Lanjutan Materi Barisan dan Deret

 E. Barisan Geometri

Perhatikan susunan bilangan-bilangan berikut 

$\begin{array}{r}1,{\displaystyle \frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\cdots }\end{array}$.

dengan rincian

$\begin{array}{r}\underset{\begin{array}{c}\downarrow \\ u_1\end{array}}{1},\underset{\begin{array}{c}\downarrow \\ u_2\end{array}}{{\displaystyle \frac{1}{2}}},\underset{\begin{array}{c}\downarrow \\ u_3\end{array}}{{\displaystyle \frac{1}{4}}},\underset{\begin{array}{c}\downarrow \\ u_4\end{array}}{{\displaystyle \frac{1}{8}}},\underset{\begin{array}{c}\downarrow \\ u_5\end{array}}{{\displaystyle \frac{1}{16}}},\cdots \end{array}$.

Dari pola di atas kita dapat tuliskan menjadi

$\begin{array}{r}1,{\displaystyle \frac{1}{2},\frac{1}{2}\times \frac{1}{2},\frac{1}{2}\times \frac{1}{4},\frac{1}{2}\times \frac{1}{8},\cdots }\end{array}$.

Dari pola yang tersusun di atas terdapat hal yang menarik yaitu:

$\begin{array}{rl}& {\displaystyle \frac{u_2}{u_1}=\frac{u_3}{u_2}=\frac{u_4}{u_3}=\cdots =\frac{u_n}{u_{n-1}}=\frac{1}{2}}\end{array}$.

Selanjutnya perhatikan

$\begin{array}{rl}& u_1=a=1\\ & u_2=u_1\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{2}\iff u_2=a.r}}\\ & u_3=u_2\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{2}\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{4}\iff u_3=a.r^2}}}}\\ & u_4=u_3\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{2}\times {\displaystyle \frac{1}{2}\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{8}\iff u_4=a.r^3}}}}}\\ & ...=...\\ & u_n=a.r^{n-1}\end{array}$.

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

$S_n=a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}$.

Untuk mencari besar $S_n$ adalah dengan mensiasatinya yaitu mengalikan  $r$  ke  $S_n$ sehingga menjadi bentuk

$r{S}_n=ar+a{r}^2+a{r}^3+a{r}^4+a{r}^5+\cdots +a{r}^n$.

Selanjutnya kita kondisikan sebagai berikut

$\begin{array}{rl}& S_n-r{S}_n\\ & =(a+ar+a{r}^2+a{r}^3+a{r}^4+...+a{r}^{n-2}+a{r}^{n-1})\\ & -(ar+a{r}^2+a{r}^3+a{r}^4+a{r}^5+...+a{r}^{n-1}+a{r}^n)\\ & (1-r)S_n=a-a{r}^n=a(1-r^n)\\ & S_n=\frac{a(1-r^n)}{1-r}\end{array}$.

Sebagai rangkuman dari materi barisan dan deret geometri ini, perhatikan tabel berikut

$\begin{array}{|cccc|}\hline \text{No}& \text{Barisan Geometri}& \text{Deret Geometri (Ukur)}& \text{Syarat}\\ 1& \begin{array}{rl}& U_1,U_2,U_3,U_4,...\\ & \\ & \text{Selanjutnya}\\ & U_1,U_2,U_3,\cdots \\ & \text{disebut suku-suku}\\ & \text{dan}{U}_1=a=\\ & \text{suku pertama}\end{array}& \begin{array}{rl}& U_1+U_2+U_3+U_4+...\\ & \\ & \text{Selanjutnya}\\ & U_1,U_2,U_3,\cdots \\ & \text{disebut suku-suku}\\ & \text{dan}{U}_1=a=\\ & \text{suku pertama}\end{array}& \begin{array}{rl}\text{Rasio}=r& ={\displaystyle \frac{U_2}{U_1}}\\ & ={\displaystyle \frac{U_3}{U_2}}\\ & ={\displaystyle \frac{U_4}{U_3}}\\ & =\cdots \\ & ={\displaystyle \frac{U_n}{U_{(n-1)}}}\end{array}\\ 2& U_n=a.r^{(n-1)}& U_n=a.r^{(n-1)}& \begin{array}{rl}{U}_t& ={\displaystyle \sqrt{a.U_n}}\\ & =\text{Suku tengah}\end{array}\\ \hline\end{array}$.

$\begin{array}{|cccc|}\hline \text{No}& \begin{array}{rl}& \text{Barisan}\\ & \text{Geometri}\end{array}& \begin{array}{rl}& \text{Deret }\\ & \text{Geometri (Ukur)}\end{array}& \text{Syarat}\\ 3& & \begin{array}{rl}{S}_n& ={\displaystyle \frac{a(r^n-1)}{r-1}}\\ & \mathbf{\text{atau}}\\ S_n& ={\displaystyle \frac{a(1-r^n)}{1-r}}\end{array}& \begin{array}{rl}& \text{sisipan}k\text{bilangan}\\ & \text{misal}\\ & U_1\cdots \cdots \cdots U_m\\ & \text{ingin disisipkan}k\text{bilangan}\\ & \text{Rasio baru}=r^{\prime }={\displaystyle \sqrt[k+1]{{\displaystyle \frac{U_m}{U_1}}}}\end{array}\\ \hline\end{array}$.

