Lanjutan 2 Materi Barisan dan Deret (Deret Aritmetika-Geometri tak Berhingga)

Sebelum kita membahas materi seperti judul di atas, Anda dapat mengulik materi sebelumnya tentang barisan dan deret di link berikut:

• link 1 (Barisan dan Deret)
• link 2 (Barisan dan Deret)

Deret Aritmetika dan Geometri Sekaligus

Perhatikan barisan bilangan berikut

$\begin{array}{rl}& a,(a+b)r,(a+2b)r^2,(a+3b)r^3,\cdots ,(a+(n-1)b)r^{n-1}\end{array}$.

Jika deretnya berhingga maka jumlah deretnya adalah

$\begin{array}{ll}\begin{array}{rl}{S}_n& =a+(a+b)r+(a+2b)r^2+\cdots +(a+(n-1)b)r^{n-1}\\ r{S}_n& =ar+(a+b)r^2+(a+2b)r^3+\cdots +(a+(n-1)b)r^n\end{array}& -\\ \begin{array}{rl}(1-r)S_n& =a-(a+(n-1)b)r^n+{\displaystyle \frac{br(1-r^{n-1})}{(1-r)}}\\ \iff S_n& ={\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r{)}^2}}\end{array}\end{array}$.

Jika deretnya tak berhingga, maka nilai  $S_n$  bergantung pada nilai  $\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle r^n}$

• Jika  $\left|r\right|<1$, maka deretnya konvergen (memiliki jumlah atau jumlahnya dapat ditentukan), yaitu : $S_{\mathrm{\infty }}={\displaystyle \frac{a}{1-r}+\frac{br}{(1-r{)}^2}}$ dengan suku awal deret geometrinya adalah 1.
• Jika  $\left|r\right|\ge 1$, maka deret tak memiliki jumlah yang pas (jumlahnya tidak dapat ditentukan)

Bukti :

untuk $\left|r\right|<1$. Diketahui bahwa  $\begin{array}{rl}{S}_n& ={\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r{)}^2}}\end{array}$. Karena harga  $\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle r^n=0}$, maka jumlah deretnya adalah:

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle S}& =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br-b{r}^n}{(1-r{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a-0}{(1-r)}+\frac{br-0}{(1-r{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ & ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}◼}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah jumlah nilai dari}\\ & {\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots }\\ & \text{dengan}\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=0\\ \bullet b& =U_2-U_1\\ & =1-0=1\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}{S}_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ & ={\displaystyle \frac{0}{1-\frac{1}{2}}+\frac{1\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=0+{\displaystyle \frac{\frac{1}{2}}{\frac{1}{4}}}}\\ & =0+{\displaystyle \frac{4}{2}}\\ & =2\end{array}\\ & \text{Jadi},{\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots =2}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah jumlah nilai dari}\\ & {\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 2S_{\mathrm{\infty }}=1+{\displaystyle \frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=3-1=2\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}2S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 2S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{2}}+\frac{2\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=2+4=6}\\ S_{\mathrm{\infty }}& =3\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots =3}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah jumlah nilai dari}\\ & 2+{\displaystyle \frac{5}{2}+\frac{8}{2^2}+\frac{11}{2^3}+\frac{14}{2^4}+\frac{17}{2^5}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}=2+{\displaystyle \frac{5}{2}+\frac{8}{2^2}+\frac{11}{2^3}+\frac{14}{2^4}+\frac{17}{2^5}+\cdots }\\ & \text{dengan}\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=2\\ \bullet b& =U_2-U_1=5-2=3\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}{S}_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ & ={\displaystyle \frac{2}{1-\frac{1}{2}}+\frac{3\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=4+6=10}\end{array}\\ & \text{Jadi},2+{\displaystyle \frac{5}{2}+\frac{8}{2^2}+\frac{11}{2^3}+\frac{14}{2^4}+\frac{17}{2^5}+\cdots =10}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

