PAS MATEMATIKA BLOG: November 2022

$\begin{array}{ll}\\ 31.&2+4+6+8+\cdots =\: ....\:.\\ &\textrm{A}.\quad \color{red}\displaystyle \sum_{k=1}^{n}2k\\ &\textrm{B}.\quad \displaystyle \sum_{k=1}^{n}2^{k}\\ &\textrm{C}.\quad \displaystyle \sum_{k=1}^{n}2k-1\\ &\textrm{D}.\quad \displaystyle \sum_{k=1}^{n}(2-k)\\ &\textrm{E}.\quad \displaystyle \sum_{k=1}^{n}k\\\\ &\textbf{Jawab}:\\ &\textrm{Cukup Jelas bahwa}\\ &\begin{aligned}2+4+6+8+\cdots \: \: \quad&\\ 2(1+2+3+4+\cdots )&=2\displaystyle \sum_{k=1}^{n}k=\displaystyle \sum_{k=1}^{n}\color{red}2k \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 32.&\textbf{(UN 2005)}\\ &\textrm{Seorang anak menabung di suatu bank dengan}\\ &\textrm{selisih kenaikan tabungan antarbulan tetap}\\ &\textrm{Pada bulan pertama sebesar Rp50.000,00},\\ &\textrm{bulan kedua Rp55.000,00, bulan ketiga}\\ &\textrm{Rp60.000,00, dan demikian seterusnya}.\\ &\textrm{Besar tabungan anak tersebut selama }\\ &\textrm{dua tahun adalah}\: ....\:.\\ &\textrm{A}.\quad \textrm{Rp1.315.000,00}\\ &\textrm{B}.\quad \textrm{Rp1.320.000,00}\\ &\textrm{C}.\quad \textrm{Rp2.040.000,00}\\ &\textrm{D}.\quad \color{red}\textrm{Rp2.580.000,00}\\ &\textrm{E}.\quad \textrm{Rp2.640.000,00}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui deret aritmetika dengan}\\ &\begin{aligned}&\bullet \quad a=U_{1}=\textrm{Rp}50.000,00\\ &\bullet \quad U_{2}=\textrm{Rp}55.000,00\\ &\bullet \quad b=U_{2}-U_{1}=\textrm{Rp}5.000,00\\ &\textrm{Ditanya: Besar tabungan selama 2 tahun}\\ &\begin{aligned}\color{blue}S_{n}&=\color{blue}\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ \textrm{Ka}&\textrm{rena 2 tahun = 24 bulan, maka}\\ S_{24}&=\displaystyle \frac{24}{2}(2\times 50.000+(24-1)\times 5.000)\\ &=12\left ( 100.000+115.000 \right )=\color{red}2.580.000 \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textbf{(UN 2006)}\\ &\textrm{Sebuah bola jatuh dari ketinggian 10 meter dan}\\ &\textrm{memantul kembali dengan ketinggian}\: \: \displaystyle \frac{3}{4}\: \: \textrm{dari}\\ &\textrm{tinggi semula dan begitu seterusnya hingga bola}\\ &\textrm{berhenti. Jumlah seluruh lintasan bola adalah}\: ....\:.\\ &\textrm{A}.\quad 65\: \: \textrm{meter}\\ &\textrm{B}.\quad \color{red}70\: \: \textrm{meter}\\ &\textrm{C}.\quad 75\: \: \textrm{meter}\\ &\textrm{D}.\quad 77\: \: \textrm{meter}\\ &\textrm{E}.\quad 80\: \: \textrm{meter}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Soal terkait dengan deret geometri tak hingga}\\ &\textrm{Yaitu}:\: S_{\infty }=\displaystyle \frac{a}{1-r},\: \: \textrm{dengan}\: \: \left | r \right |<1\\ &\color{blue}\begin{aligned}S&=10+2.\displaystyle \frac{3}{4}.10+2.\frac{3}{4}.\frac{3}{4}.10+2.\frac{3}{4}.\frac{3}{4}.\frac{3}{4}.10+...\\ &=10+20.\displaystyle \frac{3}{4}+20.\left (\frac{3}{4} \right )^{2}+20+\left ( \displaystyle \frac{3}{4} \right )^{3}+...\\ &=10+20\left ( \displaystyle \frac{3}{4}+\frac{9}{16}+\frac{27}{64}+... \right ) \end{aligned}\\ &\textrm{adalah deret geometri tak hingga dengan}\\ &a=r=\displaystyle \frac{3}{4},\: \: \textrm{maka}\\ &\begin{aligned}S_{\infty }&=10+20.\left (\displaystyle \frac{\displaystyle \frac{3}{4}}{1-\displaystyle \frac{3}{4}} \right )=10+20.\left ( \displaystyle \frac{\displaystyle \frac{3}{4}}{\displaystyle \frac{1}{4}} \right )\\ &=10+20.2=10+60=\color{red}70\: \: \textrm{meter} \end{aligned} \end{aligned}$.

