MATEMATIKA PEMINATAN
Kelas X
1. Fungsi eksponen
2. Fungsi logaritma
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Kelas XI
1. Persamaan trigonometri
2. Rumus jumlah dan selisih pada trigonometri
Lihat di sini lebih dari 100 contoh soal dan pembahasan
Kelas XII
1. Limit fungsi trigonometri
2. Turunan fungsi trigonometri
Lihat di sini lebih dari 90 contoh soal dan pembahasan
$\begin{array}{ll}\\ 31.&2+4+6+8+\cdots =\: ....\:.\\ &\textrm{A}.\quad \color{red}\displaystyle \sum_{k=1}^{n}2k\\ &\textrm{B}.\quad \displaystyle \sum_{k=1}^{n}2^{k}\\ &\textrm{C}.\quad \displaystyle \sum_{k=1}^{n}2k-1\\ &\textrm{D}.\quad \displaystyle \sum_{k=1}^{n}(2-k)\\ &\textrm{E}.\quad \displaystyle \sum_{k=1}^{n}k\\\\ &\textbf{Jawab}:\\ &\textrm{Cukup Jelas bahwa}\\ &\begin{aligned}2+4+6+8+\cdots \: \: \quad&\\ 2(1+2+3+4+\cdots )&=2\displaystyle \sum_{k=1}^{n}k=\displaystyle \sum_{k=1}^{n}\color{red}2k \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 32.&\textbf{(UN 2005)}\\ &\textrm{Seorang anak menabung di suatu bank dengan}\\ &\textrm{selisih kenaikan tabungan antarbulan tetap}\\ &\textrm{Pada bulan pertama sebesar Rp50.000,00},\\ &\textrm{bulan kedua Rp55.000,00, bulan ketiga}\\ &\textrm{Rp60.000,00, dan demikian seterusnya}.\\ &\textrm{Besar tabungan anak tersebut selama }\\ &\textrm{dua tahun adalah}\: ....\:.\\ &\textrm{A}.\quad \textrm{Rp1.315.000,00}\\ &\textrm{B}.\quad \textrm{Rp1.320.000,00}\\ &\textrm{C}.\quad \textrm{Rp2.040.000,00}\\ &\textrm{D}.\quad \color{red}\textrm{Rp2.580.000,00}\\ &\textrm{E}.\quad \textrm{Rp2.640.000,00}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui deret aritmetika dengan}\\ &\begin{aligned}&\bullet \quad a=U_{1}=\textrm{Rp}50.000,00\\ &\bullet \quad U_{2}=\textrm{Rp}55.000,00\\ &\bullet \quad b=U_{2}-U_{1}=\textrm{Rp}5.000,00\\ &\textrm{Ditanya: Besar tabungan selama 2 tahun}\\ &\begin{aligned}\color{blue}S_{n}&=\color{blue}\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ \textrm{Ka}&\textrm{rena 2 tahun = 24 bulan, maka}\\ S_{24}&=\displaystyle \frac{24}{2}(2\times 50.000+(24-1)\times 5.000)\\ &=12\left ( 100.000+115.000 \right )=\color{red}2.580.000 \end{aligned} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 33.&\textbf{(UN 2006)}\\ &\textrm{Sebuah bola jatuh dari ketinggian 10 meter dan}\\ &\textrm{memantul kembali dengan ketinggian}\: \: \displaystyle \frac{3}{4}\: \: \textrm{dari}\\ &\textrm{tinggi semula dan begitu seterusnya hingga bola}\\ &\textrm{berhenti. Jumlah seluruh lintasan bola adalah}\: ....\:.\\ &\textrm{A}.\quad 65\: \: \textrm{meter}\\ &\textrm{B}.\quad \color{red}70\: \: \textrm{meter}\\ &\textrm{C}.\quad 75\: \: \textrm{meter}\\ &\textrm{D}.\quad 77\: \: \textrm{meter}\\ &\textrm{E}.\quad 80\: \: \textrm{meter}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{array}$.
$.\qquad\begin{aligned}&\textrm{Soal terkait dengan deret geometri tak hingga}\\ &\textrm{Yaitu}:\: S_{\infty }=\displaystyle \frac{a}{1-r},\: \: \textrm{dengan}\: \: \left | r \right |<1\\ &\color{blue}\begin{aligned}S&=10+2.\displaystyle \frac{3}{4}.10+2.\frac{3}{4}.\frac{3}{4}.10+2.\frac{3}{4}.\frac{3}{4}.\frac{3}{4}.10+...\\ &=10+20.\displaystyle \frac{3}{4}+20.\left (\frac{3}{4} \right )^{2}+20+\left ( \displaystyle \frac{3}{4} \right )^{3}+...\\ &=10+20\left ( \displaystyle \frac{3}{4}+\frac{9}{16}+\frac{27}{64}+... \right ) \end{aligned}\\ &\textrm{adalah deret geometri tak hingga dengan}\\ &a=r=\displaystyle \frac{3}{4},\: \: \textrm{maka}\\ &\begin{aligned}S_{\infty }&=10+20.\left (\displaystyle \frac{\displaystyle \frac{3}{4}}{1-\displaystyle \frac{3}{4}} \right )=10+20.\left ( \displaystyle \frac{\displaystyle \frac{3}{4}}{\displaystyle \frac{1}{4}} \right )\\ &=10+20.2=10+60=\color{red}70\: \: \textrm{meter} \end{aligned} \end{aligned}$.
$.\qquad\begin{aligned}&\textbf{Alternatif Jawaban}\\ &\textrm{Dengan rumus praktis, yaitu}\\ &\begin{array}{|l|}\hline\\ \begin{aligned}&\color{blue}\textrm{Panjang seluruh lintasan bola}\\ & \end{aligned}\\ \begin{array}{|l|}\hline \\\begin{aligned}S&=\textrm{Jatuh 1}\times \displaystyle \frac{\textrm{Jumlah perbandingan}}{\textrm{Selisih perbandingan}}\\ &=10\times \displaystyle \frac{4+3}{4-3}=10\times \displaystyle \frac{7}{1}=\color{red}70\: \: \textrm{meter}\\\ \end{aligned}\\\hline \end{array}\\ \\\hline \end{array} \end{aligned}$.
