# Is the system is a geometric progression? I am trying to show that the following system is a geomet

Is the system is a geometric progression?
I am trying to show that the following system is a geometric progression i.e. ${a}_{2}={a}_{1}^{2}$, ${a}_{3}={a}_{1}^{3}$ etc.
${a}_{k}^{2}={a}_{k-1}{a}_{k+1}$
${a}_{k-1}^{2}={a}_{k-2}{a}_{k}$
${a}_{2}^{2}={a}_{1}{a}_{3}$
here ${a}_{k}>...>{a}_{2}>{a}_{1}$ is assumed.
Is there sufficient information to conclude that ${a}_{i}={a}_{1}^{i}$ in general? I am seeing that I would probably need to assume ${a}_{2}={a}_{1}^{2}$. Could the geometric progression be concluded otherwise?
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Step 1
$\frac{{a}_{k+1}}{{a}_{k}}=\frac{{a}_{k}}{{a}_{k-1}}=\cdots =\frac{{a}_{3}}{{a}_{2}}=\frac{{a}_{2}}{{a}_{1}}$ so $\left\{{a}_{k}\right\}$ forms a geometric sequence. let the common ratio be r.
Then the sequence is $\left\{{a}_{1},{a}_{1}r,...\right\}$
if $\left({a}_{1},r\right)=\left(1,2\right)$, then ${a}_{i}\ne {a}_{1}^{i}$.
if we assume ${a}_{2}={a}_{1}^{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}r={a}_{1}$
Step 2
Then the sequence is $\left\{{a}_{1},{a}_{1}^{2},{a}_{1}^{3},...\right\}$
i have excluded the case ${a}_{1}=0$, otherwise the sequence is constantly 0.