$.\qquad\begin{aligned}&\textbf{Alternatif Jawaban}\\ &\textrm{Dengan rumus praktis, yaitu}\\ &\begin{array}{|l|}\hline\\ \begin{aligned}&\color{blue}\textrm{Panjang seluruh lintasan bola}\\ & \end{aligned}\\ \begin{array}{|l|}\hline \\\begin{aligned}S&=\textrm{Jatuh 1}\times \displaystyle \frac{\textrm{Jumlah perbandingan}}{\textrm{Selisih perbandingan}}\\ &=10\times \displaystyle \frac{4+3}{4-3}=10\times \displaystyle \frac{7}{1}=\color{red}70\: \: \textrm{meter}\\\ \end{aligned}\\\hline \end{array}\\ \\\hline \end{array} \end{aligned}$.

$\begin{array}{ll}\\ 34.&^{4}\log 2+\: ^{4}\log 4+\: ^{4}\log 16+^{4}\log 64+...\\ &\textrm{membentuk}\: ....\:.\\ &\textrm{A}.\quad \textrm{deret aritmetika dengan beda}\: \: ^{4}\log 2\\ &\textrm{B}.\quad \textrm{deret geometri dengan pembanding}\: \: ^{4}\log 2\\ &\textrm{C}.\quad \textrm{deret aritmetika dengan beda 2}\\ &\textrm{D}.\quad \textrm{deret geometri dengan pembanding 2}\\ &\textrm{E}.\quad \color{red}\textrm{bukan deret geometri maupun matematika}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}S_{n}&=\: ^{4}\log 2+\: ^{4}\log 4+\: ^{4}\log 16+^{4}\log 64+...\\ &=\: ^{4}\log 4^{\frac{1}{2}}+\: ^{4}\log 4^{1}+\: ^{4}\log 4^{2}+^{4}\log 4^{3}+...\\ &=\displaystyle \frac{1}{2}+1+2+3+...\\ &\textrm{dengan}\quad a=U_{1}=\displaystyle \frac{1}{2},\: U_{2}=1,\: \&\: U_{3}=2\\ &\textrm{Kita perlu cek dengan ciri masing-masing}\\ &\textrm{deret, yaitu}:\\ &\begin{array}{|l|l|}\hline \textrm{Deret Aritmetika}&\textrm{Deret Geometri}\\\hline \begin{aligned}&2U_{2}=U_{1}+U_{3}\\ &\qquad\color{blue}\textrm{atau}\\ &2U_{n+1}=U_{n}+U_{n+2} \end{aligned}&\begin{aligned}&U_{2}^{2}=U_{1}\times U_{3}\\ &\qquad\color{blue}\textrm{atau}\\ &U_{n+1}^{2}=U_{n}\times U_{n+2} \end{aligned}\\\hline \begin{aligned}&\color{red}2.(1)\neq \displaystyle \frac{1}{2}+2\\ & \end{aligned}&\begin{aligned}&(1)^{2}= \displaystyle \frac{1}{2}\times 2\\ &\color{red}(2)^{2}\neq 1\times 2 \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Suatu modal sebesar}\: \: M\: \: \textrm{rupiah dibungakan dengan}\\ &\textrm{bunga}\: \: p\%\: \: \textrm{pertahun. Jika bunganya majmuk, maka}\\ &\textrm{setelah}\: \: n\: \: \textrm{tahun modal tersebut akan menjadi}\: ...\: .\\ &\textrm{A}.\quad M+(p/100)^{n}\\ &\textrm{B}.\quad (M+p\%.M)^{n}\\ &\textrm{C}.\quad nM.p\%\\ &\textrm{D}.\quad M(1-0,5)^{n}\\ &\textrm{E}.\quad \color{red}M(1+p\%)^{n}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Untuk kasus bunga majmuk di atas adalah}:\\ &\color{red}M,M(1+p\%),M(1+p\%)^{2},M(1+p\%)^{3},\cdots \\ &\textrm{adalah barisan geometri}\\ &\color{red}M_{n}=M_{0}(1+p\%)^{n}\: \: \color{black}\textrm{atau}\: \: M_{n}=M_{0}(1+i)^{n}\\ &\textrm{dengan}\\ &\circ \quad i=p\%\: \: \: \textrm{adalah persentase bunga}\\ &\circ \quad n =\textrm{Jangka waktu}\\ &\circ \quad M_{0}=\textrm{Modal yang diperbungakan} \end{aligned} \end{array}$.