$\begin{array}{ll}\\ 34.&^{4}\log 2+\: ^{4}\log 4+\: ^{4}\log 16+^{4}\log 64+...\\ &\textrm{membentuk}\: ....\:.\\ &\textrm{A}.\quad \textrm{deret aritmetika dengan beda}\: \: ^{4}\log 2\\ &\textrm{B}.\quad \textrm{deret geometri dengan pembanding}\: \: ^{4}\log 2\\ &\textrm{C}.\quad \textrm{deret aritmetika dengan beda 2}\\ &\textrm{D}.\quad \textrm{deret geometri dengan pembanding 2}\\ &\textrm{E}.\quad \color{red}\textrm{bukan deret geometri maupun matematika}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}S_{n}&=\: ^{4}\log 2+\: ^{4}\log 4+\: ^{4}\log 16+^{4}\log 64+...\\ &=\: ^{4}\log 4^{\frac{1}{2}}+\: ^{4}\log 4^{1}+\: ^{4}\log 4^{2}+^{4}\log 4^{3}+...\\ &=\displaystyle \frac{1}{2}+1+2+3+...\\ &\textrm{dengan}\quad a=U_{1}=\displaystyle \frac{1}{2},\: U_{2}=1,\: \&\: U_{3}=2\\ &\textrm{Kita perlu cek dengan ciri masing-masing}\\ &\textrm{deret, yaitu}:\\ &\begin{array}{|l|l|}\hline \textrm{Deret Aritmetika}&\textrm{Deret Geometri}\\\hline \begin{aligned}&2U_{2}=U_{1}+U_{3}\\ &\qquad\color{blue}\textrm{atau}\\ &2U_{n+1}=U_{n}+U_{n+2} \end{aligned}&\begin{aligned}&U_{2}^{2}=U_{1}\times U_{3}\\ &\qquad\color{blue}\textrm{atau}\\ &U_{n+1}^{2}=U_{n}\times U_{n+2} \end{aligned}\\\hline \begin{aligned}&\color{red}2.(1)\neq \displaystyle \frac{1}{2}+2\\ & \end{aligned}&\begin{aligned}&(1)^{2}= \displaystyle \frac{1}{2}\times 2\\ &\color{red}(2)^{2}\neq 1\times 2 \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 35.&\textrm{Suatu modal sebesar}\: \: M\: \: \textrm{rupiah dibungakan dengan}\\ &\textrm{bunga}\: \: p\%\: \: \textrm{pertahun. Jika bunganya majmuk, maka}\\ &\textrm{setelah}\: \: n\: \: \textrm{tahun modal tersebut akan menjadi}\: ...\: .\\ &\textrm{A}.\quad M+(p/100)^{n}\\ &\textrm{B}.\quad (M+p\%.M)^{n}\\ &\textrm{C}.\quad nM.p\%\\ &\textrm{D}.\quad M(1-0,5)^{n}\\ &\textrm{E}.\quad \color{red}M(1+p\%)^{n}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Untuk kasus bunga majmuk di atas adalah}:\\ &\color{red}M,M(1+p\%),M(1+p\%)^{2},M(1+p\%)^{3},\cdots \\ &\textrm{adalah barisan geometri}\\ &\color{red}M_{n}=M_{0}(1+p\%)^{n}\: \: \color{black}\textrm{atau}\: \: M_{n}=M_{0}(1+i)^{n}\\ &\textrm{dengan}\\ &\circ \quad i=p\%\: \: \: \textrm{adalah persentase bunga}\\ &\circ \quad n =\textrm{Jangka waktu}\\ &\circ \quad M_{0}=\textrm{Modal yang diperbungakan} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 26.&\textrm{Syarat untuk deret geometri tak hingga }\\ &\textrm{dengan suku pertama}\: \: a\: \: \textrm{konvergen dengan }\\ &\textrm{jumlah 2 adalah}\: ....\:.\\ &\textrm{A}.\quad -2< a< 0\\ &\textrm{B}.\quad -4< a< 0\\ &\textrm{C}.\quad 0< a< 2\\ &\textrm{D}.\quad \color{red}0< a< 4\\ &\textrm{E}.\quad -4< a< 4\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: S_{\infty }=2,\: \: \textrm{dengan}\\ &S_{\infty }=\displaystyle \frac{a}{1-r}\Leftrightarrow 1-r=\displaystyle \frac{a}{S_{\infty }}\Leftrightarrow r=1-\displaystyle \frac{a}{S_{\infty }}\\ &\Leftrightarrow -1< 1-\displaystyle \frac{a}{S_{\infty }}< 1\Leftrightarrow -2< -\displaystyle \frac{a}{S_{\infty }}< 0\\ &\Leftrightarrow 0< \displaystyle \frac{a}{S_{\infty }}< 2\Leftrightarrow \Leftrightarrow 0< \displaystyle \frac{a}{2}< 2\\ & \color{red}\Leftrightarrow 0< a<4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 27.&\textrm{Tiga bilangan membentuk barisan geometri}\\ &\textrm{dengan jumlah}\: \: 26\: .\: \textrm{Jika suku tengah ditambah}\\ &\textrm{4 , maka terbentuklah barisan aritmetika, suku}\\ &\textrm{suku tengah dari barisan geometri tersebut}\: ....\:.\\ &\textrm{A}.\quad 2\\ &\textrm{B}.\quad 4\\ &\textrm{C}.\quad \color{red}6\\ &\textrm{D}.\quad 10\\ &\textrm{E}.\quad 18\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Barisan Geometri}:\: \: U_{1}+U_{2}+U_{3}=26\\ &\bullet \quad U_{1}+U_{3}=26-U_{2}\\ &\bullet \quad U_{2}^{2}=U_{1}.U_{3}\\ &\textrm{Barisan Aritmetika}:\: \: U_{1},U_{2}+4,U_{3}\\ &\bullet \quad U_{1}+U_{3}=2(U_{2}+4)=2U_{2}+8\\ &\textrm{maka}\\ &26-U_{2}=2U_{2}+8\\ &\Leftrightarrow -2U_{2}-U_{2}=8-26\\ &\Leftrightarrow -3U_{2}=-18\\ &\Leftrightarrow U_{2}=\color{red}\displaystyle \frac{-18}{-3}\color{black}=\color{red}6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Selish suku tengah pada barisan aritmetika}\\ &\textrm{dengan suku pertama dan terakhir masing-}\\ &\textrm{masing 1 dan 25 dengan barisan geometri}\\ &\textrm{yang suku-sukunya positif dengan suku-suku}\\ &\textrm{pertama dan terakhir juga 1 dan 25 adalah}\: ....\:.\\ &\textrm{A}.\quad 5\\ &\textrm{B}.\quad \textrm{sekitar}\: \: 7,1\\ &\textrm{C}.\quad \color{red}8\\ &\textrm{D}.\quad 13\\ &\textrm{E}.\quad 18\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&U_{t}=\textrm{Suku tengah}\\ &\textrm{Barisan Aritmetika (BA)}:\: \: U_{t_{BA}}=\displaystyle \frac{1}{2}(U_{1}+U_{n})\\ &\Leftrightarrow U_{t_{BA}}=\displaystyle \frac{1}{2}(1+25)=13\\ &\textrm{Barisan Geometri (BG)}:\: \: U_{t}^{2}=U_{1}.U_{n}\\ &\Leftrightarrow U_{t_{BG}}=\sqrt{U_{1}.U_{n}}=\sqrt{1\times 25}=5\\ &\qquad\qquad(\textrm{ambil nilai yang positif})\\ &\textrm{maka}\\ &U_{t_{BA}}-U_{t_{BG}}=13-5=\color{red}8 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 29.&\textbf{UM UGM}\\ &\textrm{Jumlah deret geometri tak hingga adalah 6}\\ & \textrm{Jika tiap suku dikuadratkan, maka jumlahnya}\\ &\textrm{adalah}\: \: 4\: .\: \textrm{Suku pertama deret ini adalah}\: ....\\ &\textrm{A}.\quad \displaystyle \frac{2}{5}\: \: \qquad\qquad\qquad\qquad\quad\: \textrm{D}.\quad \displaystyle \frac{5}{6}\\ &\textrm{B}.\quad \displaystyle \frac{3}{5}\qquad\qquad \color{black}\textrm{C}.\quad \displaystyle \frac{4}{5}\qquad\quad \color{black}\textrm{E}.\quad \color{red}\displaystyle \frac{6}{5}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{DG}=\textrm{Deret Geometri}\\ &a+ar+ar^{2}+\cdots =S_{\infty }=\displaystyle \frac{a}{1-r}=\color{blue}6\\ &\Leftrightarrow a=6(1-r)=6-6r\: ............(1)\\ &\textrm{Saat dikuadratkan masing-masing sukunya}\\ &a^{2}+a^{2}r^{2}+a^{2}r^{4}+\cdots =S_{\infty }=\displaystyle \frac{a^{2}}{1-r^{2}}=\color{blue}4\\ &\Leftrightarrow a^{2}=4(1-r^{2})=4-4r^{2}\: .......(2)\\ &\textrm{Substitusi (1) ke (2), maka} \end{aligned}\\ &\begin{aligned}&a^{2}=a^{2}\\ &\Leftrightarrow (6-6r)^{2}=4-4r^{2}\\ &\Leftrightarrow 36-72r+36r^{2}=4-4r^{2}\\ &\Leftrightarrow 40r^{2}-72r+32=0\\ &\Leftrightarrow (5r-4)(r-1)=0\\ &\Leftrightarrow r=\displaystyle \frac{4}{5}\: (memenuhi)\: \: \textbf{atau}\: \: r=1\: (tidak)\\ &\textrm{Selanjutnya kita tentukan nilai}\: \: a,\\ &a=6-6\left ( \displaystyle \frac{4}{5} \right )=6\left ( \displaystyle \frac{1}{5} \right )=\color{red}\displaystyle \frac{6}{5} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textbf{Soal Mat SNMPTN}\\ &\textrm{Agar deret geometri}\: \: \displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}\\ & \textrm{jumlahnya memiliki limit, maka nilai}\: \: x\\ &\textrm{harus memenuhi}\: ....\\ &\textrm{A}.\quad x>0\\ &\textrm{B}.\quad x<1\\ &\textrm{C}.\quad 0<x<1\\ &\textrm{D}.\quad x>2\\ &\textrm{E}.\quad \color{red}x<0\: \: \textrm{atau}\: \: x>2\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Deret Geometri (DG)}:\: \displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}\\ &r=\displaystyle \frac{\frac{1}{x}}{\frac{x-1}{x}}=\displaystyle \frac{1}{x-1}\\ &\textrm{Syarat DG memiliki limit (konvergen)}:\color{blue}\left | r \right |<1\\ &\Leftrightarrow -1<r<1\\ &\Leftrightarrow -1<\displaystyle \frac{1}{x-1}<1 \end{aligned}\\ &\begin{aligned}&\textrm{Selesaian 1}\\ &-1<\displaystyle \frac{1}{x-1}\Leftrightarrow \displaystyle \frac{1}{x-1}+1>0\\ &\Leftrightarrow \displaystyle \frac{1}{x-1}+\frac{x-1}{x-1}>0\Leftrightarrow \frac{x}{x-1}>0\\ &\textrm{Selesaian 2}\\ &\displaystyle \frac{1}{x-1}<1\Leftrightarrow \displaystyle \frac{1}{x-1}-1<0\\ &\Leftrightarrow \displaystyle \frac{1}{x-1}-\frac{x-1}{x-1}<0\Leftrightarrow \frac{-x+2}{x-1}<0\\ &\\ &\textrm{HP}:\left \{ x<0\: \: \textrm{atau}\: \: x>2 \right \} \end{aligned}\\ &\textbf{Berikut ilustrasi garis bilangannya}\\ &\begin{array}{ccc|ccccc|cccc|c}\\ (1)&\color{red}+&\color{red}+&-&-&-&-&-&\color{red}+&\color{red}+&\color{red}+&\color{red}+&\color{red}+\\\hline &&0&&&&&1&&&&\\\\ (2)&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&+&+&+&+&\color{red}-\\\hline &&&&&&&1&&&&2&\\\\ \end{array} \end{array}$.
