Prove a sequence is geometric

Question

Suppose I am given that the sum of the first $2n$ ($n$ is a positive integer) terms of a sequence $u_{1},u_{2},...$ is given by $\frac{3}{10}-\frac{1}{10(3)^{2n-1}}$ and I need to show that the sequence is geometric.

My question : Is it possible to recover the sum of $n$ terms by replacing the $2n$ by $n$? Because if yes, then I can work out my common ratio from there.

Answer 1

hint:$$s_{2n}=\frac{3}{10}-\frac{1}{10(3)^{2n-1}}\\s_{2n}=\frac{u1_(1-q^{2n})}{1-q}\\ \to\begin{cases}\frac{u1}{1-q}=\frac{3}{10}\\\frac{-u1q^{2n}}{1-q} =-\frac{1}{10(3)^{2n-1}}\end{cases}\\\div$$ $$\frac{\frac{-u1q^{2n}}{1-q}}{\frac{u1}{1-q}}=\frac{-\frac{1}{10(3)^{2n-1}}}{\frac{3}{10}}\to -q^{2n}=-\frac{1}{3^{2n}}\\q=\frac{1}{3}\text{ other possible -1/3} $$suppose $q=\frac{1}{3}$so $$\frac{u_1}{1-q}=\frac {3}{10}\to u_1=\frac{1}{5}$$

Answer 2

Is it possible to recover the sum of $n$ terms by replacing the $2n$ by $n$?

No. Consider for example that the sequences $\,u_k\,$ and $\,u_k+(-1)^k\,$ have the same sum for the first $2n$ terms for any $\forall n$, but they can't both be geometric.