Lanjutan 6 Contoh Soal Barisan dan Deret

$\begin{array}{ll}\\ 26.&\textrm{Syarat untuk deret geometri tak hingga }\\ &\textrm{dengan suku pertama}\: \: a\: \: \textrm{konvergen dengan }\\ &\textrm{jumlah 2 adalah}\: ....\:.\\ &\textrm{A}.\quad -2< a< 0\\ &\textrm{B}.\quad -4< a< 0\\ &\textrm{C}.\quad 0< a< 2\\ &\textrm{D}.\quad \color{red}0< a< 4\\ &\textrm{E}.\quad -4< a< 4\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: S_{\infty }=2,\: \: \textrm{dengan}\\ &S_{\infty }=\displaystyle \frac{a}{1-r}\Leftrightarrow 1-r=\displaystyle \frac{a}{S_{\infty }}\Leftrightarrow r=1-\displaystyle \frac{a}{S_{\infty }}\\ &\Leftrightarrow -1< 1-\displaystyle \frac{a}{S_{\infty }}< 1\Leftrightarrow -2< -\displaystyle \frac{a}{S_{\infty }}< 0\\ &\Leftrightarrow 0< \displaystyle \frac{a}{S_{\infty }}< 2\Leftrightarrow \Leftrightarrow 0< \displaystyle \frac{a}{2}< 2\\ & \color{red}\Leftrightarrow 0< a<4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 27.&\textrm{Tiga bilangan membentuk barisan geometri}\\ &\textrm{dengan jumlah}\: \: 26\: .\: \textrm{Jika suku tengah ditambah}\\ &\textrm{4 , maka terbentuklah barisan aritmetika, suku}\\ &\textrm{suku tengah dari barisan geometri tersebut}\: ....\:.\\ &\textrm{A}.\quad 2\\ &\textrm{B}.\quad 4\\ &\textrm{C}.\quad \color{red}6\\ &\textrm{D}.\quad 10\\ &\textrm{E}.\quad 18\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Barisan Geometri}:\: \: U_{1}+U_{2}+U_{3}=26\\ &\bullet \quad U_{1}+U_{3}=26-U_{2}\\ &\bullet \quad U_{2}^{2}=U_{1}.U_{3}\\ &\textrm{Barisan Aritmetika}:\: \: U_{1},U_{2}+4,U_{3}\\ &\bullet \quad U_{1}+U_{3}=2(U_{2}+4)=2U_{2}+8\\ &\textrm{maka}\\ &26-U_{2}=2U_{2}+8\\ &\Leftrightarrow -2U_{2}-U_{2}=8-26\\ &\Leftrightarrow -3U_{2}=-18\\ &\Leftrightarrow U_{2}=\color{red}\displaystyle \frac{-18}{-3}\color{black}=\color{red}6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Selish suku tengah pada barisan aritmetika}\\ &\textrm{dengan suku pertama dan terakhir masing-}\\ &\textrm{masing 1 dan 25 dengan barisan geometri}\\ &\textrm{yang suku-sukunya positif dengan suku-suku}\\ &\textrm{pertama dan terakhir juga 1 dan 25 adalah}\: ....\:.\\ &\textrm{A}.\quad 5\\ &\textrm{B}.\quad \textrm{sekitar}\: \: 7,1\\ &\textrm{C}.\quad \color{red}8\\ &\textrm{D}.\quad 13\\ &\textrm{E}.\quad 18\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&U_{t}=\textrm{Suku tengah}\\ &\textrm{Barisan Aritmetika (BA)}:\: \: U_{t_{BA}}=\displaystyle \frac{1}{2}(U_{1}+U_{n})\\ &\Leftrightarrow U_{t_{BA}}=\displaystyle \frac{1}{2}(1+25)=13\\ &\textrm{Barisan Geometri (BG)}:\: \: U_{t}^{2}=U_{1}.U_{n}\\ &\Leftrightarrow U_{t_{BG}}=\sqrt{U_{1}.U_{n}}=\sqrt{1\times 25}=5\\ &\qquad\qquad(\textrm{ambil nilai yang positif})\\ &\textrm{maka}\\ &U_{t_{BA}}-U_{t_{BG}}=13-5=\color{red}8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 29.&\textbf{UM UGM}\\ &\textrm{Jumlah deret geometri tak hingga adalah 6}\\ & \textrm{Jika tiap suku dikuadratkan, maka jumlahnya}\\ &\textrm{adalah}\: \: 4\: .