If $\sum n, \frac{\sqrt{10}}{3}\sum n^2, \sum n^3$ are in geometric progression, then find the value of n?

Question

If $\sum n, \frac{\sqrt{10}}{3}\sum n^2, \sum n^3$ are in geometric progression, then find the value of n?

I had initially typed the question incorrectly which was the sole reason why I couldn't solve it. Have corrected it now.

The options given for 'n'are: 3,4,2,6

Answer 1

I suppose that the problem is to find $n$ such that $$a_1=\sqrt{\sum_{i=1}^n i}\qquad a_2=\frac {10}{\sqrt 3}\qquad a_3=\sqrt{\sum_{i=1}^n i^2}\qquad a_4=\sqrt{\sum_{i=1}^n i^3}$$ be in geometric progression.

Considering $$\frac{a_2}{a_1}=\frac{a_4}{a_3}\implies \frac{10 \sqrt{\frac{2}{3}}}{\sqrt{n (n+1)}}=\frac{\sqrt{\frac{3}{2}} n (n+1)}{\sqrt{n (n+1) (2 n+1)}}$$ that is to say $${10 \sqrt{\frac{2}{3}}}=\sqrt{\frac{3}{2}}\frac{n(n+1)}{\sqrt{2n+1} }$$ which is effectively satisfied for $n=4$.

Answer 2

Use the notation $\displaystyle\sigma_m=\sum_{r=1}^n r^m$.

Given the following GP: $$\begin{align} a_1&=\sigma_1\tag{1}\\ a_2&=\frac{\sqrt{10}}3\sigma_2=\frac{\sqrt{10}}3\bigg(\frac {2n+1}3\sigma_1\bigg)\tag{2}\\ a_3&=\sigma_3=\sigma_1^2\tag{3}\\ r^2=\frac {(3)}{(1)}=\bigg(\frac {(2)}{(1)}\bigg)^2:\\ r^2=\quad\sigma_1=\frac {n(n+1)}2&=\frac {10}{81}(2n+1)^2\\ 81n(n+1)&=20(4n^2+4n+1)\\ n^2+n-20&=0\\ (n+5)(n-4)&=0\\ \color{red}n&\color{red}{=4} \end{align}$$

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The GP is $$\begin{align} a_1&=\qquad \;\; 1\; +2\; +3\; +4&&=10\\ a_2&=\frac {\sqrt{10}}3\big(1^2+2^2+3^2+4^2\big)=\frac{\sqrt{10}}3 (30)&&=10\sqrt{10}\\ a_3&=\qquad\ 1^3+2^3+3^3+4^3&&=100 \end{align}$$ with common ratio $r=\sqrt{10}$.