E. Barisan Geometri
Perhatikan susunan bilangan-bilangan berikut
$\begin{aligned}1,\displaystyle \frac{1}{2},\: \frac{1}{4},\: \frac{1}{8},\: \frac{1}{16},\: \cdots \end{aligned}$.
dengan rincian
$\begin{aligned}\underset{\begin{matrix}\downarrow\\ u_{1} \end{matrix}}{1},\underset{\begin{matrix}\downarrow\\ u_{2} \end{matrix}}{\displaystyle \frac{1}{2}},\: \underset{\begin{matrix}\downarrow\\ u_{3} \end{matrix}}{\displaystyle \frac{1}{4}},\: \underset{\begin{matrix}\downarrow\\ u_{4} \end{matrix}}{\displaystyle \frac{1}{8}},\: \underset{\begin{matrix}\downarrow\\ u_{5} \end{matrix}}{\displaystyle \frac{1}{16}},\: \cdots \end{aligned}$.
Dari pola di atas kita dapat tuliskan menjadi
$\begin{aligned}1,\displaystyle \frac{1}{2},\: \frac{1}{2}\times \frac{1}{2},\: \frac{1}{2}\times \frac{1}{4},\: \frac{1}{2}\times \frac{1}{8},\: \cdots \end{aligned}$.
Dari pola yang tersusun di atas terdapat hal yang menarik yaitu:
$\begin{aligned}&\displaystyle \frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}=\frac{u_{4}}{u_{3}}=\cdots =\frac{u_{n}}{u_{n-1}}=\color{red}\frac{1}{2} \end{aligned}$.
Selanjutnya perhatikan
$\begin{aligned}&u_{1}=a=1\\ &u_{2}=u_{1}\times \displaystyle \frac{1}{2}=1\times \displaystyle \frac{1}{2}\: \Leftrightarrow \: u_{2}=a.r\\ &u_{3}=u_{2}\times \displaystyle \frac{1}{2}=1\times \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}=1\times \displaystyle \frac{1}{4}\: \Leftrightarrow \: u_{3}=a.r^{2}\\ &u_{4}=u_{3}\times\displaystyle \frac{1}{2}=1\times \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}\times \displaystyle \frac{1}{2}=1\times \displaystyle \frac{1}{8}\: \Leftrightarrow \: u_{4}=a.r^{3} \\ &...=...\\ &u_{n}=a.r^{n-1} \end{aligned}$.
Selanjutnya pembanding yang selalu tetap dinamakan rasio atau disingkat dengan huruf $r$.
F. Deret Geometri
Perhatikan bahwa pada barisan suku-suku barisan geometri jika dijumlahkan akn terbentuk deret geometri atau deret ukur
Misalkan
$S_{n}=a+ar+ar^{2}+ar^{3}+ar^{4}+\cdots +ar^{n-1}$.
Untuk mencari besar $S_{n}$ adalah dengan mensiasatinya yaitu mengalikan $r$ ke $S_{n}$ sehingga menjadi bentuk
$rS_{n}=ar+ar^{2}+ar^{3}+ar^{4}+ar^{5}+\cdots +ar^{n}$.
Selanjutnya kita kondisikan sebagai berikut
$\begin{aligned}&\color{red}S_{n}-rS_{n}\\ &=\left ( a+ar+ar^{2}+ar^{3}+ar^{4}+...+ar^{n-2}+ar^{n-1} \right )\\ &\quad-\left ( ar+ar^{2}+ar^{3}+ar^{4}+ar^{5}+...+ar^{n-1}+ar^{n} \right )\\ &\color{red}\left ( 1-r \right )S_{n}\color{black}=a-ar^{n}=a\left ( 1-r^{n} \right )\\ &\color{red}S_{n}\color{black}=\frac{a\left ( 1-r^{n} \right )}{1-r} \end{aligned}$.
Sebagai rangkuman dari materi barisan dan deret geometri ini, perhatikan tabel berikut
$\begin{array}{|c|c|c|c|}\hline \textrm{No}&\textrm{Barisan Geometri}&\textrm{Deret Geometri (Ukur)}&\textrm{Syarat}\\\hline 1&\begin{aligned}&U_{1},U_{2},U_{3},U_{4},...\\ &\\ &\textrm{Selanjutnya}\\ &U_{1},U_{2},U_{3},\cdots \\ &\textrm{disebut suku-suku}\\ &\textrm{dan}\: \: U_{1}=a=\\ &\textrm{suku pertama} \end{aligned}&\begin{aligned}&U_{1}+U_{2}+U_{3}+U_{4}+...\\ &\\ &\textrm{Selanjutnya}\\ &U_{1},U_{2},U_{3},\cdots \\ &\textrm{disebut suku-suku}\\ &\textrm{dan}\: \: U_{1}=a=\\ &\textrm{suku pertama} \end{aligned} &\begin{aligned}\textrm{Rasio}=r&=\displaystyle \frac{U_{2}}{U_{1}}\\ &=\displaystyle \frac{U_{3}}{U_{2}}\\ &=\displaystyle \frac{U_{4}}{U_{3}}\\ &=\cdots \\ &=\displaystyle \frac{U_{n}}{U_{(n-1)}} \end{aligned}\\\hline 2&U_{n}=a.r^{(n-1)}&U_{n}=a.r^{(n-1)}&\begin{aligned}U_{t}&=\displaystyle \sqrt{a.U_{n}}\\ &=\textrm{Suku tengah} \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|c|c|c|}\hline \textrm{No}&\begin{aligned}&\textrm{Barisan}\\ &\textrm{Geometri} \end{aligned}&\begin{aligned}&\textrm{Deret }\\ &\textrm{Geometri (Ukur)} \end{aligned}&\textrm{Syarat}\\\hline 3&&\begin{aligned}S_{n}&=\displaystyle \frac{a(r^{n}-1)}{r-1}\\ &\textbf{atau}\\ S_{n}&=\displaystyle \frac{a(1-r^{n})}{1-r} \end{aligned}&\begin{aligned}&\textrm{sisipan}\: k\: \textrm{bilangan}\\ &\textrm{misal}\\ &U_{1}\cdots \cdots \cdots U_{m}\\ &\textrm{ingin disisipkan}\: k\: \textrm{bilangan}\\ &\textrm{Rasio baru}=r'=\displaystyle \sqrt[k+1]{\displaystyle \frac{U_{m}}{U_{1}}} \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|c|c|l|}\hline \textrm{No}&\begin{aligned}&\textrm{Barisan}\\ &\textrm{Geometri} \end{aligned}&\begin{aligned}&\textrm{Deret }\\ &\textrm{Geometri (Ukur)} \end{aligned}&\qquad\textrm{Syarat}\\\hline &\textrm{Deret tak Hingga}&\textrm{Deret tak Hingga}&\textrm{Hubungan}\\ 4&\textrm{Konvergen}&\textrm{Divergen}&\textrm{suku dan jumlah}\\ &S_{\infty }=\displaystyle \frac{a}{1-r},\: \left | r \right |< 1&r\leq -1\: \textrm{atau}\: r\geq 1&\begin{aligned}U_{1}&=S_{1}=a\\ U_{n}&=S_{n}-S_{(n-1)} \end{aligned}\\\hline \end{array}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukan suku ke}-12\: \: \textrm{dari barisan} \\ &\textrm{berikut}\\ &\qquad \qquad 4,\: 1,\: \displaystyle \frac{1}{4},\: \frac{1}{16},\: \cdots \\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\begin{cases} u_{1} & =a=4 \\ r & =\displaystyle \frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}=\cdots =\frac{1}{4} \end{cases}\\ &\textrm{Untuk mencari suku ke}-12,\: \textrm{maka}\\ &\begin{aligned}u_{12}&=a.r^{(12-1)}=ar^{11}\\ &=4.\left ( \displaystyle \frac{1}{4} \right )^{11}\\ &=4^{1}.4^{-11}=4^{1-11}=4^{-10}\\ &=\color{red}\displaystyle \frac{1}{4^{10}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukan jumlah 12 suku pertama} \\ &\textrm{dari}\\ &\qquad \qquad 4+\: 1+\: \displaystyle \frac{1}{4}+\: \frac{1}{16}+\: \cdots +\frac{1}{4^{10}}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}S_{n}&=\displaystyle \frac{a(1-r^{n})}{1-r}\\ S_{12}&=\displaystyle \frac{4\left ( 1-\left (\displaystyle \frac{1}{4} \right )^{12} \right )}{1-\displaystyle \frac{1}{4}}\\ &=\displaystyle \frac{4\left ( 1-\left (\displaystyle \frac{1}{4} \right )^{12} \right )}{\displaystyle \frac{3}{4}}\\ &=\color{red}\displaystyle \frac{16}{3}\left ( 1-\left ( \displaystyle \frac{1}{4} \right )^{12} \right ) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Suatu deret geometri dengan jumlah} \\ &S_{n}=3.2^{n}-1,\: \textrm{maka suku ke}-2022\\ &\textrm{dari deret tersebut adalah}\: ...\: .\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: S_{n}=3.2^{n}-1,\: \: \textrm{maka}\\ &u_{2022}=S_{2022}-S_{2021}\\ &\textrm{Sehingga}\\ &\begin{aligned}u_{2022}&=S_{2022}-S_{2021}\\ &=\left (3.2^{2022}-1 \right )-\left ( 3.2^{2021}-1 \right )\\ &=3.2^{2022}-3.2^{2021}=3.2^{2021}(2-1)\\ &=\color{red}2.3^{2021} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Diketahui deret geometri dengan}\: \: \displaystyle \frac{u_{4}}{u_{6}}=k \\ &\textrm{dan}\: \: u_{2}\times u_{8}=\displaystyle \frac{1}{k},\: \: \textrm{maka suku pertama}\\ &\textrm{deret geometri ini adalah}\: ...\: .\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\begin{cases} &\displaystyle \frac{u_{4}}{u_{6}}=k \\ &u_{2}\times u_{8}=\displaystyle \frac{1}{k} \end{cases}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}&\displaystyle \frac{u_{6}}{u_{4}}=\frac{ar^{5}}{ar^{3}}=r^{2}=\frac{1}{k}\: \: \textrm{dan}\\ &u_{2}\times u_{8}\\ &=ar\times ar^{7}=a^{2}r^{8}=\left ( ar^{4} \right )^{2}=\frac{1}{k}\\ &\Leftrightarrow \left (u_{5} \right )^{2}=\displaystyle \frac{1}{k}\Leftrightarrow u_{5}=\sqrt{\frac{1}{k}}\\ &\textrm{sehingga}\\ &u_{5}=ar^{4}=a\left ( r^{2} \right )^{2}\Leftrightarrow \: \sqrt{\frac{1}{k}}=a\left ( \frac{1}{k} \right )^{2}\\ &\begin{aligned}a&=k^{2}\times \sqrt{\displaystyle \frac{1}{k}}\\ &=k\times k\times k^{-\frac{1}{2}}\\ &=k\times k^{^{.^{\frac{1}{2}}}}=\color{red}k\sqrt{k} \end{aligned} \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Waji, J., Linggih, S., Syahrudin,Y.R. 1981. Ringkasan Materi IPA. Bandung: GANECA EXACT.
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