$\begin{array}{ll}\\ 26.&\textrm{Syarat untuk deret geometri tak hingga }\\ &\textrm{dengan suku pertama}\: \: a\: \: \textrm{konvergen dengan }\\ &\textrm{jumlah 2 adalah}\: ....\:.\\ &\textrm{A}.\quad -2< a< 0\\ &\textrm{B}.\quad -4< a< 0\\ &\textrm{C}.\quad 0< a< 2\\ &\textrm{D}.\quad \color{red}0< a< 4\\ &\textrm{E}.\quad -4< a< 4\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: S_{\infty }=2,\: \: \textrm{dengan}\\ &S_{\infty }=\displaystyle \frac{a}{1-r}\Leftrightarrow 1-r=\displaystyle \frac{a}{S_{\infty }}\Leftrightarrow r=1-\displaystyle \frac{a}{S_{\infty }}\\ &\Leftrightarrow -1< 1-\displaystyle \frac{a}{S_{\infty }}< 1\Leftrightarrow -2< -\displaystyle \frac{a}{S_{\infty }}< 0\\ &\Leftrightarrow 0< \displaystyle \frac{a}{S_{\infty }}< 2\Leftrightarrow \Leftrightarrow 0< \displaystyle \frac{a}{2}< 2\\ & \color{red}\Leftrightarrow 0< a<4 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 27.&\textrm{Tiga bilangan membentuk barisan geometri}\\ &\textrm{dengan jumlah}\: \: 26\: .\: \textrm{Jika suku tengah ditambah}\\ &\textrm{4 , maka terbentuklah barisan aritmetika, suku}\\ &\textrm{suku tengah dari barisan geometri tersebut}\: ....\:.\\ &\textrm{A}.\quad 2\\ &\textrm{B}.\quad 4\\ &\textrm{C}.\quad \color{red}6\\ &\textrm{D}.\quad 10\\ &\textrm{E}.\quad 18\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Barisan Geometri}:\: \: U_{1}+U_{2}+U_{3}=26\\ &\bullet \quad U_{1}+U_{3}=26-U_{2}\\ &\bullet \quad U_{2}^{2}=U_{1}.U_{3}\\ &\textrm{Barisan Aritmetika}:\: \: U_{1},U_{2}+4,U_{3}\\ &\bullet \quad U_{1}+U_{3}=2(U_{2}+4)=2U_{2}+8\\ &\textrm{maka}\\ &26-U_{2}=2U_{2}+8\\ &\Leftrightarrow -2U_{2}-U_{2}=8-26\\ &\Leftrightarrow -3U_{2}=-18\\ &\Leftrightarrow U_{2}=\color{red}\displaystyle \frac{-18}{-3}\color{black}=\color{red}6 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Selish suku tengah pada barisan aritmetika}\\ &\textrm{dengan suku pertama dan terakhir masing-}\\ &\textrm{masing 1 dan 25 dengan barisan geometri}\\ &\textrm{yang suku-sukunya positif dengan suku-suku}\\ &\textrm{pertama dan terakhir juga 1 dan 25 adalah}\: ....\:.\\ &\textrm{A}.\quad 5\\ &\textrm{B}.\quad \textrm{sekitar}\: \: 7,1\\ &\textrm{C}.\quad \color{red}8\\ &\textrm{D}.\quad 13\\ &\textrm{E}.\quad 18\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&U_{t}=\textrm{Suku tengah}\\ &\textrm{Barisan Aritmetika (BA)}:\: \: U_{t_{BA}}=\displaystyle \frac{1}{2}(U_{1}+U_{n})\\ &\Leftrightarrow U_{t_{BA}}=\displaystyle \frac{1}{2}(1+25)=13\\ &\textrm{Barisan Geometri (BG)}:\: \: U_{t}^{2}=U_{1}.U_{n}\\ &\Leftrightarrow U_{t_{BG}}=\sqrt{U_{1}.U_{n}}=\sqrt{1\times 25}=5\\ &\qquad\qquad(\textrm{ambil nilai yang positif})\\ &\textrm{maka}\\ &U_{t_{BA}}-U_{t_{BG}}=13-5=\color{red}8 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 29.&\textbf{UM UGM}\\ &\textrm{Jumlah deret geometri tak hingga adalah 6}\\ & \textrm{Jika tiap suku dikuadratkan, maka jumlahnya}\\ &\textrm{adalah}\: \: 4\: .\: \textrm{Suku pertama deret ini adalah}\: ....