Barisan dan Deret (KTSP Kelas XII)

Barisan dan Deret Bilangan

A. Pengertian

Silahkan lihat materi tentang barisan dan deret di sini (1) , di sini (2) , dan di sini (3).

Barisan Bilangan adalah urutan bilangan yang mengikuti aturan (pola) tertentu.

Sebagai misal

\begin{array}{lll}\\ .&\textcircled{1}&1,2,3,4,5,...\\ &\textcircled{2}&3,5,7,9,11,...\\ &\textcircled{3}&2,6,18,54,...\\ &\textcircled{4}&4,2,1,\displaystyle \frac{1}{2},\displaystyle \frac{1}{4},...\\ &\textcircled{5}&8,-4,2,-1,\displaystyle \frac{1}{2},...\\ &&\vdots \\ &&dst \end{array}.

Sedangkan deret bilangan adalah jumlah suku-suku terurut pada suatu barisan.

Misal

\begin{array}{lll}\\ .&\textcircled{1}&1+2+3+4+5+...\\ &\textcircled{2}&3+5+7+9+11+...\\ &\textcircled{3}&2+6+18+54+...\\ &\textcircled{4}&4+2+1+\displaystyle \frac{1}{2}+\displaystyle \frac{1}{4}+...\\ &\textcircled{5}&8-4+2-1+\displaystyle \frac{1}{2}-...\\ &&\vdots \\ &&dst \end{array}.

B. Barisan dan Deret Aritmetika ( Hitung )

Perhatikanlah tabel berikut;

\begin{array}{|c|l|l|l|}\hline \textrm{No}&\textrm{Barisan Aritmetika}&\textrm{Deret Aritmetika (Hitung)}&\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{Beda}=b&=U_{2}-U_{1}\\ &=U_{3}-U_{2}\\ &=U_{4}-U_{3}\\ &=\cdots \\ &=U_{n}-U_{(n-1)}\\ &\\ & \end{aligned}\\\hline 2&\multicolumn{2}{|c|}{\begin{aligned}\textrm{suku ke}-n&=U_{n}.\\ U_{n}&=a+(n-1)b \end{aligned}}&\begin{aligned}U_{t}&=\displaystyle \frac{U_{1}+U_{n}}{2}\\ &=\textrm{Suku tengah} \end{aligned}\\\hline 3&\multicolumn{2}{|c|}{\begin{aligned}\textrm{Jumlah}&\:n\: \textrm{suku pertama}=S_{n}\\ S_{n}&=\displaystyle \frac{1}{2}n\left ( a+U_{n} \right )\\ &=\displaystyle \frac{1}{2}n\left ( 2a+(n-1)b \right )\\ &\\ \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{beda baru}=b'=\displaystyle \frac{U_{m}-U_{1}}{k+1} \end{aligned}\\\hline \end{array}.

C. Barisan dan Deret Geometri 

Perhatikan pula tabel berikut;

\begin{array}{|c|c|c|l|}\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&\multicolumn{2}{|c|}{\begin{aligned}\textrm{suku ke}-n&=U_{n}.\\ U_{n}&=a.r^{(n-1)} \end{aligned}}&\begin{aligned}U_{t}&=\displaystyle \sqrt{a.U_{n}}\\ &=\textrm{Suku tengah} \end{aligned}\\\hline 3&\multicolumn{2}{|c|}{\begin{aligned}\textrm{Jumlah}&\:n\: \textrm{suku pertama}=S_{n}\\ S_{n}&=\displaystyle \frac{a(r^{n}-1)}{r-1}\\ &\\ & \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 &\multicolumn{2}{|c|}{\textrm{Deret tak Hingga}}&\textrm{Hubungan}\\\cline{2-3} 4&\textrm{Konvergen}&\textrm{Divergen}&\textrm{suku dan jumlah}\\\cline{2-3} &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}.

Catatan :

Deret Geometri divergen berarti Deret Geometri tersebut memiliki jumlah sedang divergen tidak memiliki jumlah.

\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}.

\begin{array}{ll}\\ \fbox{1}&\textrm{Tentukanlah Jenis barisan atau deretnya},\: U_{100},\: S_{100},\: S_\infty \textrm{(bagi yang memiliki)}\\ &\begin{array}{ll}\\ \textrm{a}.&1,3,5,7,11,...\\ \textrm{b}.&1+3+5+7+11+...\\ \textrm{c}.&1,2,4,8,16,...\\ \textrm{d}.&1+2+4+8+16+...\\ \textrm{e}.&1,\displaystyle \frac{1}{2},\displaystyle \frac{1}{4},\displaystyle \frac{1}{8},\displaystyle \frac{1}{16},...\\ \textrm{f}.&1+\displaystyle \frac{1}{2}+\displaystyle \frac{1}{4}+\displaystyle \frac{1}{8}+\displaystyle \frac{1}{16}+...\\ \textrm{g}.&1,-1,1,-1,1,-1,1,-1,...\\ \textrm{h}.&1-1+1-1+1-1+1-1+... \end{array}\end{array}.

Jawab:

\textrm{Untuk jenis barisan atau deretnya, maka}\\ \begin{array}{ll}\\ .&\begin{array}{ll}\\ \textrm{a}.&1,3,5,7,11,...\textrm{(Barisan aritmetika)}\\ \textrm{b}.&1+3+5+7+11+...\textrm{(Deret Aritmetika)}\\ \textrm{c}.&1,2,4,8,16,...\textrm{(Barisan Geometri)}\\ \textrm{d}.&1+2+4+8+16+...\textrm{(Deret Geometri Divergen)}\\ \textrm{e}.&1,\displaystyle \frac{1}{2},\displaystyle \frac{1}{4},\displaystyle \frac{1}{8},\displaystyle \frac{1}{16},...\textrm{(Barisan Geometri)}\\ \textrm{f}.&1+\displaystyle \frac{1}{2}+\displaystyle \frac{1}{4}+\displaystyle \frac{1}{8}+\displaystyle \frac{1}{16}+...\textrm{(Deret Geometri Konvergen)}\\ \textrm{g}.&1,-1,1,-1,1,-1,1,-1,...\textrm{(Barisan Geometri)}\\ \textrm{h}.&1-1+1-1+1-1+1-1+...\textrm{(Deret Geometri Divergen)} \end{array} \end{array}.

\textrm{Untuk}\: \: U_{100},\: S_{100},\: \textrm{dan}\: S_{\infty }\textrm{(jika ada)},\: \textrm{adalah}:\\\\ \begin{array}{|l|l|c|l|}\hline U_{100}(\textrm{suku ke-100})&S_{100}(\textrm{jumlah 100 suku pertama})&S_{\infty }(\textrm{jumlah tak hingga})&\textrm{Keterangan}\\\hline \begin{aligned}\textrm{a}.\quad U_{n}&=a+(n-1).b\\ U_{100}&=1+(100-1).2\\ &=1+198\\ &=199\\ & \end{aligned}&\begin{aligned}S_{n}&=\displaystyle \frac{1}{2}n\left ( a+U_{n} \right )\\ S_{100}&=\displaystyle \frac{1}{2}.100(a+U_{100})\\ &=50(1+199)\\ &=10000 \end{aligned}&&\begin{cases} a & =U_{1}=1 \\ b & = U_{2}-U_{1}=3-1=2 \end{cases}\\\cline{1-2} \textrm{b}.\: \: U_{100}=199&S_{100}=10000&\textrm{tidak ada}&\\\cline{1-2}\cline{4-4} \begin{aligned}\textrm{c}.\quad U_{n}&=a.r^{(n-1)}\\ U_{100}&=1.2^{(100-1)}\\ &=2^{99}\\ & \end{aligned}&\begin{aligned}S_{n}&=\displaystyle \frac{a\left ( r^{n}-1 \right )}{r-1}\\ S_{100}&=\displaystyle \frac{1.\left ( 2^{100}-1 \right )}{2-1}\\ &=2^{100}-1 \end{aligned}&&\begin{cases} a & =U_{1}=1 \\ r & =\displaystyle \frac{U_{2}}{U_{1}}=\frac{2}{1}=2 \end{cases}\\\cline{1-2} \textrm{d}.\: \: U_{100}=2^{99}&S_{100}=2^{100}-1&&\\\hline \end{array}.

\begin{array}{|l|l|c|l|}\hline \begin{aligned}\textrm{e}.\quad U_{n}&=a.r^{(n-1)}\\ U_{100}&=1.\left ( \displaystyle \frac{1}{2} \right )^{(100-1)}\\ &=\left ( \displaystyle \frac{1}{2} \right )^{99}\\ &=\displaystyle \frac{1}{2^{99}}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}S_{n}&=\displaystyle \frac{1.\left ( 1-r^{n} \right )}{1-r}\\ S_{100}&=\displaystyle \frac{1\left ( 1-\left ( \displaystyle \frac{1}{2} \right )^{100} \right )}{1-\displaystyle \frac{1}{2}}\\ &=2\left ( 1-\displaystyle \frac{1}{2^{100}} \right )\\ &=2\left ( \displaystyle \frac{2^{100}-1}{2^{100}} \right )\\ &=\left ( \displaystyle \frac{2^{100}-1}{2^{99}} \right ) \end{aligned}&\begin{aligned}S_{\infty } &=\displaystyle \frac{a}{1-r}= \frac{1}{1-\frac{1}{2}}\\ &=\displaystyle \frac{1}{\left ( \frac{1}{2} \right )}\\ &=2 \end{aligned}&\begin{cases} a & =U_{1}=1 \\ r & =\displaystyle \frac{U_{2}}{U_{1}}=\frac{\left ( \frac{1}{2} \right )}{1}=\frac{1}{2} \end{cases}\\\cline{1-2} \textrm{f}.\: \: U_{100}=\displaystyle \frac{1}{2^{99}}&S_{100}=\left ( \displaystyle \frac{2^{100}-1}{2^{99}} \right )&&\\\hline \begin{aligned}\textrm{g}.\quad U_{n}&=a.r^{(n-1)}\\ U_{100}&=1.\left (-1 \right )^{(100-1)}\\ &=\left ( -1 \right )^{99}\\ &=-1\\ &\\ & \end{aligned}&\begin{aligned}S_{n}&=\displaystyle \frac{1.\left ( 1-r^{n} \right )}{1-r}\\ S_{100}&=\displaystyle \frac{1\left ( 1-\left ( -1 \right )^{100} \right )}{1-(-1)}\\ &=\displaystyle \frac{1(1-1)}{1+1}\\ &=0 \end{aligned}&\begin{aligned}&\textrm{Lihat Syaratnya}\\ &\textrm{bahwa}\\ &\left | r \right |< 1\\ &\textrm{sehingga}\\ &r=-1\\ &\textrm{(tidak memenuhi)} \end{aligned}&\begin{cases} a & =U_{1}=1 \\ r & =\displaystyle \frac{U_{2}}{U_{1}}=\frac{-1}{1}=-1 \end{cases}\\\cline{1-2} \textrm{h}.\: \: U_{100}=-1&S_{100}=0&\begin{aligned}&\textrm{atau}\: S_\infty\: \textrm{tak ada} \end{aligned}&\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Perhatikanlah Contoh Soal No.1 a dan c}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \textrm{Jika pada No.1 a di antara 2 buah suku disisipkan sebuah suku}\\ \textrm{b}.\quad \textrm{Jika pada No.1 a di antara 2 buah suku disisipkan tiga suku}\\ \textrm{c}.\quad \textrm{Jika pada No.1 c di antara 2 buah suku disisipkan sebuah suku}\\ \textrm{d}.\quad \textrm{Jika pada No.1 c di antara 2 buah suku disisipkan tiga suku} \end{array} \end{array}.

Jawab:

\begin{array}{|c|c|c|c|c|}\hline &\multicolumn{4}{c|}{\textrm{Sisipan untuk Barisan}}\\\cline{2-5} \raisebox{1.5ex}[0cm][0cm]{No} &\textrm{Aritmetika}&\textrm{Aritmetika}&\textrm{Geometri}&\textrm{Geometri}\\\hline &\multicolumn{2}{|c|}{1\underbrace{\cdots }3\underbrace{\cdots }5\underbrace{\cdots }7\cdots} &\multicolumn{2}{|c|}{1\underbrace{\cdots }2\underbrace{\cdots }4\underbrace{\cdots }8\cdots }\\\cline{2-5} 1&\textrm{disisipkan sebuah suku}&\textrm{disisipkan tiga buah suku}&\textrm{disisipkan sebuah suku}&\textrm{disisipkan tiga buah suku}\\\cline{2-5} &2.\: a&2.\: b&2.\: c&2.\: d\\\hline 2&\begin{aligned}b'&=\displaystyle \frac{U_{2}-U_{1}}{k+1}\\ &=\displaystyle \frac{3-1}{1+1}=\frac{2}{2}\\ &=1 \end{aligned}&\begin{aligned}b'&=\displaystyle \frac{U_{2}-U_{1}}{k+1}\\ &=\displaystyle \frac{3-1}{3+1}=\frac{2}{4}\\ &=\frac{1}{2} \end{aligned}&\begin{aligned}r'&=\displaystyle \sqrt[k+1]{\displaystyle \frac{U_{2}}{U_{1}}}\\ &=\displaystyle \sqrt[1+1]{\displaystyle \frac{2}{1}}=\sqrt{2}\\ & \end{aligned}&\begin{aligned}r'&=\displaystyle \sqrt[k+1]{\displaystyle \frac{U_{2}}{U_{1}}}\\ &=\displaystyle \sqrt[3+1]{\displaystyle \frac{2}{1}}=\sqrt[4]{2}\\ & \end{aligned}\\\hline 3&\multicolumn{4}{|c|}{\textrm{Setelah sisipan dimasukkan, maka barisan sukunya akan menjadi}}\\\cline{2-5} &1,2,3,4,5,6,7,...&1,1\frac{1}{2},2,2\frac{1}{2},3,3\frac{1}{2},4,...&1,\sqrt{2},2,2\sqrt{2},4,4\sqrt{2},...&1,\sqrt[4]{2},\sqrt[4]{4},\sqrt[4]{8},2,2\sqrt[4]{2},... \\\hline \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui sebuah barisan aritmetika 5,8,11,...,131. Jika banyaknya suku adalah ganjil, carilah}\\ &\textrm{a}.\quad \textrm{suku tengahnya}\\ &\textrm{b}.\quad \textrm{suku keberapakah suku tengahnya tersebut}\\ &\textrm{c}.\quad \textrm{banyak suku barisan tersebut adalah} \end{array}.

Jawab:

\begin{array}{|c|c|c|}\hline \textrm{Suku tengah}=U_{t}&\textrm{Posisi suku tengah}&\textrm{Banyak suku barisan tersebut}\\\hline \begin{aligned}\textrm{Diketahui}&\begin{cases} U_{1} & =a=5,\begin{cases} U_{2} & =8 \\ b & =3 \end{cases} \\ U_{n} & =U_{2t-1}=131 \end{cases}\\ \textrm{maka},\: \: U_{t}&=\displaystyle \frac{1}{2}\left ( a+U_{2t-1} \right )\\ &=\displaystyle \frac{1}{2}\left ( 5+131 \right )\\ &=\displaystyle \frac{136}{2}\\ U_{t}&=68\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}U_{n}&=a+(n-1)b\\ \textrm{ma}&\textrm{ka}\\ U_{t}&=a+(t-1)b\\ 68&=5+(t-1)3\\ 63&=(t-1)3\\ \displaystyle \frac{63}{3}&=t-1\\ 21&=t-1\\ 22&=t\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\left (2t-1 \right )&\: \: \textrm{adalah banyak suku}\\ \left (2t-1 \right )&=2(22)-1\\ &=44-1\\ &=43\\ \textrm{atau}&\\ U_{2t-1}=U_{n}&=131\\ a+(n-1)b&=131\\ 5+(n-1)3&=131\\ (n-1)3&=126\\ n-1&=\displaystyle \frac{126}{3}\\ n-1&=42\\ n&=43\end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textbf{(EBTANAS 1994)}\textrm{Diketahui sebuah deret bilangan 10+11+12+13+...+99}.\\ & {\textrm{Dari deret bilangan tersebut, jumlah bilangan yang habis dibagi 2 tetapi tidak habis digai oleh 5 adalah...}}\\ &\textrm{a}.\quad 950\\ &\textrm{b}.\quad 1480\\ &\textrm{c}.\quad 1930\\ &\textrm{d}.\quad 1980\\ &\textrm{e}.\quad 2430\\ \end{array}

Jawab:

\begin{array}{|l|l|l|}\hline (1 )\textrm{Deret mula-mula}&(2 )\textrm{Deret yang habis dibagi 2}&(3)\textrm{Habis dibagi 5 dari deret yang habis dibagi 2}\\\hline 10+11+12+13+\cdots +99&10+12+14+16+\cdots +98&10+20+30+40+50+60+70+80+90\\ &\begin{cases} \bullet \quad U_{1} & =a=10\\ \bullet \quad U_{2} & =12 \\ \bullet \quad b&=U_{2}-U_{1}=2\\ \bullet \quad U_{n} & =a+(n-1)b\\ \qquad 98&=10+(n-1)2\\ \qquad 45&=n \end{cases}&\begin{aligned}&=10(1+2+3+4+5+6+7+8+9)\\ &=10\left ( 45 \right )\\ &=450\\ &\\ &\\ & \end{aligned}\\\cline{2-2} &\begin{aligned}S_{45}&=\frac{1}{2}45\left ( 10+98 \right )\\ &=2430 \end{aligned}&\\\cline{2-3} &\multicolumn{2}{|l|}{\begin{aligned}&\textrm{Sehingga deret bilangan yang habis dibagi 2 tetapi tidak habis dibagi 5 adalah}\\ &=\textrm{kolom(2)-kolom(3)}\\ &=2430-450\\ &=1980 \end{aligned}}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\: \: \textrm{adalah akar-akar dari persamaan kuadrat}\: \: x^{2}+bx+c=0.\\ &\textrm{Jika}\: \: 12,\: x_{1},\: x_{2}\: \: \textrm{adalah tiga suku pertama dari barisan aritmetika dan}\: \: x_{1},\: x_{2},\: 4\: \: \textrm{adalah}\\ &\textrm{tiga suku pertama barisan geometri. Maka diskriminan persamaan kuadrat tersebut adalah ...}\\ \end{array}.

Jawab:

\begin{array}{|c|c|c|c|c|}\hline \textrm{Barisan}&\multicolumn{2}{c|}{\textrm{Sintak/Tahapan}}&\textrm{Diskriminan(D)}\\\hline \begin{cases} Aritmetika(BA) & :\: 12,\: x_{1},\: x_{2} \\ Geometri(BG) & :\: x_{1},\: x_{2},\: 4 \end{cases}&\begin{aligned}&\textrm{pada BA berlaku}\\ &2U_{2}=U_{1}+U_{3}\\ &2x_{1}=12+x_{2}.......\textcircled{1}\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{pada BG berlaku}\\ &U_{2}^{2}=U_{1}\times U_{3}\\ &x_{2}^{2}=4x_{1}.........\textcircled{2}\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Persamaan Kuadrat(PK)}\\ &\textrm{untuk}\: x_{1}=9\: \textrm{dan}\: x_{2}=6,\: \textrm{maka}\\ &x^{2}+bx+c=0\\ &x^{2}+\left ( -\left ( x_{1}+x_{2} \right ) \right )x+\left ( x_{1}.x_{2} \right )=0\\ &x^{2}-15x+54=0\\ &\\ &D=b^{2}-4ac\\ &D=(-15)^{2}-4.1.54=9\\ & \end{aligned}\\\cline{1-3} \multicolumn{3}{|l|}{\begin{aligned}&\textrm{Proses penyelesaian : pers}\: \textcircled{1}\: \textrm{disubstitusikan ke pers}\: \textcircled{2},\: \textrm{yaitu};\\ &x_{2}^{2}=4x_{1}=2\left ( 2x_{1} \right )=2\left ( 12+x_{2} \right )=24+2x_{2}\qquad (\textrm{menjadi pers kuadrat})\\ &x_{2}^{2}-2x_{2}-24=0\\ &\left ( x_{2}-6 \right )\left ( x_{2}+4 \right )=0\\ &x_{2}=6\: \: \textrm{V}\: \: x_{2}=-4\\ &\textrm{untuk}\: x_{2}=6\: \Rightarrow \: x_{1}=9\\ &\textrm{untuk}\: x_{2}=-4\: \Rightarrow \: x_{1}=4\\ & \end{aligned}}&\begin{aligned}&\textrm{Persamaan Kuadrat(PK)}\\ &\textrm{untuk}\: x_{1}=4\: \textrm{dan}\: x_{2}=-4,\: \textrm{maka}\\ &x^{2}+bx+c=0\\ &x^{2}+\left ( -\left ( x_{1}+x_{2} \right ) \right )x+\left ( x_{1}.x_{2} \right )=0\\ &x^{2}-0x+(-16)=0\\ &x^{2}-16=0\\ &\\ &D=b^{2}-4ac\\ &D=0^{2}-4.1.(-16)=64 \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textbf{(SPMB 2007)}\textrm{Tiga buah bilangan merupakan suku-suku dari deret aritmetika. Jika suku pertama dikurangi 2} \\ &\textrm{dan suku ketiga ditambah 6, maka akan menjadi barisan geometri dengan rasio 2. Hasil kali ketiga bilangan pada}\\ &\textrm{barisan geometri tersebut adalah}...\\ &\textrm{a}.\quad 128\quad\quad\quad\quad\quad\quad\quad\: \textrm{d}.\quad 480\\ &\textrm{b}.\quad 240\qquad \textrm{c}.\quad 256\qquad \textrm{e}.\quad 512\\ \end{array}.

Jawab:

\begin{array}{|l|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Barisan}}&\textrm{Jadi},\\\cline{1-2} \textrm{Aritmetika}&\textrm{Geometri}&\\\cline{1-2} U_{1},U_{2},U_{3}&U_{1}-2,U_{2},U_{3}+6&\\\cline{1-2} 2U_{2}=U_{1}+U_{3}.....\textcircled{1}&\textrm{Rasio}=2=\displaystyle \frac{U_{2}}{U_{1}-2}=\frac{U_{3}+6}{U_{2}}&\\\cline{2-2} &\begin{aligned}\textrm{maka},&\\ 2=\: &\displaystyle \frac{U_{2}}{U_{1}-2}\Leftrightarrow U_{1}=\displaystyle \frac{1}{2}U_{2}+2\: ..........\textcircled{2}\\ 2=\: &\displaystyle \frac{U_{3}+6}{U_{2}}\Leftrightarrow U_{3}=\displaystyle 2U-6\: ............\textcircled{3} \end{aligned}&\begin{aligned}&\left ( U_{1}-2 \right )\times U_{2}\times \left ( U_{3}+6 \right )\\ &=(6-2)(8)(10+6)\\ &=4.8.16=512 \end{aligned}\\\cline{1-2} \textrm{Pers}\: \textcircled{2}\: \textrm{dan}\: \textcircled{3}\: \textrm{disubstitusi ke}\: \textcircled{1}&&\\\cline{1-1} \multicolumn{2}{|l|}{\begin{aligned}2U_{2}&=U_{1}+U_{3}\\ 2U_{2}&=\left (\displaystyle \frac{1}{2}U_{2}+2 \right )+\left ( 2U_{2}-6 \right )\\ -\displaystyle \frac{1}{2}U_{2}&=-4\\ U_{2}&=8......\textcircled{4}\\ \textcircled{2}\Rightarrow U_{1}&=\displaystyle \frac{1}{2}U_{2}+2\\ &=\displaystyle \frac{1}{2}(8)+2=6\\ \textcircled{2}\Rightarrow U_{1}&=\displaystyle 2U_{2}-6\\ &=2(8)-6=10 \end{aligned} }&\\\hline \end{array}.

\LARGE{\fbox{\LARGE{\fbox{LATIHAN SOAL}}}}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah lima suku pertama dari barisan-barisan berikut}\\ &\textrm{a}.\quad U_{n}=4n+3\\ &\textrm{b}.\quad U_{n}=2^{n-1}\\ &\textrm{c}.\quad U_{n}=2^{n}+3\\ &\textrm{d}.\quad U_{n}=\displaystyle \frac{1}{n^{2}}-n\\ &\textrm{e}.\quad U_{n}=5n^{2}+2n-1\\ &\textrm{f}.\quad U_{n}=\left ( n-1 \right )^{2}+3\\ &\textrm{g}.\quad U_{n}=5^{n}+n\\ &\textrm{h}.\quad U_{n}=\displaystyle \frac{n^{2}-1}{n+1} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukanlah rumus suku ke-\textit{n} dari barisan-barisan berikut}\\ &\textrm{a}.\quad 2,\: 8,\: 14,\: 20,\: 26,...\\ &\textrm{b}.\quad 1,\: 2,\: 4,\: 8,\: 16,...\\ &\textrm{c}.\quad 5,\: -1,\: -7,\: -13,\: -19,...\\ &\textrm{d}.\quad 1,\: -2,\: 4,\: -8,\: 16,...\\ &\textrm{e}.\quad 6,\: 11,\: 16,\: 21,\: 26,...\\ &\textrm{f}.\quad 1,\: \displaystyle \frac{1}{3},\: \frac{1}{9},\: \frac{1}{27},\: \frac{1}{81},...\\ &\textrm{g}.\quad -15,\: -7,\: 1,\: 9,\: 17,...\\ &\textrm{h}.\quad \sqrt{3},\: \sqrt{6},\: 2\sqrt{3},\: 2\sqrt{6},\: 4\sqrt{3},... \end{array}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Jika}\: \: a+2,\: 3a-1,\: \textrm{dan}\: 10a+3\: \: \textrm{membentuk barisan geometri, maka hasil kali}\\ &\textrm{semua nilai}\: \: a\: \: \textrm{yang memenuhi adalah}\: ....\\ &\textrm{A}.\quad 5\qquad \textrm{B}.\quad 8\qquad \textrm{C}.\quad 10\qquad \textrm{D}.\quad 15\\ &\qquad\qquad\qquad\qquad (\textrm{\textbf{Seleksi Penerimaan Mahasiswa Unnes 2010 Kode Soal 4928}}) \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Penataan kursi dalam suatu gedung pertunjukan membentuk deret aritmetika. Baris terdepan memuat}\\ &\textrm{20 kursi dan baris kedua belas (baris terakhir) memuat 86 kursi. Berapakah banyak kursi dalam gedung} \\ &\textrm{pertunjukan tersebut?}\\ &\textrm{A}.\quad 466\qquad \textrm{B}.\quad 586\qquad \textrm{C}.\quad 606\qquad \textrm{D}.\quad 636\\ &\qquad\qquad\qquad\qquad (\textrm{\textbf{Seleksi Penerimaan Mahasiswa Unnes 2010 Kode Soal 4928}}) \end{array}.

Sumber Referensi

  1. Kuntarti, Sulistiyono dan Sri Kurnianingsih. 2007. Matematika  SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: esis.

 

 

 

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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