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Hint: Use the formulas of sequences and series to find the sum of $n$ natural numbers and their squares. Substitute these formulas in the equation which can be obtained by reading the question and then find the value of $n$.

Before proceeding with the question, we must know all the formulas of the sequences and series which will be required to solve this question.

We have a formula from which, the sum of first $r$ natural numbers (denoted by $\sum{r}$) is given by,

$\sum{r=\dfrac{r\left( r+1 \right)}{2}}....................\left( 1 \right)$

Also, we have a formula from which, the sum of the squares first $r$ natural number (denoted by $\sum{{{r}^{2}}}$) is given by,

$\sum{{{r}^{2}}=\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}................\left( 2 \right)}$

In the question, it is given that the sum of the first $n$ natural number is one-fifth of the sum of their squares.

$\Rightarrow $ sum of first $n$ natural numbers $=$ $\dfrac{1}{5}\times $ sum of squares of first $n$ natural numbers

$\Rightarrow $$\sum{n}=\dfrac{1}{5}\sum{{{n}^{2}}.................\left( 3 \right)}$

Substituting $r=n$ in formula $\left( 1 \right)$, the sum of first $n$ natural numbers is equal to,

$\sum{n=\dfrac{n\left( n+1 \right)}{2}}....................\left( 4 \right)$

Substituting $r=n$ in formula $\left( 2 \right)$, the sum of the squares of first $n$ natural numbers is equal to,

$\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}................\left( 5 \right)}$

Substituting $\sum{n}$ from equation $\left( 4 \right)$ and $\sum{{{n}^{2}}}$ from equation $\left( 5 \right)$ in equation $\left( 3 \right)$, we get,

$\dfrac{n\left( n+1 \right)}{2}=\dfrac{1}{5}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)$

Cancelling $n\left( n+1 \right)$ on both the sides of the above equation, we get,

$\begin{align}

& \dfrac{1}{2}=\dfrac{\left( 2n+1 \right)}{30} \\

& \Rightarrow 15=2n+1 \\

& \Rightarrow 2n=14 \\

& \Rightarrow n=7 \\

\end{align}$

Hence, the answer is option (c).

Note: There is a possibility that one may make a mistake while applying the formula for sum of the first $n$ natural number. It is a very common mistake that one uses the formula as $\sum{n=\dfrac{n\left( n-1 \right)}{2}}$ instead of the formula $\sum{n=\dfrac{n\left( n+1 \right)}{2}}$. This may lead us to an incorrect answer.

__Complete step-by-step answer:__Before proceeding with the question, we must know all the formulas of the sequences and series which will be required to solve this question.

We have a formula from which, the sum of first $r$ natural numbers (denoted by $\sum{r}$) is given by,

$\sum{r=\dfrac{r\left( r+1 \right)}{2}}....................\left( 1 \right)$

Also, we have a formula from which, the sum of the squares first $r$ natural number (denoted by $\sum{{{r}^{2}}}$) is given by,

$\sum{{{r}^{2}}=\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}................\left( 2 \right)}$

In the question, it is given that the sum of the first $n$ natural number is one-fifth of the sum of their squares.

$\Rightarrow $ sum of first $n$ natural numbers $=$ $\dfrac{1}{5}\times $ sum of squares of first $n$ natural numbers

$\Rightarrow $$\sum{n}=\dfrac{1}{5}\sum{{{n}^{2}}.................\left( 3 \right)}$

Substituting $r=n$ in formula $\left( 1 \right)$, the sum of first $n$ natural numbers is equal to,

$\sum{n=\dfrac{n\left( n+1 \right)}{2}}....................\left( 4 \right)$

Substituting $r=n$ in formula $\left( 2 \right)$, the sum of the squares of first $n$ natural numbers is equal to,

$\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}................\left( 5 \right)}$

Substituting $\sum{n}$ from equation $\left( 4 \right)$ and $\sum{{{n}^{2}}}$ from equation $\left( 5 \right)$ in equation $\left( 3 \right)$, we get,

$\dfrac{n\left( n+1 \right)}{2}=\dfrac{1}{5}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)$

Cancelling $n\left( n+1 \right)$ on both the sides of the above equation, we get,

$\begin{align}

& \dfrac{1}{2}=\dfrac{\left( 2n+1 \right)}{30} \\

& \Rightarrow 15=2n+1 \\

& \Rightarrow 2n=14 \\

& \Rightarrow n=7 \\

\end{align}$

Hence, the answer is option (c).

Note: There is a possibility that one may make a mistake while applying the formula for sum of the first $n$ natural number. It is a very common mistake that one uses the formula as $\sum{n=\dfrac{n\left( n-1 \right)}{2}}$ instead of the formula $\sum{n=\dfrac{n\left( n+1 \right)}{2}}$. This may lead us to an incorrect answer.

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