$\begin{array}{ll}\\ 21.&\textrm{Jika jumlah}\: \: n\: \: \textrm{suku pertama suatu barisan}\\ &\textrm{adalah}\: \: S_{n}=n^{3}+2n\: ,\: \textrm{maka suku keempat}\\ &\textrm{adalah}\: ....\\ &\textrm{A}.\quad 33 \: \: \qquad\qquad\qquad\qquad\quad\: \, \, \textrm{D}.\quad 63\\ &\textrm{B}.\quad \color{red}39\qquad\qquad \color{black}\textrm{C}.\quad 49\qquad\quad \color{black}\textrm{E}.\quad 72\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui jumlah dari suatu barisan bilangan}\\ &\textrm{adalah}\: \: S_{n}=n^{3}+2n,\: \: \textrm{maka}\\ &\begin{aligned}U_{n}&=S_{n}-S_{n-1}\\ U_{4}&=\left ( 4^{3}+2(4) \right )-\left ( 3^{3}+2(3) \right )\\ &=(64+8)-(27+6)\\ &=72-33=\color{red}39 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 22.&\textrm{Dari suatu deret diketahui}\, \: S_{n}=3n^{2}-15n\\ &U_{n}=0\: \: \textrm{saat}\: n=\: ....\\ &\textrm{A}.\quad 1 \: \: \qquad\qquad\qquad\qquad\quad \textrm{D}.\quad 4\\ &\textrm{B}.\quad 2\qquad\qquad \textrm{C}.\quad \color{red}3\qquad\quad \color{black}\textrm{E}.\quad 5\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan hal yang diketahui di atas}\\ &\begin{aligned}U_{n}&=S_{n}-S_{n-1}\\ 0&=\left ( 3n^{2}-15n \right )-\left ( 3(n-1)^{2}-15(n-1) \right )\\ 0&=3\left ( n^{2}-(n-1)^{2} \right )+15(n-1-n)\\ 0&=3(2n-1)(1)+15(-1)\\ 0&=6n-3-15\\ 0&=6n-18\\ \color{red}3&=n \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 23.&\textrm{Diketahui sebuah deret}\: \: U_{n}=2an+b+4\\ &\textrm{dan}\: \: S_{n}=3bn^{2}+an\: ,\: \textrm{maka nilai}\: \: a\: \: \textrm{dan}\\ &b\: \: \: \textrm{adalah}\: ....\\ &\textrm{A}.\quad 12\: \: \textrm{dan}\: 4 \: \: \qquad\qquad\qquad\qquad \\ &\textrm{B}.\quad -12\: \textrm{dan}\: 4\quad \\ &\textrm{C}.\quad 12\: \: \textrm{dan}\: -4\quad\quad \\ &\textrm{D}.\quad \color{red}-12\: \: \textrm{dan}\: -4\\ &\textrm{E}.\quad -4\: \: \textrm{dan}\: -12\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: U_{n}=2an+b+4\: \: \textrm{dan}\\ &S_{n}=3bn^{2}+an,\: \: \textrm{maka}\\ &\begin{aligned}&U_{n}=S_{n}-S_{n-1}\\ &U_{2}=S_{2}-S_{1}\\ &2a(2)+b+4=(3b.2^{2}+a.2)-(3b.1^{2}+a.1)\\ &\Leftrightarrow \: \: 4a+b+4=9b+a\\ &\Leftrightarrow \: \: 3a-8b=-4\: ........(1) \end{aligned}\\ &\textrm{Dan juga}\\ &\begin{aligned}&U_{1}=S_{1}\\ &\Leftrightarrow \: \: 2a(1)+b+4=3b.1^{2}+a.1\\ &\Leftrightarrow \: \: 2a+b+4=3b+a\\ &\Leftrightarrow \: \: a-2b=-4\\ &\Leftrightarrow \: \: 3a-6b=-12\: ........(2) \end{aligned}\\ &\begin{aligned}&\textrm{Persamaan (2) disubstitusikan ke (1)}\\ &3a-8b=-4\\ &\Leftrightarrow \: \: 3a-6b-2b=4\\ &\Leftrightarrow \: \: (-12)-2b=-4\\ &\Leftrightarrow \: \: -2b=-4+12=8\\ &\Leftrightarrow \: \: b=\color{red}-4\: ........(3)\\ &\textrm{Selanjutnya dikembalikan ke (1), maka}\\ &3a-8b=-4\\ &\Leftrightarrow \: \: 3a-8(-4)=-4\\ &\Leftrightarrow \: \: 3a+32=-4\\ &\Leftrightarrow \: \: 3a=-4-32=-36\\ &\Leftrightarrow \: \: a=\color{red}-12 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 24.&\textrm{Jumlah}\: \: n\: \: \textrm{suku pertam sebuah barisan}\\ &\textrm{adalah}\: \: S_{n}=\displaystyle \frac{1}{6}(4n^{3}-63n^{2}-n)\: ,\: \textrm{suku ke}-n\\ &\textrm{akan mempunyai nilai terkecil untuk}\: \: n=\: ....\\ &\textrm{A}.\quad 3 \: \: \qquad\qquad\qquad\qquad \textrm{D}.\quad \color{red}6\\ &\textrm{B}.\quad 4\: \: \qquad\qquad\qquad\qquad \textrm{E}.\quad 7\\ &\textrm{C}.\quad 5\quad\quad \\\\ &\textbf{Jawab}:\\ &\textrm{Dengan menggunakan rumus}\\ &U_{n}=S_{n}-S_{n-1}\qquad \textrm{dengan}\: \: \: \: U_{1}=S_{1},\: \textrm{maka}\\ &\textrm{akan didapatkan nilai}\\ &U_{1}=-10,\: \: U_{2}=-27,\: \: U_{3}=-40,\: \: U_{4}=-49\\ &U_{5}=-54,\: \: U_{\color{red}6}=\color{red}-55\color{black},\: \: U_{7}=-52\\ &\textrm{Kesemuanya membentuk barisan aritmetika}\\ &\textrm{tingkat ke-2}.\: \textrm{Berikut ilustrasinya}\\ &\begin{aligned}&\underset{+4\qquad +4\qquad+4\qquad +4\qquad +4}{\underset{\underbrace{}\qquad\underbrace{}\quad\underbrace{}\quad\underbrace{}\qquad\underbrace{}}{\underset{-17\qquad -13\qquad -9\qquad -5\qquad -1\qquad +3}{\underset{\: \: \underbrace{}\: \quad\underbrace{}\: \qquad\underbrace{}\: \quad\underbrace{}\: \: \quad\underbrace{}\: \qquad\underbrace{}}{-10\quad -27\quad -40\quad -49\quad -54\quad -55\quad -52}}}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 25.&\textrm{Jika suku pertama dan kedua sebuah deret}\\ &\textrm{geometri masing-masing adalah}\: \: a^{-4}\: \: \textrm{dan}\: \: a^{x}\\ &\textrm{serta suku kedelapan ialah}\: \: a^{52},\: \: \textrm{maka nilai}\\ &x\: \: \textrm{adalah}\: ....\\ &\textrm{A}.\quad -32 \: \: \qquad\qquad\qquad\qquad\quad\: \, \textrm{D}.\quad 8\\ &\textrm{B}.\quad -16\qquad\qquad \color{black}\textrm{C}.\quad 12\qquad\quad \textrm{E}.\quad \color{red}4\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&U_{8}=ar^{7}=U_{1}r^{7}=a^{52}\\ &\Leftrightarrow \: \: U_{8}=a^{-4}r^{7}=a^{52}\\ &\Leftrightarrow \: \: r^{7}=\displaystyle \frac{a^{52}}{a^{-4}}=a^{52+4}=a^{56}\\ &\Leftrightarrow \: \: r=a^{.^{\frac{56}{7}}}=a^{8}\\ &\textrm{Maka nilai}\: \: x-\textrm{nya adalah}\\ &U_{2}=U_{1}r=a^{x}\\ &\Leftrightarrow \: \: (a^{-4})(a^{8})=a^{-4+8}=a^{x}\\ &\Leftrightarrow \: \: x=\color{red}4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 16.