(10)
Now, from equation (5) and equation (10), we get
\[\Rightarrow {{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\] ……………………………………………(11)
So, the last term of \[{{n}^{th}}\] row is \[{{\left\{ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right\}}^{th}}\] term of the A.P, 1, 9, 27, ……………………………………..(12)
Now, from equation (11) and equation (12), we get
\[\begin{align}
& \Rightarrow {{L}_{n}}=1+\left[ \left\{ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right\}-1 \right]2 \\
& \Rightarrow {{L}_{n}}=1+\left\{ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)-6}{6} \right\}2 \\
& \Rightarrow {{L}_{n}}=1+\left\{ \dfrac{2{{n}^{3}}+3{{n}^{2}}+n-6}{3} \right\} \\
\end{align}\]
\[{{L}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] ………………………………………..(13)
So, the last term of the \[{{n}^{th}}\] row is \[{{L}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] ……………………………………(14)
Now, our \[{{n}^{th}}\] row will look like,
\[\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right),.......................,\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] ………………………………….(15)
From equation (1), equation (2), and equation (3), we can say that every row is in Arithmetic Progression which has a common difference of 2.
We know the formula of summation of all terms of an Arithmetic Progression is \[\dfrac{N}{2}\left( First\,term+Last\,term \right)\] . Here, N is the total number of terms ………………………………….(16)
From equation (4), and equation (15), we have the total number of terms, the first term, and the last term of \[{{n}^{th}}\] row.
Now, from equation (4), equation (15), and equation (16), we have
Summation of all terms of \[{{n}^{th}}\] row = \[\dfrac{{{n}^{2}}}{2}\left\{ \dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)+\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right) \right\}=\dfrac{{{n}^{2}}}{2}\left\{ \dfrac{2}{3}\left( 2{{n}^{3}}+n \right) \right\}=\dfrac{{{n}^{2}}}{3}\left( 2{{n}^{3}}+n \right)\] ……………………………..(17)
We know the formula of mean, \[Mean=\dfrac{summation\,of\,all\,terms}{total\,number\,of\,terms}\] …………………………………..(18)
Now, from equation (4), equation (17), and equation (18), we get
Mean of all numbers of \[{{n}^{th}}\] row = \[\dfrac{\dfrac{{{n}^{2}}}{3}\left( 2{{n}^{3}}+n \right)}{{{n}^{2}}}=\dfrac{1}{3}\left( 2{{n}^{3}}+n \right)=\dfrac{n}{3}\left( 2{{n}^{2}}+1 \right)\] .
Therefore, the mean of \[{{n}^{th}}\] row is \[\dfrac{n}{3}\left( 2{{n}^{2}}+1 \right)\].
Hence, the correct option is (D).
Note: In this question, one might get confused because the number of terms in any row is not given. Here, the hidden information is that the first row contains 1 term, the second row contains 4 terms, and the third row contains 9 terms. So, the number of terms for the \[{{n}^{th}}\] row is \[{{n}^{2}}\] .
Complete step-by-step solution:
According to the question, we have the sequence of odd positive integers written as 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27.
In the \[{{1}^{st}}\] row, we have only one term that is 1 ……………………………………(1)
In the \[{{2}^{nd}}\] row, we have four terms that are 3, 5, 7, and 9 …………………………………………….(2)
In the \[{{3}^{rd}}\] row, we have nine terms that are 11, 13, 15, 17, 19, 21, 23, 25, and 27 …………………………………….(3)
It means we can say that for \[{{n}^{th}}\] row, we have a total of \[{{n}^{2}}\] terms ………………………………………….(4)
Now, from equation (1), equation (2), and equation (3), we have
The first row is starting with the number 1. So, the first number of the first row is 1.
Similarly, we can observe that the second and the third row are starting with the numbers 3 and 11 respectively. So, the first number of the second and the third row is 3 and 11 respectively.
Let us assume that \[{{T}_{n}}\] is the first number of the \[{{n}^{th}}\] row.