\\ &\textrm{A}.\quad \displaystyle \frac{2}{5}\: \: \qquad\qquad\qquad\qquad\quad\: \textrm{D}.\quad \displaystyle \frac{5}{6}\\ &\textrm{B}.\quad \displaystyle \frac{3}{5}\qquad\qquad \color{black}\textrm{C}.\quad \displaystyle \frac{4}{5}\qquad\quad \color{black}\textrm{E}.\quad \color{red}\displaystyle \frac{6}{5}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{DG}=\textrm{Deret Geometri}\\ &a+ar+ar^{2}+\cdots =S_{\infty }=\displaystyle \frac{a}{1-r}=\color{blue}6\\ &\Leftrightarrow a=6(1-r)=6-6r\: ............(1)\\ &\textrm{Saat dikuadratkan masing-masing sukunya}\\ &a^{2}+a^{2}r^{2}+a^{2}r^{4}+\cdots =S_{\infty }=\displaystyle \frac{a^{2}}{1-r^{2}}=\color{blue}4\\ &\Leftrightarrow a^{2}=4(1-r^{2})=4-4r^{2}\: .......(2)\\ &\textrm{Substitusi (1) ke (2), maka} \end{aligned}\\ &\begin{aligned}&a^{2}=a^{2}\\ &\Leftrightarrow (6-6r)^{2}=4-4r^{2}\\ &\Leftrightarrow 36-72r+36r^{2}=4-4r^{2}\\ &\Leftrightarrow 40r^{2}-72r+32=0\\ &\Leftrightarrow (5r-4)(r-1)=0\\ &\Leftrightarrow r=\displaystyle \frac{4}{5}\: (memenuhi)\: \: \textbf{atau}\: \: r=1\: (tidak)\\ &\textrm{Selanjutnya kita tentukan nilai}\: \: a,\\ &a=6-6\left ( \displaystyle \frac{4}{5} \right )=6\left ( \displaystyle \frac{1}{5} \right )=\color{red}\displaystyle \frac{6}{5} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textbf{Soal Mat SNMPTN}\\ &\textrm{Agar deret geometri}\: \: \displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}\\ & \textrm{jumlahnya memiliki limit, maka nilai}\: \: x\\ &\textrm{harus memenuhi}\: ....\\ &\textrm{A}.\quad x>0\\ &\textrm{B}.\quad x<1\\ &\textrm{C}.\quad 0<x<1\\ &\textrm{D}.\quad x>2\\ &\textrm{E}.\quad \color{red}x<0\: \: \textrm{atau}\: \: x>2\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Deret Geometri (DG)}:\: \displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}\\ &r=\displaystyle \frac{\frac{1}{x}}{\frac{x-1}{x}}=\displaystyle \frac{1}{x-1}\\ &\textrm{Syarat DG memiliki limit (konvergen)}:\color{blue}\left | r \right |<1\\ &\Leftrightarrow -1<r<1\\ &\Leftrightarrow -1<\displaystyle \frac{1}{x-1}<1 \end{aligned}\\ &\begin{aligned}&\textrm{Selesaian 1}\\ &-1<\displaystyle \frac{1}{x-1}\Leftrightarrow \displaystyle \frac{1}{x-1}+1>0\\ &\Leftrightarrow \displaystyle \frac{1}{x-1}+\frac{x-1}{x-1}>0\Leftrightarrow \frac{x}{x-1}>0\\ &\textrm{Selesaian 2}\\ &\displaystyle \frac{1}{x-1}<1\Leftrightarrow \displaystyle \frac{1}{x-1}-1<0\\ &\Leftrightarrow \displaystyle \frac{1}{x-1}-\frac{x-1}{x-1}<0\Leftrightarrow \frac{-x+2}{x-1}<0\\ &\\ &\textrm{HP}:\left \{ x<0\: \: \textrm{atau}\: \: x>2 \right \} \end{aligned}\\ &\textbf{Berikut ilustrasi garis bilangannya}\\ &\begin{array}{ccc|ccccc|cccc|c}\\ (1)&\color{red}+&\color{red}+&-&-&-&-&-&\color{red}+&\color{red}+&\color{red}+&\color{red}+&\color{red}+\\\hline &&0&&&&&1&&&&\\\\ (2)&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&\color{red}-&+&+&+&+&\color{red}-\\\hline &&&&&&&1&&&&2&\\\\ \end{array} \end{array}$.
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