&\textrm{Carilah semua barisan bilangan yang berupa }\\ &\textrm{barisan aritmetika dan sekaligus juga barisan }\\ &\textrm{geometri}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah bentuk barisan bilangan berikut}:\\ &\underset{\underset{\displaystyle a}{\updownarrow}}{\overset{\overset{\displaystyle a}{\updownarrow}}{U_{1}}},\underset{\underset{\displaystyle ar}{\updownarrow}}{\overset{\overset{\displaystyle (a+b)}{\updownarrow}}{U_{2}}},\underset{\underset{\displaystyle ar^{2}}{\updownarrow}}{\overset{\overset{\displaystyle (a+2b)}{\updownarrow}}{U_{3}}},\underset{\underset{\displaystyle ar^{3}}{\updownarrow}}{\overset{\overset{\displaystyle (a+3b)}{\updownarrow}}{U_{4}}},\underset{\underset{\displaystyle ar^{4}}{\updownarrow}}{\overset{\overset{\displaystyle (a+4b)}{\updownarrow}}{U_{5}}},\quad \cdots\quad ,\underset{\underset{\displaystyle ar^{n-1}}{\updownarrow}}{\overset{\overset{\displaystyle (a+(n-1)b)}{\updownarrow}}{U_{n}}} \end{aligned}\\ &\begin{aligned}&\textrm{Misalkan}\\ &\textrm{Pada BA berlaku}\: \: 2U_{2}=U_{1}+U_{3}\\ &\textrm{yaitu}\begin{cases} 2(a+b)=a+(a+2b),&\textrm{atau}\\ 2(ar)=a+(ar^{2}) \end{cases}\\ &2ar=a+ar^{2},\quad \textrm{dibagi a masing-masing ruas}\\ &r^{2}-2r+1=0\\ &(r^{2}-2r+1)=0\\ &(r-1)^{2}=0\\ &r=1\\ &\textrm{Sehingga barisannya akan menjadi}\\ &a,a,a,a,\cdots \\ &\textrm{Pada BG juga berlaku}\: \: U_{2}^{2}=U_{1}\times U_{3}\\ &\textrm{yaitu}\begin{cases} (a+b)^{2}=a\times (a+2b),&\textrm{atau}\\ (ar)^{2}=a\times (ar^{2}) \end{cases}\\ &\textrm{ambil saja}\\ &(a+b)^{2}=a\times (a+2b)\\ &a^{2}+2ab+b^{2}=a^{2}+2ab\\ &b^{2}=0\\ &b=0\\ &\textrm{dan barisannya juga sama, yaitu}\\ &a,a,a,a,\cdots \\ &\textrm{Jadi, semua bilangan memenuhi}\: \: a\neq 0\\ &\textrm{saat}\: \: r=1\: \: atau\: \: b=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Pada waktu yang sama Anton mulai }\\ &\textrm{menabung Rp}10.000.000,00\: \: \textrm{dan Budi }\\ &\textrm{menabung}\: \: \textrm{Rp}8.000.000,00.\: \textrm{Selanjutnya}\\ &\textrm{Anton menabung Rp}100.000,00\: \: \textrm{tiap}\\ &\textrm{bulan dan Budi menabung Rp}150.000,00\\ &\textrm{Setelah berapa bulan tabungan keduanya}\\ &\textrm{tepat sama}\: ....\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Soal di atas adalah aplikasi dari deret}\\ &\textrm{aritmetika}\\ &\bullet \quad \textrm{barisan pertama}\\ &\qquad u_{1}=a=10.000.000,\: \: b=100.000\\ &\bullet \quad \textrm{barisan kedua}\\ &\qquad u_{1}=a=8.000.000,\: \: b=150.000\\ &\color{blue}\textrm{Selanjutnya adalah}\\ &u_{1}+(n-1)b=u_{1}^{'}+(n-1)b'\\ &\Leftrightarrow 10.000.000+(n-1)\times 100.000\\ &\quad =8.000.000+(n-1)\times 150.000\\ &\Leftrightarrow (100.000-150.000)\times (n-1)\\ &\quad =8.000.000-10.000.000\\ &\Leftrightarrow -50.000(n-1)=-2.000.000\\ &\Leftrightarrow (n-1)=\displaystyle \frac{-2.000.000}{-50.000}=40\\ &n=40+1=\color{red}41 \end{aligned}\\ &\textrm{Jadi, tabungan keduanya akan sama setelah}\\ &\color{red}\textrm{40 bulan} \end{array}$.
$\begin{array}{ll}\\ 18.&\textrm{Tentukan jumlah semua bilangan asli}\\ &\textrm{antara 1 dan 150 yang habis dibagi 4}\\ &\textrm{tetapi tidak habis dibagi 7}?\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Soal di atas adalah aplikasi dari deret}\\ &\textrm{aritmetika}\\ &\bullet \quad \textrm{barisan pertama adalah bilangan asli}\\ &\qquad \textrm{yang habis dibagi 4}\\ &\qquad u_{1}=a=4,\: \: b=4,\: \: U_{n}=148\\ &\qquad U_{n}=a+(n-1)b=4+(n-1)4=148\\ &\qquad \textrm{didapatkan nilai}\: \: n=37\\ &\qquad S_{n}=\displaystyle \frac{n}{2}\left ( a+u_{n} \right )=\frac{37}{2}(4+148)=\color{red}2812\\ &\bullet \quad \textrm{barisan kedua adalah bilangan asli}\\ &\qquad \textrm{yang habis dibagi}=4\times 7=28\\ &\qquad u_{1}=a=28,\: \: b=28,\: \: U_{n}=140\\ &\qquad \textrm{didapatkan nilai}\: \: n=5\\ &\qquad S_{n}=\displaystyle \frac{n}{2}\left ( a+u_{n} \right )=\frac{5}{2}(28+140)=\color{red}420\\ \end{aligned}\\ &\textrm{Jadi, jumlah bilangan yang dimaksud}\\ &\textrm{adalah}\: \: \color{red}\textrm{2812}-420=2392 \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Sebuah tali dibagi menjadi 6 bagian}\\ &\textrm{dengan panjang membentuk suatu}\\ &\textrm{barisan geometri. Jika panjang tali}\\ &\textrm{terpendek adalah 3 cm dan yang}\\ &\textrm{terpanjang 96 cm, berapakah panjang}\\ &\textrm{tali sebelum terpotong}?\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Misalkan panjang tali yang dimaksud}\\ &a,\: ar,\: ar^{2},\: ar^{3},\: ar^{4},\: ar^{5}\\ &\textrm{Dan diketahui juga bahwa}\\ &\begin{cases} a & =\textrm{3 cm} \\ ar^{5} & =\textrm{96 cm} \end{cases}\\ &\textrm{maka}\\ &\begin{aligned}ar^{5}&=96\\ \Leftrightarrow &\: \: 3.r^{5}=96\\ \Leftrightarrow &\: \: r^{5}=32\\ \Leftrightarrow &\: \: r^{5}=2^{5}\\ \Leftrightarrow &\: \: r=2\\ \color{blue}S_{n}&=\displaystyle \frac{a\left ( r^{n}-1 \right )}{r-1},\: \: r\geq 1\\ &=\displaystyle \frac{3\left (2^{6}-1 \right )}{2-1}=3\times \left ( 64-1 \right )\\ &=3\times 63=\color{red}189\: \textrm{cm} \end{aligned} \end{aligned}\\ &\textrm{Jadi, panjang talinya sebelum dipotong}\\ &\textrm{adalah}\: \: \color{red}\textrm{189 cm} \end{array}$.