$\begin{array}{|cccl|}\hline \text{No}& \begin{array}{rl}& \text{Barisan}\\ & \text{Geometri}\end{array}& \begin{array}{rl}& \text{Deret }\\ & \text{Geometri (Ukur)}\end{array}& \text{Syarat}\\ & \text{Deret tak Hingga}& \text{Deret tak Hingga}& \text{Hubungan}\\ 4& \text{Konvergen}& \text{Divergen}& \text{suku dan jumlah}\\ & S_{\mathrm{\infty }}={\displaystyle \frac{a}{1-r},\left|r\right|<1}& r\le -1\text{atau}r\ge 1& \begin{array}{rl}{U}_1& =S_1=a\\ U_n& =S_n-S_{(n-1)}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan suku ke}-12\text{dari barisan}\\ & \text{berikut}\\ & 4,1,{\displaystyle \frac{1}{4},\frac{1}{16},\cdots }\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \{\begin{array}{ll}{u}_1& =a=4\\ r& ={\displaystyle \frac{u_2}{u_1}=\frac{u_3}{u_2}=\cdots =\frac{1}{4}}\end{array}\\ & \text{Untuk mencari suku ke}-12,\text{maka}\\ & \begin{array}{rl}{u}_{12}& =a.r^{(12-1)}=a{r}^{11}\\ & =4.{\left({\displaystyle \frac{1}{4}}\right)}^{11}\\ & =4^1{.4}^{-11}=4^{1-11}=4^{-10}\\ & ={\displaystyle \frac{1}{4^{10}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan jumlah 12 suku pertama}\\ & \text{dari}\\ & 4+1+{\displaystyle \frac{1}{4}+\frac{1}{16}+\cdots +\frac{1}{4^{10}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}{S}_n& ={\displaystyle \frac{a(1-r^n)}{1-r}}\\ S_{12}& ={\displaystyle \frac{4(1-{\left({\displaystyle \frac{1}{4}}\right)}^{12})}{1-{\displaystyle \frac{1}{4}}}}\\ & ={\displaystyle \frac{4(1-{\left({\displaystyle \frac{1}{4}}\right)}^{12})}{{\displaystyle \frac{3}{4}}}}\\ & ={\displaystyle \frac{16}{3}(1-{\left({\displaystyle \frac{1}{4}}\right)}^{12})}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Suatu deret geometri dengan jumlah}\\ & S_n={3.2}^n-1,\text{maka suku ke}-2022\\ & \text{dari deret tersebut adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}{S}_n={3.2}^n-1,\text{maka}\\ & u_{2022}=S_{2022}-S_{2021}\\ & \text{Sehingga}\\ & \begin{array}{rl}{u}_{2022}& =S_{2022}-S_{2021}\\ & =({3.2}^{2022}-1)-({3.2}^{2021}-1)\\ & ={3.2}^{2022}-{3.2}^{2021}={3.2}^{2021}(2-1)\\ & ={2.3}^{2021}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui deret geometri dengan}{\displaystyle \frac{u_4}{u_6}=k}\\ & \text{dan}{u}_2\times u_8={\displaystyle \frac{1}{k},\text{maka suku pertama}}\\ & \text{deret geometri ini adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \{\begin{array}{ll}& {\displaystyle \frac{u_4}{u_6}=k}\\ & u_2\times u_8={\displaystyle \frac{1}{k}}\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}& {\displaystyle \frac{u_6}{u_4}=\frac{a{r}^5}{a{r}^3}=r^2=\frac{1}{k}\text{dan}}\\ & u_2\times u_8\\ & =ar\times a{r}^7=a^2{r}^8={\left(a{r}^4\right)}^2=\frac{1}{k}\\ & \iff {\left(u_5\right)}^2={\displaystyle \frac{1}{k}\iff u_5=\sqrt{\frac{1}{k}}}\\ & \text{sehingga}\\ & u_5=a{r}^4=a{\left(r^2\right)}^2\iff \sqrt{\frac{1}{k}}=a{\left(\frac{1}{k}\right)}^2\\ & \begin{array}{rl}a& =k^2\times \sqrt{{\displaystyle \frac{1}{k}}}\\ & =k\times k\times k^{-\frac{1}{2}}\\ & =k\times k^{{}^{{.}^{\frac{1}{2}}}}=k\sqrt{k}\end{array}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Waji, J., Linggih, S., Syahrudin,Y.R. 1981. Ringkasan Materi IPA. Bandung: GANECA EXACT.