Tentukan besar jumlah dari deret berikut

$\begin{array}{rl}1.& 1+{\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\frac{6}{243}+\cdots }\\ 2.& 1+{\displaystyle \frac{3}{3}+\frac{5}{9}+\frac{7}{27}+\frac{9}{81}+\frac{11}{243}+\cdots }\\ 3.& 1+{\displaystyle \frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+\frac{5}{5^4}+\frac{6}{5^5}+\cdots }\\ 4.& 1+{\displaystyle \frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+\frac{11}{7^5}+\cdots }\\ 5.& {\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\frac{6}{243}+\frac{7}{729}+\cdots }\\ 6.& {\displaystyle \frac{3}{3}+\frac{5}{9}+\frac{7}{27}+\frac{9}{81}+\frac{11}{243}+\frac{13}{729}+\cdots }\\ 7.& {\displaystyle \frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+\frac{5}{5^4}+\frac{6}{5^5}+\frac{7}{5^6}+\cdots }\\ 8.& {\displaystyle \frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+\frac{11}{7^5}+\frac{13}{7^6}+\cdots }\end{array}$.

DAFTAR PUSTAKA

1. Thohir, A. 2013. Barisan dan Deret Materi Pendamping Olimpiade Matematika MA/SMA. Grobogan: MA Futuhiyah.

SUMBER INTERNET

https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence 

Lanjutan 3 (Barisan & Deret)

$\begin{array}{l}\\ 11.& \text{Hasil penjumlahan dari}\\ & {\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\cdots +\frac{1}{97.99}=\cdots }\\ & \begin{array}{l}\\ \text{A}.& {\displaystyle \frac{98}{99}}& & & \text{D}.& {\displaystyle \frac{48}{99}}\\ \\ \text{B}.& {\displaystyle \frac{50}{99}}& \text{C}.& {\displaystyle \frac{49}{99}}& \text{E}.& {\displaystyle \frac{47}{99}}\end{array}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& \text{Bentuk di atas memenuhi bentuk}\\ & {\displaystyle \frac{1}{x(x+2)}={\displaystyle \frac{1}{2}\left({\displaystyle \frac{1}{x}-\frac{1}{(x+2)}}\right).\text{Bentuk ini pada}}}\\ & \text{bilangan dengan pola tertentu seperti di atas akan}\\ & \text{menghabiskan dengan bilangan sebelahnya}\\ & \text{atau lazim dikenal dengan}\mathbf{\text{prinsip teleskoping}}\\ & \text{Sebagaimana bentuk penjumlahan dengan pola di atas}\\ & \text{maka}\\ & ={\displaystyle \frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\cdots +\frac{1}{97}-\frac{1}{99})}\\ & ={\displaystyle \frac{1}{2}(1-\frac{1}{99})={\displaystyle \frac{1}{2}.\frac{98}{99}={\displaystyle \frac{49}{99}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Diketahui}\\ & x={\displaystyle \frac{1+p+p^2+p^3+\cdots +p^{n-1}}{1+p+p^2+p^3+\cdots +p^{n-2}+p^{n-1}+p^n}}\\ & y={\displaystyle \frac{1+q+q^2+q^3+\cdots +q^{n-1}}{1+q+q^2+q^3+\cdots +q^{n-2}+q^{n-1}+q^n}}\\ \\ & \text{dan}p>q>0\\ \\ & \text{Tunjukkan bahwa}x<y\\ \\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}:p>q>0\\ & \text{sehingga}\\ & {\displaystyle \frac{1}{p}<\frac{1}{q},{\displaystyle \frac{1}{p^2}<\frac{1}{q^2},\cdots ,{\displaystyle \frac{1}{p^n}<\frac{1}{q^n}}}}\\ & \text{Jika bentuk di atas dijumlahkan, maka}\\ & {\displaystyle \frac{1}{p}+\frac{1}{p^2}+\cdots +\frac{1}{p^n}<\frac{1}{q}+\frac{1}{q^2}+\cdots +\frac{1}{q^n}}\\ & \iff {\displaystyle \frac{p^{n-1}+\cdots +p^2+p+1}{p^n}<{\displaystyle \frac{q^{n-1}+\cdots +q^2+q+1}{q^n}}}\\ & \iff {\displaystyle \frac{p^n}{1+p+p^2+\cdots +p^{n-1}}>{\displaystyle \frac{q^n}{1+q+q^2+\cdots +q^{n-1}}}}\\ & \iff {\displaystyle \frac{p^n}{1+p+p^2+\cdots +p^{n-1}}+1>{\displaystyle \frac{q^n}{1+q+q^2+\cdots +q^{n-1}}+1}}\\ & \iff {\displaystyle \frac{1+p+p^2+\cdots +p^{n-1}+p^n}{1+p+p^2+\cdots +p^{n-1}}>{\displaystyle \frac{1+q+q^2+\cdots +q^{n-1}+q^n}{1+q+q^2+\cdots +q^{n-1}}}}\\ & \iff {\displaystyle \frac{1+p+p^2+\cdots +p^{n-1}}{1+p+p^2+\cdots +p^{n-1}+p^n}<{\displaystyle \frac{1+q+q^2+\cdots +q^{n-1}}{1+q+q^2+\cdots +q^{n-1}+q^n}}}\\ & \iff x<y◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Aziz, A. 2016. Rahasia Juara Olimpiade Matematika SMA. Yogyakarta: ANDI.
2. Baskoro, B.D. 2012. Aljabar dan Trigonometri Cespleng Olimpiade Matematika. Yogyakarta: BERLIAN.
3. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Nasional & Internasional. Yogyakarta: INDONESIA CERDAS.
4. Idris, M,. Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
5. Sembiring, S. 2002. Olimpiade Matematika Untuk SMU. Bandung: YRAMA WIDYA.
6. Sembiring, S., Suparmin, S. 2015. Pena Emas OSN Matematika SMA. Bandung: YRAMA WIDYA.

