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Question: If the terms \(\sum{n}\), \(\dfrac{\sqrt{10}}{3}\sum{{{n}^{2}}}\), \(\sum{{{n}^{3}}}\) are in G.P., ...

If the terms ∑n\sum{n}, 103∑n2\dfrac{\sqrt{10}}{3}\sum{{{n}^{2}}}, ∑n3\sum{{{n}^{3}}} are in G.P., then the value of nn is:
(a) 3
(b) 4
(c) 2
(d) Does not exist

Explanation

Solution

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 nn 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 nn. We will find the roots of this equation to obtain a value of nn.

Complete step by step solution:
We know that if three numbers aa, bb, cc are in G.P., then we have ab=bc\dfrac{a}{b}=\dfrac{b}{c}. This implies that b2=a×c{{b}^{2}}=a\times c. Using this fact for the given geometric progression, we get the following,
[103∑n2]2=[∑n]×[∑n3]....(i){{\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 nn natural numbers. It is given as follows,
∑n=n(n+1)2\sum{n}=\dfrac{n\left( n+1 \right)}{2}
The sum of squares of first nn natural numbers is given by
∑n2=n(n+1)(2n+1)6\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}
The sum of cubes of first nn natural numbers is given by
∑n3=(n(n+1)2)2\sum{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}
Substituting these three formulae in equation (i)(i), we get the following,
(103(n(n+1)(2n+1)6))2=n(n+1)2×(n(n+1)2)2{{\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
109×n2(n+1)2(2n+1)236=n(n+1)2×n2(n+1)24\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 n2(n+1)2{{n}^{2}}{{\left( n+1 \right)}^{2}} from both sides of the equation, we get
109×(2n+1)236=n(n+1)2×14\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
109×(4n2+4n+1)=n(n+1)2×14×36\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,
109×(4n2+4n+1)=(n2+n)×92\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×(4n2+4n+1)=(n2+n)×8120\times \left( 4{{n}^{2}}+4n+1 \right)=\left( {{n}^{2}}+n \right)\times 81
⇒80n2+80n+20=81n2+81n ⇒81n2+81n−80n2−80n−20=0 ∴n2+n−20=0 \begin{aligned} & \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{aligned}
We can factorize the above quadratic equation in the following manner,
n2+5n−4n−20=0 ⇒n(n+5)−4(n+5)=0 ∴(n+5)(n−4)=0 \begin{aligned} & {{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{aligned}
So, we have n+5=0n+5=0 and n−4=0n-4=0. From this, we get n=−5n=-5 and n=4n=4. We can discard n=−5n=-5 since we assumed nn to be a natural number. Therefore, the value is n=4n=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 nn 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.