Q.-- If sigma n, (root10)/3 sigma n² and sigma n³ are in G.P, then find the value of n.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given} \mathrm{that} \\ \sum _{}^{}\mathrm{n}, \frac{\sqrt{10}}{3}\sum _{}^{}{\mathrm{n}}^2, \sum _{}^{} {\mathrm{n}}^3 \mathrm{are} \mathrm{in} \mathrm{G}.\mathrm{P}. \\ \Rightarrow {\left\{\frac{\sqrt{10}}{3}\sum _{}^{}{\mathrm{n}}^2\right\}}^2=\left(\sum _{}^{}\mathrm{n}\right)\left(\sum _{}^{} {\mathrm{n}}^3\right) \\ \Rightarrow \frac{10}{9}{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\right\}}^2=\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^2 \\ \Rightarrow \frac{10}{9}\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2{\left(2\mathrm{n}+1\right)}^2}{36}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}.\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4} \\ \Rightarrow \frac{10}{9}\frac{{\left(2\mathrm{n}+1\right)}^2}{36}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}.\frac{1}{4} \\ \Rightarrow \frac{80}{9\times 36}{\left(2\mathrm{n}+1\right)}^2=\mathrm{n}\left(\mathrm{n}+1\right) \\ \Rightarrow \frac{20}{81}\left(4{\mathrm{n}}^2+1+4\mathrm{n}\right)={\mathrm{n}}^2+\mathrm{n} \\ \Rightarrow 80{\mathrm{n}}^2+20+80\mathrm{n}=81{\mathrm{n}}^2+81\mathrm{n} \\ \Rightarrow {\mathrm{n}}^2+\mathrm{n}-20=0 \\ \Rightarrow {\mathrm{n}}^2+5\mathrm{n}-4\mathrm{n}-20=0 \\ \Rightarrow \mathrm{n}\left(\mathrm{n}+5\right)-4\left(\mathrm{n}+4\right)=0 \\ \Rightarrow \left(\mathrm{n}-4\right)\left(\mathrm{n}+5\right)=0 \\ \mathrm{But} \mathrm{n} \mathrm{cannot} \mathrm{be} \mathrm{negative}, \mathrm{hence} \\ \mathbf{n}\mathbf{=}\mathbf{4}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards