$\begin{array}{ll}\\ 21.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &3+6+12+24+...+\left ( 3.2^{n-1} \right )=3.\left ( 2^{n}-1 \right )\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k+1,\: \: \textrm{maka}\\ &\textrm{ruas kiri persamaan tersebut dapat dituliskan}\\ &\textrm{dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&3+6+12+24+...+ 3.2^{k+1} \\ \textrm{b}.&3+6+12+24+...+ 3.2^{k-1} \\ \color{red}\textrm{c}.&3+6+12+24+...+ 3.2^{k-1}+3.2^{k} \\ \textrm{d}.&3+6+12+24+...+ 3.2^{k-1}+3.2^{k+1} \\ \textrm{e}.&3+6+12+24+...+ 3.2^{k}+3.2^{k+1} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&3+6+12+24+...+ 3.2^{n-1} =3.\left ( 2^{n}-1 \right )\\ &\color{red}3+6+12+24+...+ 3.2^{k-1}+3.2^{k}\color{black}=\color{blue}3.\left ( 2^{k+1}-1 \right ) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 22.&\textbf{(EBTANAS 1999)}\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 30.900\\ &\textrm{b}.\quad 30.500\\ &\textrm{c}.\quad 16.250\\ &\textrm{d}.\quad 15.450\\ &\textrm{e}.\quad \color{red}15.250\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\textrm{Diketahi}\\ &\begin{aligned}\displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)&=\sum_{k=1}^{100}(5k-2k+1)\\ &=\displaystyle \sum_{k=1}^{100}(3k+1)\\ &=3\displaystyle \sum_{k=1}^{100}k+1.100\\ &=3\left ( \displaystyle \frac{100}{2}(1+100) \right )+100\\ &=3.(5.050)+100\\ &=15.150+100\\ &=\color{red}15.250\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 23.&\textbf{(EBTANAS 2000)}\\ &\textrm{Diketahui}\: \: \displaystyle \sum_{k=5}^{25}(2-pk)=0, \textrm{maka nilai}\\ & \displaystyle \sum_{k=5}^{25}pk= ... .\\ &\textrm{a}.\quad 20\\ &\textrm{b}.\quad 28\\ & \textrm{c}.\quad 30\\ &\textrm{d}.\quad \color{red}42\\ & \textrm{e}.\quad 112\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \sum_{k=5}^{25}(2-pk)=\displaystyle \sum_{k=5}^{25}2-\sum_{k=5}^{25}pk&=0\\ \displaystyle \sum_{k=5-4}^{25-4}2-\sum_{k=5}^{25}pk&=0\\ \displaystyle \sum_{k=5}^{25}pk&=\displaystyle \sum_{k=1}^{21}2\\ &=21.2\\ &=\color{red}42 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 24.&\textbf{(EBTANAS 2000)}\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{7}\left ( \displaystyle \frac{1}{2} \right )^{k+1}=... .\\ &\textrm{a}.\quad \displaystyle \frac{127}{1024}\\\\ &\textrm{b}.\quad \displaystyle \frac{127}{256}\\\\ & \textrm{c}.\quad \displaystyle \frac{255}{512}\\\\ &\textrm{d}.\quad \displaystyle \frac{127}{128}\\\\ & \textrm{e}.\quad \displaystyle \frac{255}{256}\\\\ &\textrm{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\displaystyle \sum_{k=1}^{7}\left ( \displaystyle \frac{1}{2} \right )^{k+1}\\ &=\left ( \displaystyle \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{2} \right )^{4}+\left ( \frac{1}{2} \right )^{5}+\left ( \frac{1}{2} \right )^{6}+\left ( \frac{1}{2} \right )^{7} +\left ( \displaystyle \frac{1}{2} \right )^{8}\\ &=\displaystyle \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\\ &=\displaystyle \frac{64+32+16+8+4+2+1}{256}\\ &=\color{red}\displaystyle \frac{127}{256} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 25.