Lanjutan 2 (Barisan & Deret)

$\begin{array}{l}\\ 9.& \text{Tentukan hasil dari}\\ & {\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \text{Misalkan}S={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{maka}{\displaystyle \frac{1}{2}S={\displaystyle \frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\frac{7}{32}+\frac{9}{64}+\cdots }}\\ & \text{Jika}\\ & S-{\displaystyle \frac{1}{2}S={\displaystyle \frac{1}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\cdots }}\\ & \begin{array}{rl}\iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{1}{2}+\underset{\text{deret geometri}}{\underbrace{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots )}}}\\ \iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{1}{2}+{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}+{\displaystyle \frac{\frac{1}{2}}{\frac{1}{2}}={\displaystyle \frac{1}{2}+1={\displaystyle \frac{3}{2}}}}}}}\\ \iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{3}{2}}\\ \iff S& =3\end{array}\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 2S_{\mathrm{\infty }}=1+{\displaystyle \frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=3-1=2\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}2S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 2S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{2}}+\frac{2\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=2+4=6}\\ S_{\mathrm{\infty }}& =3\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots =3}\end{array}\end{array}$.

$.\text{Untuk link materi pada pembahasan alternatif 2}$. di sini

$\begin{array}{l}\\ 10.& \text{Hasil penjumlahan dari}\\ & {\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots =\cdots }\\ & \begin{array}{l}\\ \text{A}.& {\displaystyle \frac{2}{3}}& & & \text{D}.& {\displaystyle \frac{4}{3}}\\ \\ \text{B}.& {\displaystyle \frac{3}{4}}& \text{C}.& 1& \text{E}.& {\displaystyle \frac{3}{2}}\end{array}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 3S_{\mathrm{\infty }}=1+{\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=2-1=1\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{3}}{1}={\displaystyle \frac{1}{3}}}\\ \hline\end{array}\\ & \begin{array}{rl}3S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 3S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{3}}+\frac{1\times \frac{1}{3}}{(1-\frac{1}{3}{)}^2}={\displaystyle \frac{3}{2}+\frac{3}{4}={\displaystyle \frac{9}{4}}}}\\ S_{\mathrm{\infty }}& ={\displaystyle \frac{3}{4}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots ={\displaystyle \frac{3}{4}}}\end{array}\end{array}$ .

