Question

A geometric sequence with a fifth term of 1/3 and a contant ratio of 1/3

Answer 1

`\bf n^(th)\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^(n-1)\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=(1)/(3)\\ n=5\\ a_5=(1)/(3) \end{cases}\implies a_5=a_1\cdot \left( (1)/(3) \right)^(5-1)`


`\bf \cfrac{1}{3}=a_1\cdot \left( (1)/(3) \right)^(5-1)\implies \cfrac{1}{3}=a_1\cdot \cfrac{1^4}{3^4}\implies \cfrac{1}{3}=\cfrac{a_1}{81}\implies \cfrac{81}{3}=a_1 \\\\\\ 27=a_1\qquad \qquad thus\qquad \qquad \boxed{a_n=27 \left( (1)/(3) \right)^(n-1)}`