Saya merupakan lulusan Institut Teknologi Bandung jurusan Matematika. Saat ini saya bekerja di salah satu Perusahaan Asuransi Umum pada bagian Digital Development.

Diketahui	\({a_{\left. {\overline {\, n \,}}\! \right\| }}\) = 10,00 ; dan \({a_{\left. {\overline {\, {3n} \,}}\! \right\| }}\) = 24,40.
Rumus yang digunakan	\({a_{\left. {\overline {\, n \,}}\! \right\| }} = \frac{{1 – {v^n}}}{i}\)
Proses pengerjaan	\({a_{\left. {\overline {\, {3n} \,}}\! \right\| }} = \frac{{1 – {v^{3n}}}}{i} = 24,40\) \({v^{3n}} = 1 – 24,4i\) \({a_{\left. {\overline {\, n \,}}\! \right\| }} = \frac{{1 – {v^n}}}{i} = 10\) \({v^n} = 1 – 10i\) \(\frac{{{a_{\left. {\overline {\, {3n} \,}}\! \right\| }}}}{{{a_{\left. {\overline {\, n \,}}\! \right\| }}}} = \frac{{\frac{{1 – {v^{3n}}}}{i}}}{{\frac{{1 – {v^n}}}{i}}} =\frac{{1 – {v^{3n}}}}{{1 – {v^n}}} = \frac{{24,40}}{{10}} = 2,44\) \(1 – {v^{3n}} = 2,44(1 – {v^n})\) \(1 – {v^{3n}} = 2,44 – 2,44{v^n}\) \({v^{3n}} – 2,44{v^n} + 1,44 = 0\) Misalkan \(x = {v^n}\) \({x^3} – 2,44x + 1,44 = 0\) Nilai x yang memenuhi adalah \(x = 1,{\rm{ }}x = – 1,8,{\rm{ }}x = 0,8\) Karena \(x = 1\) dan \(x = – 1,8\) tidak memenuhi maka \(x = 0,8\) \({v^n} = 0,8\) \({v^{3n}} = 0,512\) \({a_{\left. {\overline {\, {4n} \,}}\! \right\| }} = {a_{\left. {\overline {\, {3n} \,}}\! \right\| }} + {v^{3n}}{a_{\left. {\overline {\, n \,}}\! \right\| }} = 24,40 + 0,512(10) = 29,52\)
Jawaban	d. 29,52