$\begin{array}{ll}\\ 20.&\textrm{Jumlah penduduk suatu kota setiap }\\ &\textrm{10 tahun menjadi dua kali lipat}.\\ &\textrm{Jika menurut perhitungan pada tahun}\\ &\textrm{2030 nanti akan mencapai 3,2 juta jiwa}\\ &\textrm{berapakah jumlah penduduk kota tersebut}\\ &\textrm{pada tahun 1980}?\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari soal diketahui bahwa}\\ &\underset{a}{\begin{matrix} 1980\\ \downarrow \end{matrix}},\: \underset{ar}{\begin{matrix} 1990\\ \downarrow \end{matrix}},\: \underset{ar^{2}}{\begin{matrix} 2000\\ \downarrow \end{matrix}},\: \underset{ar^{3}}{\begin{matrix} 2010\\ \downarrow \end{matrix}},\: \underset{ar^{4}}{\begin{matrix} 2020\\ \downarrow \end{matrix}},\: \underset{ar^{5}}{\begin{matrix} 2030\\ \downarrow \end{matrix}}\\ &\quad \color{red}?\qquad\qquad\qquad\qquad\qquad\qquad 3,2\: \textrm{juta}\\ &\textrm{dengan}\\ &\begin{cases} r &=2 \\ n & =6 \end{cases}\\ &\textrm{maka}\\ &\begin{aligned}\: U_{6}=ar^{5}&=3.200.000&\\ a.2^{5}&=3.200.000\\ a.32&=3.200.000\\ a&=\displaystyle \frac{3.200.000}{32}\\ &=\color{red}100.000 \end{aligned} \end{aligned}\\ &\textrm{Jadi, jumlah penduduk pada tahun 1980}\\ &\textrm{sejumlah}\: \: \color{red}\textrm{100.000 jiwa} \end{array}$.
DAFTAR PUSTAKA
$\begin{array}{ll}\\ 11.&(\textbf{UN 2014})\\ &\textrm{Diketahui seutas kawat dipotong menjadi 5 }\\ &\textrm{bagian dan hasil potongannya membentuk }\\ &\textrm{deret geometri. Jika panjang kawat terpendek}\\ &\textrm{16 cm dan terpanjang 81 cm, maka panjang }\\ &\textrm{kawat semula adalah}\: ....\:\textrm{cm}\\ &\textrm{A}.\quad 121 \qquad\qquad\qquad\qquad\qquad\quad\: \: \textrm{D}.\quad \color{red}211\\ &\textrm{B}.\quad 130\qquad\qquad \textrm{C}.\quad 133\qquad\qquad \textrm{E}.\quad 242\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \begin{cases} a & =U_{1}=16 \\ U_{n} & =a.r^{n-1} \\ S_{n} & =\displaystyle \frac{a(r^{n}-1)}{r-1} \end{cases} \end{aligned}\\ &\begin{aligned}\displaystyle \frac{U_{5}}{U_{1}}=\frac{ar^{4}}{a}&=\displaystyle \frac{81}{16}\\ r^{4}&=\displaystyle \frac{3^{4}}{2^{4}}\\ r&=\displaystyle \frac{3}{2},\qquad \left (r>1 \right )\\ \textrm{Sehingga},\: &\\ S_{5}&=\displaystyle \frac{16\left ( \left ( \displaystyle \frac{3}{2} \right )^{5}-1 \right )}{\displaystyle \frac{3}{2}-1}\\ &=\displaystyle \frac{16\left ( \frac{3^{5}-2^{5}}{2^{5}} \right )}{\displaystyle \frac{1}{2}}\\ &=\displaystyle 32\left ( \frac{243-32}{32} \right )\\ &=\color{red}211 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 12.&(\textbf{UMPTN 2001})\\ &\textrm{Diketahui sepotong kawat dengan panjang }\\ &\textrm{124 cm akan dipotong menjadi 5 bagian }\\ &\textrm{dan hasil potongan kawatnya membentuk }\\ &\textrm{barisan geometri. Jika pajang potongan }\\ &\textrm{kawat yang terpendek adalah 4 cm, maka}\\ &\textrm{potongan kawat yang terpanjang adalah}\: ....\:\textrm{cm}\\ &\textrm{A}.\quad 60 \qquad\qquad\qquad\qquad\qquad\quad \textrm{D}.\quad 72\\ &\textrm{B}.\quad \color{red}64\qquad\qquad \color{black}\textrm{C}.\quad 68\qquad\qquad \textrm{E}.\quad 76\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan hal yang diketahui di atas}\\ &\begin{aligned}S_{5}&=U_{1}+U_{2}+U_{3}+U_{4}+U_{5}\\ S_{5}&=a+ar+ar^{2}+ar^{3}+ar^{4}\\ 124&=4+4r+4r^{2}+4r^{3}+4r^{4}\\ 31&=1+r+r^{2}+r^{3}+r^{4}\\ 30&=r+r^{2}+r^{3}+r^{4}\\ &\: \quad r^{4}+r^{3}+r^{2}+r-30=0 \end{aligned} \\ &\begin{aligned}&\textrm{Perhatikanlah bahwa pada polinom}\: \\ &r^{4}+r^{3}+r^{2}+r-30=0\\ &\textrm{faktor prima dari 30 adalah 2, 3, dan 5.}\\ &\textrm{Dan faktor yang memenuhi adalah r = 2}\\ &\textrm{Sehingga},\: U_{5}\: \: \textrm{sebagai potongan kawat terpanjang;}\\ &U_{5}=ar^{4}=4.2^{4}=4.16=\color{red}64 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 13.&(\textbf{UMPTN 2001})\\ &\textrm{Diketahui rasio sebuah deret geometri tak hingga }\\ &\textrm{adalah}\: \: ^{3}\log (2x-1).\: \: \textrm{Jika deret tersebut memiliki }\\ &\textrm{jumlah (konvergen), maka nilai}\: \: x\: \: \textrm{yang memenuhi }\\ &\textrm{adalah}\: ....\:.\\ &\textrm{A}.\quad \displaystyle \frac{1}{2}<x<\frac{2}{3} \\\\ &\textrm{B}.\quad \displaystyle \frac{1}{2}<x<2\\\\ &\color{red}\textrm{C}.\quad \displaystyle \frac{2}{3}<x<2\\\\ &\textrm{D}.\quad \displaystyle \frac{2}{3}\leq x\leq 2\\\\ &\textrm{E}.\quad \displaystyle \frac{1}{2}\leq x\leq \frac{3}{2}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Syarat}\: \textbf{konvergen adalah}\: \: \color{blue}\left | r \right |< 1\\ &\textrm{atau}\: :\: \, \: \: -1< r< 1\\ &\textrm{maka}\\ &-1< ^{3}\log (2x-1)< 1\\ &\Leftrightarrow \: (-1).\: \left (^{3}\log 3 \right )< \: ^{3}\log (2x-1)< 1.\left ( ^{3}\log 3 \right )\\ &\Leftrightarrow \: ^{3}\log 3^{-1}< \: ^{3}\log (2x-1)< \: ^{3}\log 3^{1}\\ &\Leftrightarrow \: 3^{-1}< (2x-1)<3\\ &\Leftrightarrow \: \displaystyle \frac{1}{3}<2x-1<3\\ &\Leftrightarrow \: \displaystyle \frac{1}{3}+1<2x-1+1<3+1\\ &\Leftrightarrow \: \displaystyle \frac{4}{3}<2x<4\\ &\Leftrightarrow \: \color{red}\displaystyle \frac{2}{3}<x<2 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 14.&(\textbf{UN 2010})\\ &\textrm{Tiga buah bilangan membentuk barisan aritmetika }\\ &\textrm{dengan beda 3. Jika suku kedua dikurangi 1, maka }\\ &\textrm{terbentuklah deret geometri dengan jumlah 14.}\\ &\textrm{Rasio barisan tersebut adalah ... .}\\ &\textrm{A}.\quad 4 \qquad\qquad\qquad\qquad\qquad\quad\: \: \textrm{D}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{B}.\quad \color{red}2\qquad\qquad \color{black}\textrm{C}.\quad \displaystyle \frac{1}{2}\qquad\qquad\: \: \: \, \textrm{E}.\quad -2\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Barisan Aritmetika (BA)}\: \begin{cases} U_{1} & =a \\ U_{2} & =a+3 \\ U_{3} & =a+6 \end{cases}\\ &\textrm{Barisan Geometri (BG)}\: \begin{cases} U_{1} & =a \\ U_{2} & =a+3-1=a+2 \\ U_{3} & =a+6 \end{cases}\\ \end{aligned}\\ &\begin{aligned}&\textrm{dan untuk deret geometri (DG)}\\ &U_{1}+U_{2}+U_{3}=14\\ &a+(a+2)+(a+6)=14\\ &3a+8=14\\ &3a=6\\ &a=2\\ &\textrm{sehingga},\\ &r=\displaystyle \frac{U_{2}}{U_{1}}=\frac{a+2}{a}=\frac{4}{2}=\color{red}2 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 15.