Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Jika diketahui :
(i) \({a_{\left. {\overline {\, n \,}}\! \right| }}\) = 10,00 ; dan
(ii) \({a_{\left. {\overline {\, {3n} \,}}\! \right| }}\) = 24,40.
Hitunglah nilai dari \({a_{\left. {\overline {\, {4n} \,}}\! \right| }}\)?
- 28,74
- 29,00
- 29,26
- 29,52
- 29,78
| Diketahui |
|
| 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 |