设等比数列\(\left\{ {{a_n}} \right\}\)的公比 \(q = \frac{1}{2}\) ，前 \(n\) 项和为 \({S_n}\) ，则 \(\frac {{S}_{4}} {{a}_{4}}=\) （       ）

等比数列中 \(q = \frac{1}{2}\) ，设首项是 \({a_1}\) ，

根据通项公式和前 \(n\) 项和公式得 \({S_4} = \frac{{{a_1}\left[ {1 - {{\left( {\frac{1}{2}} \right)}^4}} \right]}}{{1 - \frac{1}{2}}} = \frac{{15}}{8}{a_1}\) ，

\({a_4} = {a_1}{\left( {\frac{1}{2}} \right)^3} = \frac{1}{8}{a_1}\) ，

\(\therefore \frac{{{S_4}}}{{{a_4}}} = \frac{{\frac{{15{a_1}}}{8}}}{{\frac{{{a_1}}}{8}}} = 15\)  

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