&(\textbf{UN 2009})\\ &\textrm{Jumlah tiga bilangan barisan aritmetika adalah}\\ &\textrm{45. Jika suku kedua dikurangi 1 dan suku ketiga}\\ &\textrm{ditambah 5, maka barisan tersebut menjadi }\\ &\textrm{barisan geometri. Rasio barisan geometri tersebut}\\ &\textrm{adalah}\: ...\: .\\ &\textrm{A}.\quad \displaystyle \frac{1}{2}\qquad\qquad\qquad\qquad\qquad\quad\: \textrm{D}.\quad 2\\\\ &\textrm{B}.\quad \displaystyle \frac{3}{4}\qquad\qquad \textrm{C}.\quad 1\displaystyle \frac{1}{2}\qquad\qquad \textrm{E}.\quad 3\\\\ &\textbf{Jawab}:\textbf{Ada 2 pilihan, yaitu A dan D}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\begin{cases} U_{1}+U_{2}+U_{3} =45 \qquad (DA)\\ \Leftrightarrow \: \: \color{blue}a+(a+b)+(a+2b)=45 \\ \Leftrightarrow a+b=15\quad \Rightarrow a=15-b\\ U_{1},\left (U_{2}-1 \right ),\left (U_{3}+5 \right ) \qquad (BG)\\ \Rightarrow \: \: \color{blue}a,(a+b-1),(a+2b+5)\\ \Rightarrow a,(14),(20+b) \end{cases}\end{aligned}\\ &\begin{aligned}&\textrm{Pada BG berlaku}\\ &14^{2}=a.(20+b)\\ &196=(15-b)(20+b)\\ &196=300-5b-b^{2}\\ &b^{2}+5b-104=0\\ &(b+13)(b-8)=0\\ &b=-13\: \: \textrm{atau}\: \: b=8\\ &\textrm{untuk}\: \: b=-13\: \Rightarrow \: a=15-(-13)=\color{blue}28\\ &\qquad\qquad\begin{cases} BA_{1}: & 28,15,2 \\ BG_{1}: & 28,14,7 \end{cases}\\ &\textrm{untuk}\: \: b=8\: \Rightarrow \: a=15-(8)=\color{blue}7\\ &\qquad\qquad\begin{cases} BA_{2}: & 7,15,23 \\ BG_{2}: & 7,14,28 \end{cases}\\ &\textrm{Jadi, rasio dari barisan geometrianya ada 2 yaitu}:\\ &r_{1}=\color{red}\displaystyle \frac{1}{2},\: \: \textrm{dan}\: \: r_{2}=2 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&(\textbf{EBTANAS 2000})\\ &\textrm{Diketahui}\: \: \displaystyle \sum_{k=5}^{25}(2-pk)=0, \textrm{maka nilai}\\ &\sum_{k=5}^{25}pk= ... .\\ &\textrm{A}.\quad 20\qquad\qquad\qquad\qquad\qquad\quad \textrm{D}.\quad \color{red}42\\ &\textrm{B}.\quad 28\qquad\qquad \textrm{C}.\quad 30\qquad\qquad \textrm{E}.\quad 112\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\displaystyle \sum_{k=5}^{25}(2-pk)=\displaystyle \sum_{k=5}^{25}2-\sum_{k=5}^{25}pk&=0\\ \displaystyle \sum_{k=5-4}^{25-4}2-\sum_{k=5}^{25}pk&=0\\ \displaystyle \sum_{k=5}^{25}pk&=\displaystyle \sum_{k=1}^{21}2\\ &=21.2\\ &=\color{red}42 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 7.&(\textbf{EBTANAS 2000})\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{7}\left ( \displaystyle \frac{1}{2} \right )^{k+1}=... .\\ &\textrm{A}.\quad \displaystyle \frac{127}{1024}\qquad\qquad\qquad\qquad\quad \textrm{D}.\quad \displaystyle \frac{127}{128}\\\\ &\textrm{B}.\quad \color{red}\displaystyle \frac{127}{256}\qquad \color{black}\textrm{C}.\quad \displaystyle \frac{255}{512}\qquad\qquad \textrm{E}.\quad \displaystyle \frac{255}{256}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \sum_{k=1}^{7}\left ( \displaystyle \frac{1}{2} \right )^{k+1}\\ &=\left ( \displaystyle \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{4}+\left ( \frac{1}{2} \right )^{5}+\left ( \frac{1}{2} \right )^{6}+\left ( \frac{1}{2} \right )^{7} +\left ( \displaystyle \frac{1}{2} \right )^{8}\\ &=\displaystyle \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\\ &=\displaystyle \frac{64+32+16+8+4+2+1}{256}\\ &=\color{red}\displaystyle \frac{127}{256} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 8.&(\textbf{UN 2013})\\ &\textrm{Diketahui suatu barisan aritmetika dengan suku }\\ &\textrm{ketiga adalah 4 dan suku ketujuhnya adalah 16}.\\ &\textrm{Jumlah 10 suku pertama dari deret tersebut }\\ &\textrm{adalah}\: ...\: .\\ &\textrm{A}.\quad \color{red}115\qquad\qquad\qquad\qquad\qquad\quad\, \, \, \color{black}\textrm{D}.\quad 135\\ &\textrm{B}.\quad 125\qquad\qquad \textrm{C}.\quad 130\qquad\qquad \textrm{E}.\quad \displaystyle 140\\\\ &\textbf{Jawab}:\\ &\textrm{Dari soal diketahui bahwa}\\ &\begin{aligned}U_{3}=a+2b&=4\\ U_{7}=a+6b&=16\quad _{-}\\ ------&--\\ -4b&=-12\\ b&=3\\ \textrm{Sehingga}&\: \textrm{didapatkan}\\ \textrm{nilai}\: \: a=&4-2b\\ =&4-2.3\\ =&-2 \end{aligned}\\ &\begin{aligned}\textrm{Maka}&\: \textrm{jumlah 10 suku pertama deret tersebut adalah}\\ S_{n}&=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ S_{10}&=\displaystyle \frac{10}{2}\left ( 2.(-2)+(10-1).3 \right )\\ &=5\left ( -4+27 \right )\\ &=5(23)\\ &=\color{red}115\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 9.&(\textbf{UN 2014})\\ &\textrm{Diketahui tempat duduk gedung pertunjukan}\\ &\textrm{film diatur mulai dari baris depan ke belakang }\\ &\textrm{dengan banyak banyak baris dibelakang lebih }\\ &\textrm{4 kursi dari baris di depannya. Jika dalam }\\ &\textrm{gedung pertunjukan terdapat 15 baris kursi dan}\\ &\textrm{baris terdepan ada 20 kursi, maka kapasitas }\\ &\textrm{gedung pertunjukan tersebut adalah}\: ...\: .\: \textrm{kursi}\\ &\textrm{A}.\quad 1200\qquad\qquad\qquad\qquad\qquad\quad \textrm{D}.\quad 600\\ &\textrm{B}.\quad 800\qquad\qquad \textrm{C}.\quad \color{red}720\qquad\qquad \color{black}\textrm{E}.\quad 300\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\:\begin{cases} a &=U_{1}=20 \\ b & =4 \\ n & =15 \\ S_{n} & =\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right ) \end{cases} \end{aligned}\\ &\begin{aligned}S_{n}&=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &=\displaystyle \frac{15}{2}\left ( 2(20)+(15-1).4 \right )\\ &=15(20+28)\\ &=15(48)\\ &=\color{red}750 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 10.&(\textbf{UN 2015})\\ &\textrm{Diketahui suatu barisan aritmetika dengan suku}\\ &\textrm{ke-3 adalah 2 dan suku ke-8 adalah -13. Jumlah}\\ &\textrm{20 suku pertama dari deret tersebut adalah}\: ...\: .\\ &\textrm{A}.\quad -580\qquad\qquad\qquad\qquad\quad\quad\: \: \: \textrm{D}.\quad \color{red}-410\\ &\textrm{B}.\quad -490\qquad \textrm{C}.\quad -440\qquad\qquad \textrm{E}.\quad -380\\\\ &\textbf{Jawab}:\\ &\textrm{Dari soal di atas diketahui bahwa}\\ &\begin{aligned}U_{3}=a+2b&=2\\ U_{8}=a+7b&=-13\quad _{-}\\ ------&---\\ -5b&=15\\ b&=-3\\ \textrm{Sehingga}&\: \textrm{didapatkan}\\ \textrm{nilai}\: \: a=&2-2b\\ =&2-2.(-3)\\ =&8 \end{aligned}\\ &\begin{aligned}\textrm{Maka}&\: \textrm{jumlah 20 suku pertama deret tersebut adalah}\\ S_{n}&=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ S_{20}&=\displaystyle \frac{20}{2}\left ( 2.