&\textbf{EBTANAS 1999}\\ &\textrm{Diketahui jumlah n suku pertama }\\ &\textrm{deret aritmetika dinyatakan sebagai}\\ &S_{n}=n^{2}+2n.\: \textrm{Beda dari deret tersebut }\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad 3 \\ &\textrm{b}.\quad \color{red}2\\ & \textrm{c}.\quad 1\\ &\textrm{d}.\quad -2\\ & \textrm{e}.\quad -3\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Diketahui bahwa}\: \: S_{n}=n^{2}+2n,\\ &\textrm{dengan}\: \begin{cases} S_{1} & =U_{1}=a \\ S_{2} & =U_{1}+U_{2} \\ S_{3} & =U_{1}+U_{2}+U_{3} \\ &\vdots \\ S_{n} & =U_{1}+U_{2}+U_{3}+\cdots +U_{n} \end{cases}\\ &\begin{aligned}\textrm{Beda}=b&=U_{2}-U_{1}\\ &=(S_{2}-S_{1})-S_{1}\\ &=S_{2}-2S_{1}\\ &=\left ( 2^{2}+2.(2) \right )-2\left ( 1^{2}+2.(1) \right )\\ &=\left ( 4+4 \right )-2\left ( 1+2 \right )=8-6\\ &=\color{red}2\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 26.&\textbf{(UMPTN 1994)}\\ &\textrm{Diketahui jumlah n suku pertama suatu }\\ &\textrm{deret dinyatakan sebagai}\quad S_{n}=12n-n^{2}.\\ & \textrm{Suku kelima dari deret tersebut adalah}\: ....\\ &\textrm{a}.\quad -1 \\ &\textrm{b}.\quad 1\\ & \textrm{c}.\quad -3\\ &\textrm{d}.\quad \color{red}3\\ & \textrm{e}.\quad 0\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\textrm{Diketahui bahwa}\\ &S_{n}=12n-n^{2}\\ &\begin{aligned}U_{5}&=S_{5}-S_{4}\\ &=\left ( 12.(5)-(5)^{2} \right )-\left ( 12.(4)-(4)^{2} \right )\\ &=\left ( 60-25 \right )-\left ( 48-16 \right )\\ &=\color{red}3\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 27.&\textbf{(UMPTN 1997)}\\ &\textrm{Diketahui}\: \: U_{n}\: \: \textrm{adalah suku ke - n }\\ &\textrm{deret aritmetika dengan}\\ &U_{1}+U_{2}+U_{3}=-9\: \: \textrm{dan}\\ & \: U_{3}+U_{4}+U_{5}=15.\\ & \textrm{Maka jumlah lima suku pertama}\\ &\textrm{deret aritmetika tersebut adalah}\: ....\\ &\textrm{a}.\quad 4\\ &\textrm{b}.\quad \color{red}5\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 15\\ &\textrm{e}.\quad 24\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&U_{1}+U_{2}+U_{3}=-9,\\ &\Leftrightarrow a+(a+b)+(a+2b)=-9\\ &\Leftrightarrow \color{blue}3a+3b=-9\\ &U_{3}+U_{4}+U_{5}=15,\\ &\Leftrightarrow (a+2b)+(a+3b)+(a+4b)=15\\ &\Leftrightarrow \color{blue}3a+9b=15\quad _{-}\\ & -----------------\\ &\, \qquad\qquad\qquad -6b=-24\Rightarrow b=\color{red}4\\ &\, \qquad \textrm{sehingga akan diperoleh}\: \: a=\color{red}-7\\ &\textrm{Selanjutnya}\\ &S_{5}=\displaystyle \frac{5}{2}\left ( U_{1}+U_{5} \right )\\ &=\displaystyle \frac{5}{2}\left ( a+a+(5-1)b \right )\\ &=\displaystyle \frac{5}{2}\left ( -7-7+4.4 \right )\\ &=\displaystyle \frac{5}{2}(2)\\ &=\color{red}5\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 28.&\textbf{(UN 2013)}\\ &\textrm{Diketahui suatu barisan aritmetika dengan }\\ &\textrm{suku ketiga adalah 4 dan suku ketujuhnya }\\ &\textrm{adalah 16. Jumlah 10 suku pertama dari }\\ &\textrm{deret tersebut adalah}\: ...