Lanjutan (Barisan & Deret)

$\begin{array}{l}\\ 5.& \mathbf{\text{(Lomba Matematika Nasional}}\\ & \mathbf{\text{HIMATIKA UGM 2006)}}\\ & \text{Jika bilangan}\\ & A={\displaystyle \frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+\cdots +\frac{1}{1+100}}\\ & B={\displaystyle \frac{1}{1+1}+\frac{1}{1+{\displaystyle \frac{1}{2}}}+\frac{1}{1+{\displaystyle \frac{1}{3}}}+\cdots +\frac{1}{1+{\displaystyle \frac{1}{100}}}}\\ & \text{maka}A+B\text{sama dengan}....\\ & \text{A}.202\text{D}.100\\ & \text{B}.200\text{C}.101\text{E}.99\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{l}\\ \begin{array}{rl}A& ={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{101}}\\ B& ={\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots +\frac{100}{101}}\end{array}& \\ & +\\ \\ A+B=\underset{100}{\underbrace{1+1+1+\cdots +1}}& =100\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \mathbf{\text{(OSN tk. Kota/Kab 2002)}}\\ & \text{Misalkan}\\ & a={\displaystyle \frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}+\cdots +\frac{{1001}^2}{2001}}\\ \\ & b={\displaystyle \frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}+\cdots +\frac{{1001}^2}{2003}}\\ & \text{Tentukanlah bilangan bulat yang}\\ & \text{nilainya paling dekat ke}a-b\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{rl}a-b& =\left({\displaystyle \frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}+\cdots +\frac{{1001}^2}{2001}}\right)\\ & -\left({\displaystyle \frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}+\cdots +\frac{{1001}^2}{2003}}\right)\\ & ={\displaystyle \frac{1^2}{1}+\left({\displaystyle \frac{2^2}{3}-{\displaystyle \frac{1^2}{3}}}\right)+\left({\displaystyle \frac{3^2}{5}-{\displaystyle \frac{2^2}{5}}}\right)}\\ & +\left({\displaystyle \frac{4^2}{7}-{\displaystyle \frac{3^2}{7}}}\right)+\cdots +\left({\displaystyle \frac{{1001}^2}{2001}-{\displaystyle \frac{{1000}^2}{2001}}}\right)\\ & -{\displaystyle \frac{{1001}^2}{2003}}\\ & =\underset{1001}{\underbrace{1+1+1+\cdots +1}}-{\displaystyle \frac{{1001}^2}{2003}}\\ & =1001-{\displaystyle \frac{{1001}^2}{2003}=1001\left({\displaystyle \frac{2003-1001}{2003}}\right)}\\ & ={\displaystyle \frac{1001\times 1002}{2003}>{\displaystyle \frac{1001}{2}=500,5}}\end{array}\\ & \text{Jadi bilangan bulat yang paling dekat}\\ & \text{ke}a-b\text{adalah}501\end{array}$.

 $\begin{array}{l}\\ 7.& \text{Misalkan}\\ & a_n={\displaystyle \frac{n(n+1)}{2},\text{tentukanlah jumlah}}\\ & \text{dari}{\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2023}}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}{a}_n& ={\displaystyle \frac{n(n+1)}{2},\text{maka}}\\ {\displaystyle \frac{1}{a_n}}& ={\displaystyle \frac{2}{n(n+1)}}\\ & =2\left({\displaystyle \frac{1}{n}-\frac{1}{n+1}}\right)\\ & \text{lihat pembahasan no.3 di atas}\\ & \end{array}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2023}}}\\ & =2((1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\cdots +\left({\displaystyle \frac{1}{2022}-\frac{1}{2023}}\right))\\ & =2(1-{\displaystyle \frac{1}{2023}})\\ & =2\left({\displaystyle \frac{2022}{2023}}\right)={\displaystyle \frac{4044}{2023}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Misalkan}n\text{adalah bilangan asli}\\ & \text{dan}\left\{a_n\right\}\text{adalah barisan bilangan real}\\ & \text{dengan}{a}_n={\displaystyle \frac{2^n}{2^{2n+1}-2^{n+1}-2^n+1}}\\ \\ & \text{Tunjukkan bahwa untuk setiap bilangan}\\ & \text{asli}n,\text{berlaku}{a}_1+a_2+\cdots +a_n<1\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}{a}_n& ={\displaystyle \frac{2^n}{2^{2n+1}-2^{n+1}-2^n+1}}\\ & ={\displaystyle \frac{2^n}{(2^{n+1}-1)(2^n-1)}}\\ & =\left({\displaystyle \frac{1}{2^{n+1}-1}-\frac{1}{2^n-1}}\right)\\ a_1& ={\displaystyle \frac{1}{2^1-1}-\frac{1}{2^2-1}=1-{\displaystyle \frac{1}{3}}}\\ a_2& ={\displaystyle \frac{1}{2^2-1}-\frac{1}{2^3-1}=\frac{1}{3}-\frac{1}{7}}\\ a_3& ={\displaystyle \frac{1}{2^3-1}-\frac{1}{2^4-1}=\frac{1}{7}-\frac{1}{15}}\\ & ⋮⋮\\ a_n& ={\displaystyle \frac{1}{2^n-1}-\frac{1}{2^{n+1}-1}+}\\ a_1& +a_2+a_3+\cdots +a_n\\ & =1-{\displaystyle \frac{1}{2^{n+1}-1}<1◼}\end{array}\end{array}$.