(8)+(20-1).(-3) \right )\\ &=10\left ( 16-57 \right )\\ &=10(-41)\\ &=\color{red}-410 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 1.&(\textbf{EBTANAS 1999})\\ &\textrm{Diketahui jumlah n suku pertama deret}\\ &\textrm{aritmetika dinyatakan sebagai}\: \: S_{n}=n^{2}+2n\\ &\textrm{Beda dari deret tersebut adalah}\: ....\\ &\textrm{a}.\quad 3\: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad -2\\ &\textrm{b}.\quad \color{red}2\qquad\qquad \color{black}\textrm{c}.\quad 1\qquad\qquad \textrm{e}.\quad -3\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: S_{n}=n^{2}+2n,\\ &\textrm{dengan}\: \begin{cases} S_{1} & =U_{1}=a \\ S_{2} & =U_{1}+U_{2} \\ S_{3} & =U_{1}+U_{2}+U_{3} \\ &\vdots \\ S_{n} & =U_{1}+U_{2}+U_{3}+\cdots +U_{n} \end{cases}\\ &\begin{aligned}\textrm{Beda}=b&=U_{2}-U_{1}\\ &=(S_{2}-S_{1})-S_{1}\\ &=S_{2}-2S_{1}\\ &=\left ( 2^{2}+2.(2) \right )-2\left ( 1^{2}+2.(1) \right )\\ &=\left ( 4+4 \right )-2\left ( 1+2 \right )=8-6\\ &=\color{red}2\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&(\textbf{UMPTN 1994})\\ &\textrm{Diketahui jumlah n suku pertama suatu}\\ & \textrm{deret dinyatakan sebagai}\: \: S_{n}=12n-n^{2}.\\ &\textrm{Suku kelima dari deret tersebut adalah}\: ....\\ &\textrm{a}.\quad -1\: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad \color{red}3\\ &\textrm{b}.\quad 1\qquad\qquad \textrm{c}.\quad -3\qquad\qquad \textrm{e}.\quad 0\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: S_{n}=12n-n^{2}\\ &\begin{aligned}U_{5}&=S_{5}-S_{4}\\ &=\left ( 12.(5)-(5)^{2} \right )-\left ( 12.(4)-(4)^{2} \right )\\ &=\left ( 60-25 \right )-\left ( 48-16 \right )\\ &=\color{red}3\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&(\textbf{EBTANAS 2000})\\ &\textrm{Diketahui suku tengah suatu deret }\\ &\textrm{aritmetika adalah 32. Jika jumlah n}\\ &\textrm{suku pertama deret itu adalah 672, }\\ &\textrm{maka banyak suku deret itu adalah}\: ....\\ &\textrm{a}.\quad 17\: \: \: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 23\\ &\textrm{b}.\quad 19\qquad\qquad \textrm{c}.\quad \color{red}21\qquad\qquad \color{black}\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui}\\ &\textrm{Suku tengah}=U_{t}=\displaystyle \frac{U_{1}+U_{n}}{2}=32\: \: \textrm{dan}\\ &\begin{aligned}S_{n}&=\displaystyle \frac{n}{2}\left ( U_{1}+U_{n} \right )\\ &=672\\ n\left ( \displaystyle \frac{U_{1}+U_{2}}{2} \right )&=672\\ 32n&=672\\ n&=\color{red}21 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 1997})\\ &\textrm{Diketahui}\: \: U_{n}\: \: \textrm{adalah suku ke - n }\\ &\textrm{deret aritmetika dengan}\\ &U_{1}+U_{2}+U_{3}=-9\: \: \textrm{dan}\: \: U_{3}+U_{4}+U_{5}=15\\ & \textrm{Maka jumlah lima suku pertama}\\ &\textrm{deret aritmetika tersebut adalah}\: ....\\ &\textrm{a}.\quad 4\: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 15\\ &\textrm{b}.\quad \color{red}5\qquad\qquad \color{black}\textrm{c}.\quad 6\qquad\qquad \textrm{e}.\quad 24\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&U_{1}+U_{2}+U_{3}=-9,\\ &\Rightarrow a+(a+b)+(a+2b)=3a+3b=-9\\ &U_{3}+U_{4}+U_{5}=15,\\ & \Rightarrow \: \: \qquad(a+2b)+(a+3b)+(a+4b)=3a+9b=15\quad _{-}\\ & -----------------\\ &\, \, -6b=-24\Rightarrow b=4\\ &\, a=-7\\ &\textrm{Maka}\\ &\begin{aligned} S_{5}&=\displaystyle \frac{5}{2}\left ( U_{1}+U_{5} \right )\\ &=\displaystyle \frac{5}{2}\left ( a+a+(5-1)b \right )\\ &=\displaystyle \frac{5}{2}\left ( -7-7+4.4 \right )\\ &=\displaystyle \frac{5}{2}(2)\\ &=\color{red}5\end{aligned} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&(\textbf{EBTANAS 1999})\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 30.900\: \: \: \: \: \: \quad\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 15.450\\ &\textrm{b}.\quad 30.500\qquad\qquad \textrm{c}.\quad 16.250\qquad\qquad \textrm{e}.\quad \color{red}15.250\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahi}\\ &\begin{aligned}\displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)&=\sum_{k=1}^{100}(5k-2k+1)\\ &=\displaystyle \sum_{k=1}^{100}(3k+1)\\ &=3\displaystyle \sum_{k=1}^{100}k+1.100\\ &=3\left ( \displaystyle \frac{100}{2}(1+100) \right )+100\\ &=3.(5.050)+100\\ &=15.150+100\\ &=\color{red}15.250\end{aligned} \end{array}$.
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
A. Pola Bilangan
Pola bilangan dalam kaitannya dengan matematika adalah suatu susunan bilangan dengan susunan tertentu.
Perhatikan ilustrasi berikut
Misalkan beberapa kue donat dalam disusun dan dikelompokkan berbentuk persegi sebagaimana ilustrasi berikut
$\begin{array}{ll}\\ 6.&\textrm{Bentuk sederhana dari}\: \: \: \\ &\left (\displaystyle \frac{1}{2} \right )^{-\left ( \frac{1}{2} \right )^{-1}}+\left (\displaystyle \frac{1}{3} \right )^{-\left ( \frac{1}{3} \right )^{-1}}+\left (\displaystyle \frac{1}{4} \right )^{-\left ( \frac{1}{4} \right )^{-1}}+\left (\displaystyle \frac{1}{5} \right )^{-\left ( \frac{1}{5} \right )^{-1}}\\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3142\\ \textrm{b}.&287\\ \color{red}\textrm{c}.&3412\\ \textrm{d}.&4116\\ \textrm{e}.&4096 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\left (\displaystyle \frac{1}{2} \right )^{-\left ( \frac{1}{2} \right )^{-1}}+\left (\displaystyle \frac{1}{3} \right )^{-\left ( \frac{1}{3} \right )^{-1}}+\left (\displaystyle \frac{1}{4} \right )^{-\left ( \frac{1}{4} \right )^{-1}}+\left (\displaystyle \frac{1}{5} \right )^{-\left ( \frac{1}{5} \right )^{-1}}\\ &=\left (\displaystyle \frac{1}{2} \right )^{-2}+\left (\displaystyle \frac{1}{3} \right )^{-3}+\left (\displaystyle \frac{1}{4} \right )^{-4}+\left (\displaystyle \frac{1}{5} \right )^{-5}\\ &=2^{2}+3^{3}+4^{4}+5^{5}\\ &=4+27+256+3125\\ &=\color{red}3412 \end{aligned}\\ \end{array}$.
$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\\ &\underset{89}{\underbrace{(-7)(-7)(-7)\cdots (-7)}}-(-7)^{89}\\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \textrm{b}.&7\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&0\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\underset{89}{\underbrace{(-7)(-7)(-7)\cdots (-7)}}-(-7)^{89}\\ &=(-7)^{89}-(-7)^{89}\\ &=\color{red}0 \end{aligned}\\ \end{array}$.