\: \textrm{Suku pertama deret ini adalah}\: ....\\ &\textrm{A}.\quad \displaystyle \frac{2}{5}\: \: \qquad\qquad\qquad\qquad\quad\: \textrm{D}.\quad \displaystyle \frac{5}{6}\\ &\textrm{B}.\quad \displaystyle \frac{3}{5}\qquad\qquad \color{black}\textrm{C}.\quad \displaystyle \frac{4}{5}\qquad\quad \color{black}\textrm{E}.\quad \color{red}\displaystyle \frac{6}{5}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{DG}=\textrm{Deret Geometri}\\ &a+ar+ar^{2}+\cdots =S_{\infty }=\displaystyle \frac{a}{1-r}=\color{blue}6\\ &\Leftrightarrow a=6(1-r)=6-6r\: ............(1)\\ &\textrm{Saat dikuadratkan masing-masing sukunya}\\ &a^{2}+a^{2}r^{2}+a^{2}r^{4}+\cdots =S_{\infty }=\displaystyle \frac{a^{2}}{1-r^{2}}=\color{blue}4\\ &\Leftrightarrow a^{2}=4(1-r^{2})=4-4r^{2}\: .......(2)\\ &\textrm{Substitusi (1) ke (2), maka} \end{aligned}\\ &\begin{aligned}&a^{2}=a^{2}\\ &\Leftrightarrow (6-6r)^{2}=4-4r^{2}\\ &\Leftrightarrow 36-72r+36r^{2}=4-4r^{2}\\ &\Leftrightarrow 40r^{2}-72r+32=0\\ &\Leftrightarrow (5r-4)(r-1)=0\\ &\Leftrightarrow r=\displaystyle \frac{4}{5}\: (memenuhi)\: \: \textbf{atau}\: \: r=1\: (tidak)\\ &\textrm{Selanjutnya kita tentukan nilai}\: \: a,\\ &a=6-6\left ( \displaystyle \frac{4}{5} \right )=6\left ( \displaystyle \frac{1}{5} \right )=\color{red}\displaystyle \frac{6}{5} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textbf{Soal Mat SNMPTN}\\ &\textrm{Agar deret geometri}\: \: \displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}\\ & \textrm{jumlahnya memiliki limit, maka nilai}\: \: x\\ &\textrm{harus memenuhi}\: ....\\ &\textrm{A}.\quad x>0\\ &\textrm{B}.\quad x<1\\ &\textrm{C}.\quad 0<x<1\\ &\textrm{D}.\quad x>2\\ &\textrm{E}.\quad \color{red}x<0\: \: \textrm{atau}\: \: x>2\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Deret Geometri (DG)}:\: \displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}\\ &r=\displaystyle \frac{\frac{1}{x}}{\frac{x-1}{x}}=\displaystyle \frac{1}{x-1}\\ &\textrm{Syarat DG memiliki limit (konvergen)}:\color{blue}\left | r \right |<1\\ &\Leftrightarrow -1<r<1\\ &\Leftrightarrow -1<\displaystyle \frac{1}{x-1}<1 \end{aligned}\\ &\begin{aligned}&\textrm{Selesaian 1}\\ &-1<\displaystyle \frac{1}{x-1}\Leftrightarrow \displaystyle \frac{1}{x-1}+1>0\\ &\Leftrightarrow \displaystyle \frac{1}{x-1}+\frac{x-1}{x-1}>0\Leftrightarrow \frac{x}{x-1}>0\\ &\textrm{Selesaian 2}\\ &\displaystyle \frac{1}{x-1}<1\Leftrightarrow \displaystyle \frac{1}{x-1}-1<0\\ &\Leftrightarrow \displaystyle \frac{1}{x-1}-\frac{x-1}{x-1}<0\Leftrightarrow \frac{-x+2}{x-1}<0\\ &\\ &\textrm{HP}:\left \{ x<0\: \: \textrm{atau}\: \: x>2 \right \} \end{aligned}\\ &\textbf{Berikut ilustrasi garis bilangannya}\\ &\begin{array}{ccc|ccccc|cccc|c}\\ (1)&\color{red}+&\color{red}+&-&-&-&-&-&\color{red}+&\color{red}+&\color{red}+&\color{red}+&\color{red}+\\\hline &&0&&&&&1&&&&\\\\ (2)&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&+&+&+&+&\color{red}-\\\hline &&&&&&&1&&&&2&\\\\ \end{array} \end{array}$.