Question

If the terms $\sum{n}$, $\dfrac{\sqrt{10}}{3}\sum{{{n}^{2}}}$, $\sum{{{n}^{3}}}$ are in G.P., then the value of $n$ is:
(a) 3
(b) 4
(c) 2
(d) Does not exist

Answer 1

Hint: We will use the geometric progression to get an expression that involves all the given three terms. We will then use the formulae for the sum of the first $n$ natural numbers, the sum of their squares and cubes. After that, we will simplify the obtained equation. We will end up with a single equation in the variable $n$. We will find the roots of this equation to obtain a value of $n$.

Complete step by step solution:
We know that if three numbers $a$, $b$, $c$ are in G.P., then we have $\dfrac{a}{b}=\dfrac{b}{c}$. This implies that ${{b}^{2}}=a\times c$. Using this fact for the given geometric progression, we get the following,
${{\left[ \dfrac{\sqrt{10}}{3}\sum{{{n}^{2}}} \right]}^{2}}=\left[ \sum{n} \right]\times \left[ \sum{{{n}^{3}}} \right]....(i)$
Now, we know the formula to find the sum of first $n$ natural numbers. It is given as follows,
$\sum{n}=\dfrac{n\left( n+1 \right)}{2}$
The sum of squares of first $n$ natural numbers is given by
$\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
The sum of cubes of first $n$ natural numbers is given by
 $\sum{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$
Substituting these three formulae in equation $(i)$, we get the following,
 ${{\left( \dfrac{\sqrt{10}}{3}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right) \right)}^{2}}=\dfrac{n\left( n+1 \right)}{2}\times {{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$
Simplifying this equation, we get
$\dfrac{10}{9}\times \dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}}}{36}=\dfrac{n\left( n+1 \right)}{2}\times \dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}$
Cancelling ${{n}^{2}}{{\left( n+1 \right)}^{2}}$ from both sides of the equation, we get
$\dfrac{10}{9}\times \dfrac{{{\left( 2n+1 \right)}^{2}}}{36}=\dfrac{n\left( n+1 \right)}{2}\times \dfrac{1}{4}$
Expanding the square term on the LHS and shifting the 36 in the denominator to the RHS, we get
$\dfrac{10}{9}\times \left( 4{{n}^{2}}+4n+1 \right)=\dfrac{n\left( n+1 \right)}{2}\times \dfrac{1}{4}\times 36$
Now, we will simplify the RHS in the following manner,
$\dfrac{10}{9}\times \left( 4{{n}^{2}}+4n+1 \right)=\left( {{n}^{2}}+n \right)\times \dfrac{9}{2}$
Simplifying the above equation, we get
$20\times \left( 4{{n}^{2}}+4n+1 \right)=\left( {{n}^{2}}+n \right)\times 81$
$\begin{align}
  & \Rightarrow 80{{n}^{2}}+80n+20=81{{n}^{2}}+81n \\
 & \Rightarrow 81{{n}^{2}}+81n-80{{n}^{2}}-80n-20=0 \\
 & \therefore {{n}^{2}}+n-20=0 \\
\end{align}$
We can factorize the above quadratic equation in the following manner,
$\begin{align}
  & {{n}^{2}}+5n-4n-20=0 \\
 & \Rightarrow n\left( n+5 \right)-4\left( n+5 \right)=0 \\
 & \therefore \left( n+5 \right)\left( n-4 \right)=0 \\
\end{align}$
So, we have $n+5=0$ and $n-4=0$. From this, we get $n=-5$ and $n=4$. We can discard $n=-5$ since we assumed $n$ to be a natural number. Therefore, the value is $n=4$. Hence, the correct option is (b).

Note: In this type of question, we should be familiar with the formulae for the sum of the first $n$ natural numbers, the sum of their squares and cubes. It is essential that we understand the concept of a geometric progression since we used it to form an equation of the given terms. To find the roots of a quadratic equation, we can also use the quadratic formula.