$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \left ( \sqrt{3\times \sqrt[4]{27\times \sqrt[3]{81}}} \right )^{\displaystyle \frac{24}{25}}\\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \color{red}\textrm{c}.&3\\ \textrm{d}.&4\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}& \left ( \sqrt{3\times \sqrt[4]{27\times \sqrt[3]{81}}} \right )^{\displaystyle \frac{24}{25}}\\ &=\left ( 3^{\frac{1}{2}}.(3^{3})^{\frac{1}{2.4}}.(3^{4})^{\frac{1}{2.4.3}} \right )^{\displaystyle \frac{24}{25}}\\ &=\left ( 3^{\frac{1}{2}+\frac{3}{8}+\frac{4}{24}} \right )^{\displaystyle \frac{24}{25}}\\ &=3^{\frac{12}{25}+\frac{9}{25}+\frac{4}{25}}\\ &=3^{\frac{25}{25}}\\ &=\color{red}3 \end{aligned}\\ \end{array}$.
$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: \sqrt{27^{2x+3}}=\displaystyle \frac{1}{3^{x-2}.9^{3x}}\: ,\: \: \textrm{maka nilai}\\ &8x+2\: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\textbf{Alternatif 1}\\ &\color{blue}\begin{aligned}\sqrt{27^{2x+3}}&=\displaystyle \frac{1}{3^{x-2}.9^{3x}}\\ 27^{\displaystyle \frac{2x+3}{2}}&=3^{-(x-2)}.9^{-3x}\\ 3^{3.\left ( \displaystyle \frac{2x+3}{2} \right )}&=3^{2-x}.3^{2(-3x)}\\ 3^{3.\left ( \displaystyle \frac{2x+3}{2} \right )}&=3^{2-x-6x}\\ \color{red}3\color{black}^{3.\left ( \displaystyle \frac{2x+3}{2} \right )}&=\color{red}3\color{black}^{2-7x}\\ \color{black}3\frac{(2x+3)}{2}&=\color{black}2-7x\\ 6x+9&=4-14x\\ 6x+14x&=4-9\\ 20x&=-5\\ &x=-\displaystyle \frac{1}{4}\\ \textrm{maka nilai}&\\ 8x+2&=8\left ( -\displaystyle \frac{1}{4} \right )+2=-2+2=\color{red}0 \end{aligned}\\ &\textbf{Alternatif 2}\\ &\color{blue}\begin{aligned}\sqrt{27^{2x+3}}&=\displaystyle \frac{1}{3^{x-2}.9^{3x}}\\ 27^{\displaystyle \frac{2x+3}{2}}.3^{x-2}.9^{3x}&=1\\ 3^{3\left ( \displaystyle \frac{2x+3}{2} \right )}.3^{x-2}.3^{2(3x)}&=3^{0}\\ 3^{3\left ( \displaystyle \frac{2x+3}{2} \right )+x-2+6x}&=3^{0}\\ \displaystyle \frac{3(2x+3)}{2}+7x-2&=0\\ 6x+9+14x-4&=0\\ 20x+5&=0\\ 20x&=-5\\ x&=-\displaystyle \frac{1}{4}\\ \textrm{Selanjutnya sama}&\\ \textrm{dengan langkah no.1}&\: \: \textrm{di atas} \end{aligned}\\ \end{array}$.
$\begin{array}{ll}\\ 10.&\textrm{Penyelesaian persamaan}\: \: 3^{2x+1}=81^{x-2}\: \\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&4\displaystyle \frac{1}{2}\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}3^{2x+1}&=81^{x-2}\\ 3^{2x+1}&=(3^{4})^{x-2}\\ 3^{2x+1}&=3^{4x-8}\\ \color{black}2x+1&=\color{black}4x-8\\ \color{black}2x-4x&=\color{black}-8-1\\ -2x&=-9\\ x&=\displaystyle \frac{-9}{-2}\\ &=\color{red}4\displaystyle \frac{1}{2} \end{aligned}\\ \end{array}$.
$\begin{array}{ll}\\ 1.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{a^{p}.a^{q}}{a^{r}}:a^{2r} \: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&a^{p+q-3r}\\ \textrm{b}.&a^{p+3r-q}\\ \textrm{c}.&a^{p-2q+r}\\ \textrm{d}.&a^{p+q+r}\\ \textrm{e}.&a^{p-3q-r} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\displaystyle \frac{a^{p}.a^{q}}{a^{r}}:a^{2r}&=\displaystyle \frac{a^{p}.a^{q}}{a^{r}.a^{2r}}\\ &=\displaystyle \frac{a^{p+q}}{a^{r+2r}}\\ &=\displaystyle \frac{a^{p+q}}{a^{3r}}\\ &=\color{red}a^{p+q-3r} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: x=\sqrt[3]{5+\sqrt[3]{8}}\: ,\: \: \textrm{maka nilai}\\ &x^{3}\: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&6\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}x&=\sqrt[3]{5+\sqrt[3]{8}}\quad \color{black}\textrm{dipangkatkan 3}\\ &\qquad \color{black}\textrm{masing-masing ruas}\\ x^{3}&=\left ( \sqrt[3]{5+\sqrt[3]{8}} \right )^{3}\\ &=5+\sqrt[3]{8}\\ &=5+\sqrt[3]{2^{3}}=5+2=\color{red}7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: 4^{a}\times 4^{b}=64\: \: \textrm{dan}\: \: \displaystyle \frac{4^{a}}{4^{b}}=16\\ &\textrm{maka nilai dari}\: \: a:b\: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&5\\ \textrm{b}.&\displaystyle \frac{5}{4}\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&3\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\\ &\bullet \quad 4^{a}\times 4^{b}=64\color{black}\Leftrightarrow 4^{a+b}=4^{3}\Leftrightarrow a+b=3\: \: ....(1)\\ &\bullet \quad \displaystyle \frac{4^{a}}{4^{b}}=16\color{black}\Leftrightarrow 4^{a-b}=4^{2}\Leftrightarrow a-b=2\: \: ...........(2) \\ &\color{black}\textrm{Dari persamaan}\: \: (1)\: \: \&\: \: (2)\: \: \textrm{akan didapatkan}\\ &\begin{array}{|l|l|}\hline \begin{array}{llllll} a+b&=&3\\ a-b&=&2&+\\\hline 2a&=&5\\ \: \: a&=&\displaystyle \frac{5}{2} \end{array}&\begin{array}{rlllll} a+b&=&3\\ a-b&=&2&-\\\hline 2b&=&1\\ \: \: b&=&\displaystyle \frac{1}{2} \end{array}\\\hline \end{array}\\ &\textrm{Sehingga nilai}\: \: a:b=\displaystyle \frac{a}{b}=\displaystyle \frac{\displaystyle \frac{5}{2}}{\displaystyle \frac{1}{2}}=\color{red}5 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&(\textbf{SPMB 2003})\\ &\textrm{Jika}\: \: a\neq 0,\: \textrm{maka nilai}\: \: \displaystyle \frac{(-2a)^{3}(2a)^{-\frac{2}{3}}}{\left ( 16a^{4} \right )^{\frac{1}{3}}}\: \: \textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-2^{2}a\\ \color{red}\textrm{b}.&\displaystyle -2a\\ \textrm{c}.&\displaystyle 2a^{2}\\ \textrm{d}.&2^{2}a\\ \textrm{e}.&\displaystyle -2a^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{(-2a)^{3}(2a)^{-\frac{2}{3}}}{\left ( 16a^{4} \right )^{\frac{1}{3}}}&=\displaystyle \frac{-8a^{3}}{(2a)^{\frac{2}{3}}.\left ( 16a^{4} \right )^{\frac{1}{3}}}\\ &=-\displaystyle \frac{(2a)^{3}}{(2a)^{\frac{2}{3}}.(2a)^{\frac{4}{3}}}\\ &=-\displaystyle \frac{(2a)^{3}}{(2a)^{\frac{2}{3}+\frac{4}{3}}}\\ &=-\displaystyle \frac{(2a)^{3}}{(2a)^{2}}\\ &=-(2a)^{3-2}\\ &=\color{red}-2a \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \frac{4^{n+3}-4^{n+1}}{4(4^{n-1})}\: \\ &\textrm{adalah ... .}\\ &\begin{array}{llll}\\ \textrm{a}.&64\\ \color{red}\textrm{b}.&60\\ \textrm{c}.&18\\ \textrm{d}.&16\\ \textrm{e}.&15 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned} \displaystyle \frac{4^{n+3}-4^{n+1}}{4(4^{n-1})}&=\displaystyle \frac{4^{n}.4^{3}-4^{n}.4}{4.\displaystyle \frac{4^{n}}{4}}\\ &=\displaystyle \frac{4^{n}(64-4)}{4^{n}}=\color{red}60\\ \end{aligned}\\ \end{array}$.
