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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

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  • $\begingroup$ a $\Sigma$ is missing $\endgroup$ – Dr. Sonnhard Graubner Jul 1 '17 at 17:49
  • $\begingroup$ Well, exactly this question has been asked in my textbook. $\endgroup$ – Siddharth Garg Jul 1 '17 at 17:51
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    $\begingroup$ what does your summation sum over? Is the first term $\sum_{i=1}^n i$ or is it $\sum_{i=1}^n x_i$ where $x_i$ is something that you didn't type? Remark: your second term is irrational. $\endgroup$ – Siong Thye Goh Jul 1 '17 at 18:07
  • $\begingroup$ The summation means sum till n terms, like 1+2+3+4+...+n $\endgroup$ – Siddharth Garg Jul 1 '17 at 18:09
  • $\begingroup$ math.meta.stackexchange.com/questions/9959/… Please format the question properly, and do not type part of the question in the title without typing it again in the body. As pointed out already, there probably is a mistake in the question as it currently is, given that the 1st, 3rd and 4th terms are integers, while the 2nd is irrational. $\endgroup$ – Shuri2060 Jul 1 '17 at 18:17
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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 $n=4$.

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  • $\begingroup$ Why take square root of sums? $\endgroup$ – hypergeometric Jul 17 '17 at 17:05
  • $\begingroup$ @hypergeometric. My answer was for the first version of the question (which changed). $\endgroup$ – Claude Leibovici Jul 18 '17 at 2:38
  • $\begingroup$ I see. Interestingly the answer for $n$ is the same for both cases! $\endgroup$ – hypergeometric Jul 18 '17 at 15:06
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Use the notation $\displaystyle\sigma_m=\sum_{r=1}^n r^m$.

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


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 $r=\sqrt{10}$.

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