Question

Help me pleaassee its due this tuesday

Answer 1

we can always find the common ratio in a geometric sequence by simply dividing the latter value by the previous value.

`\cfrac{60}{80}\implies \cfrac{3}{4}\hspace{5em}\cfrac{45}{60}\implies \cfrac{3}{4} \\\\[-0.35em] ~\dotfill\\\\ 80~~,~~\stackrel{ 80\cdot (3)/(4) }{60}~~,~~\stackrel{ 60\cdot (3)/(4) }{45}~~,~~... \hspace{5em}\stackrel{\textit{common ratio}}{\cfrac{3}{4}} \\\\[-0.35em] \rule{34em}{0.25pt}`

`n^(th)\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^(n-1)\qquad \begin{cases} a_n=n^(th)\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ a_1=80\\ n=9\\ r=(3)/(4) \end{cases}\implies a_9=80\cdot \left( \cfrac{3}{4} \right)^(9-1) \\\\\\ a_9=80\cdot \left( \cfrac{3}{4} \right)^8\implies a_9=80\cdot \cfrac{6561}{65536}\implies a_9=\cfrac{32805}{4096}\implies a_9\approx 8.01`