\[{{T}_{1}}=1\]
\[{{T}_{2}}=3\]
\[{{T}_{3}}=11\]
Since every \[{{n}^{th}}\] row has \[{{n}^{2}}\] terms so, the first term of the \[{{n}^{th}}\] row will be
\[{{\left[ \left\{ {{1}^{2}}+{{2}^{2}}+.......+{{\left( n-1 \right)}^{2}} \right\}+1 \right]}^{th}}\] term of the Arithmetic Progression, 1, 3, 5, 7, ………… ………………………………(4)
We know the formula for the summation of the squares of \[n\] natural numbers,
\[{{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\] …………………………………………(5)
Now, from equation (4) and equation (5), we get
\[\Rightarrow {{S}_{n}}=\dfrac{\left( n-1 \right)\left( n-1+1 \right)\left( 2n-2+1 \right)}{6}\]
\[\Rightarrow {{S}_{n}}=\dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}\] ……………………………………………(6)
So, the first term of \[{{n}^{th}}\] row is \[{{\left\{ \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right\}}^{th}}\] term of the A.P, 1, 3, 5, 7, ……………………………………..(7)
Now, from equation (5) and equation (6), we get
\[\begin{align}
& \Rightarrow {{T}_{n}}=1+\left[ \left\{ \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}+1 \right\}-1 \right]2 \\
& \Rightarrow {{T}_{n}}=1+\left\{ \dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right\}2 \\
& \Rightarrow {{T}_{n}}=1+\left\{ \dfrac{2{{n}^{3}}-3{{n}^{2}}+n}{3} \right\} \\
\end{align}\]
\[{{T}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)\] ………………………………………..(8)
So, the first term of the \[{{n}^{th}}\] row is \[{{T}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)\] ……………………………………(9)
Now, again from equation (1), equation (2), and equation (3), we have
The first row is ending with the number 1. So, the last number of the first row is 1.
Similarly, we can observe that the second and the third row are ending with the numbers 9 and 27 respectively. So, the last number of the second and the third row is 9 and 27 respectively.
Similarly, let us assume that \[{{L}_{n}}\] is the last number of the \[{{n}^{th}}\] row.
\[{{L}_{1}}=1\]
\[{{L}_{2}}=9\]
\[{{L}_{3}}=27\]
Since every \[{{n}^{th}}\] row has \[{{n}^{2}}\] terms so, the last term of the \[{{n}^{th}}\] row will be
\[{{\left\{ {{1}^{2}}+{{2}^{2}}+.......+{{n}^{2}} \right\}}^{th}}\] term of the Arithmetic Progression, 1, 9, 27, ………… ………………………………(10)
Now, from equation (5) and equation (10), we get
\[\Rightarrow {{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\] ……………………………………………(11)
So, the last term of \[{{n}^{th}}\] row is \[{{\left\{ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right\}}^{th}}\] term of the A.P, 1, 9, 27, ……………………………………..(12)
Now, from equation (11) and equation (12), we get
\[\begin{align}
& \Rightarrow {{L}_{n}}=1+\left[ \left\{ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right\}-1 \right]2 \\
& \Rightarrow {{L}_{n}}=1+\left\{ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)-6}{6} \right\}2 \\
& \Rightarrow {{L}_{n}}=1+\left\{ \dfrac{2{{n}^{3}}+3{{n}^{2}}+n-6}{3} \right\} \\
\end{align}\]
\[{{L}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] ………………………………………..(13)
So, the last term of the \[{{n}^{th}}\] row is \[{{L}_{n}}=\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] ……………………………………(14)
Now, our \[{{n}^{th}}\] row will look like,
\[\dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right),.......................,\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right)\] ………………………………….(15)
From equation (1), equation (2), and equation (3), we can say that every row is in Arithmetic Progression which has a common difference of 2.
We know the formula of summation of all terms of an Arithmetic Progression is \[\dfrac{N}{2}\left( First\,term+Last\,term \right)\] . Here, N is the total number of terms ………………………………….(16)
From equation (4), and equation (15), we have the total number of terms, the first term, and the last term of \[{{n}^{th}}\] row.
Now, from equation (4), equation (15), and equation (16), we have
Summation of all terms of \[{{n}^{th}}\] row = \[\dfrac{{{n}^{2}}}{2}\left\{ \dfrac{1}{3}\left( 2{{n}^{3}}-3{{n}^{2}}+n+3 \right)+\dfrac{1}{3}\left( 2{{n}^{3}}+3{{n}^{2}}+n-3 \right) \right\}=\dfrac{{{n}^{2}}}{2}\left\{ \dfrac{2}{3}\left( 2{{n}^{3}}+n \right) \right\}=\dfrac{{{n}^{2}}}{3}\left( 2{{n}^{3}}+n \right)\] ……………………………..(17)
We know the formula of mean, \[Mean=\dfrac{summation\,of\,all\,terms}{total\,number\,of\,terms}\] …………………………………..(18)
Now, from equation (4), equation (17), and equation (18), we get
Mean of all numbers of \[{{n}^{th}}\] row = \[\dfrac{\dfrac{{{n}^{2}}}{3}\left( 2{{n}^{3}}+n \right)}{{{n}^{2}}}=\dfrac{1}{3}\left( 2{{n}^{3}}+n \right)=\dfrac{n}{3}\left( 2{{n}^{2}}+1 \right)\] .
Therefore, the mean of \[{{n}^{th}}\] row is \[\dfrac{n}{3}\left( 2{{n}^{2}}+1 \right)\].
Hence, the correct option is (D).
Note: In this question, one might get confused because the number of terms in any row is not given. Here, the hidden information is that the first row contains 1 term, the second row contains 4 terms, and the third row contains 9 terms. So, the number of terms for the \[{{n}^{th}}\] row is \[{{n}^{2}}\] .
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