\: .\\ &\textrm{a}.\quad \color{red}115\\ &\textrm{b}.\quad 125\\ & \textrm{c}.\quad 130\\ &\textrm{d}.\quad 135\\ &\textrm{e}.\quad \displaystyle 140\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}U_{3}=a+2b&=4\\ U_{7}=a+6b&=16\quad _{-}\\ ------&--\\ -4b&=-12\\ b&=3\\ \textrm{Sehingga}&\: \textrm{didapatkan}\\ \textrm{nilai}\: \: a=&4-2b\\ =&4-2.3\\ =&-2 \end{aligned}\\ &\begin{aligned}&\textrm{Maka}\: \textrm{jumlah 10 suku pertama }\\ &\textrm{deret tersebut adalah}\\ &S_{n}=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &S_{10}=\displaystyle \frac{10}{2}\left ( 2.(-2)+(10-1).3 \right )\\ &\: \: \quad=5\left ( -4+27 \right )\\ &\: \: \quad=5(23)\\ &\: \: \quad=\color{red}115 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 29.&\textbf{(UN 2014)}\\ &\textrm{Diketahui tempat duduk gedung pertunjukan }\\ &\textrm{film diatur mulai dari baris depan ke belakang }\\ &\textrm{dengan banyak banyak baris dibelakang lebih }\\ &\textrm{4 kursi dari baris di depannya.}\\ &\textrm{Jika dalam gedung pertunjukan terdapat 15}\\ &\textrm{baris kursi dan baris terdepan ada 20 kursi, }\\ &\textrm{maka kapasitas gedung pertunjukan tersebut }\\ &\textrm{adalah}\: ...\: .\: \textrm{kursi}\\ &\textrm{a}.\quad 1200\\ &\textrm{b}.\quad 800\\ & \textrm{c}.\quad \color{red}720\\ &\textrm{d}.\quad 600\\ &\textrm{e}.\quad 300\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diketahui}&\:\begin{cases} a &=U_{1}=20 \\ b & =4 \\ n & =15 \\ S_{n} & =\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right ) \end{cases}\\ &\\ & \end{aligned}\\ &\begin{aligned}S_{n}&=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &=\displaystyle \frac{15}{2}\left ( 2(20)+(15-1).4 \right )\\ &=15(20+28)\\ &=15(48)\\ &=\color{red}750 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textbf{(UN 2015)}\\ &\textrm{Diketahui suatu barisan aritmetika dengan }\\ &\textrm{suku ke-3 adalah 2 dan suku ke-8 adalah -13}.\\ &\textrm{Jumlah 20 suku pertama dari deret tersebut }\\ &\textrm{adalah}\: ...\: .\\ &\textrm{a}.\quad -580\\ &\textrm{b}.\quad -490\\ &\textrm{c}.\quad -440\\ &\textrm{d}.\quad \color{red}-410\\ &\textrm{e}.\quad -380\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}U_{3}=a+2b&=2\\ U_{8}=a+7b&=-13\quad _{-}\\ ------&---\\ -5b&=15\\ b&=-3\\ \textrm{Sehingga}&\: \textrm{didapatkan}\\ \textrm{nilai}\: \: a=&2-2b\\ =&2-2.(-3)\\ =&8 \end{aligned}\\ &\begin{aligned}&\textrm{Maka}\: \textrm{jumlah 20 suku pertama }\\ &\textrm{deret tersebut adalah}\\ &S_{n}=\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ &S_{20}=\displaystyle \frac{20}{2}\left ( 2.(8)+(20-1).(-3) \right )\\ &\: \: \quad=10\left ( 16-57 \right )\\ &\: \: \quad=10(-41)\\ &\: \: \quad=\color{red}-410\\ &\\ & \end{aligned} \end{array}$
DAFTAR PUSTAKA
- Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA.
- Thohir, A. https://ahmadthohir1089.wordpress.com/2016/01/11/insyaallah-44/
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