Question

Which formula can be used to find the nth term of a geometric sequence where the fifth term is 1/16 and the common ratio is 1/4?

Answer 1

B is the answer

Explanation:

Answer 2

`\bf 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}\\ \cline{1-1} a_5=(1)/(16)\\ r=(1)/(4)\\ n=5 \end{cases}\implies \cfrac{1}{16}=a_1\left(\cfrac{1}{4} \right)^(5-1)`

`\bf \cfrac{1}{16}=a_1\left(\cfrac{1}{4} \right)^4\implies \cfrac{1}{16}=a_1\left( \cfrac{1^4}{4^4} \right)\implies \cfrac{1}{16}=a_1\cdot \cfrac{1}{256}\implies \cfrac{1}{16}=\cfrac{a_1}{256} \\\\\\ \cfrac{256}{16}=a_1\implies 16=a_1~\hfill \boxed{a_n=16\left( (1)/(4) \right)^(n-1)}`