Selingan (Barisan & Deret)

Lanjutan contoh soal dan pembahasannya terkait barisan dan deret

$\begin{array}{l}\\ 1.& \text{Hasil kali bilangan bentuk berikut}\\ & (1-{\displaystyle \frac{1}{4}})(1-{\displaystyle \frac{1}{5}})(1-{\displaystyle \frac{1}{6}})\cdots (1-{\displaystyle \frac{1}{100}})\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{100}\text{D}.{\displaystyle \frac{4}{100}}}\\ \\ & \text{B}.{\displaystyle \frac{2}{100}\text{C}.{\displaystyle \frac{3}{100}\text{E}.{\displaystyle \frac{5}{100}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{rl}& (1-{\displaystyle \frac{1}{4}})(1-{\displaystyle \frac{1}{5}})(1-{\displaystyle \frac{1}{6}})\cdots (1-{\displaystyle \frac{1}{100}})\\ & =\left({\displaystyle \frac{3}{4}}\right)\left({\displaystyle \frac{4}{5}}\right)\left({\displaystyle \frac{5}{6}}\right)\cdots \left({\displaystyle \frac{99}{100}}\right)\\ & =\left({\displaystyle \frac{3}{\text{⧸}4}}\right)\left({\displaystyle \frac{\text{⧸}4}{\text{⧸}5}}\right)\left({\displaystyle \frac{\text{⧸}5}{\text{⧸}6}}\right)\cdots \left({\displaystyle \frac{\text{⧸}99}{100}}\right)\\ & ={\displaystyle \frac{3}{100}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Hasil kali bilangan bentuk berikut}\\ & (1-{\displaystyle \frac{2}{3}})(1-{\displaystyle \frac{2}{5}})(1-{\displaystyle \frac{2}{7}})\cdots (1-{\displaystyle \frac{2}{2023}})\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{2023}\text{D}.{\displaystyle \frac{4}{2023}}}\\ \\ & \text{B}.{\displaystyle \frac{2}{2023}\text{C}.{\displaystyle \frac{3}{2023}\text{E}.{\displaystyle \frac{5}{2023}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan cara pembahasan pada no.1 di atas, maka}\\ & \begin{array}{rl}& (1-{\displaystyle \frac{2}{3}})(1-{\displaystyle \frac{2}{5}})(1-{\displaystyle \frac{2}{7}})\cdots (1-{\displaystyle \frac{2}{2023}})\\ & =\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{3}{5}}\right)\left({\displaystyle \frac{5}{7}}\right)\cdots \left({\displaystyle \frac{2021}{2023}}\right)\\ & =\left({\displaystyle \frac{1}{\text{⧸}3}}\right)\left({\displaystyle \frac{\text{⧸}3}{\text{⧸}5}}\right)\left({\displaystyle \frac{\text{⧸}5}{\text{⧸}7}}\right)\cdots \left({\displaystyle \frac{\text{⧸}2021}{2023}}\right)\\ & ={\displaystyle \frac{1}{2023}}\end{array}\end{array}$.

 $\begin{array}{l}\\ 3.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{2022\times 2023}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& =(1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\left({\displaystyle \frac{1}{3}-\frac{1}{4}}\right)+\\ & +\cdots +\left({\displaystyle \frac{1}{2022}-\frac{1}{2023}}\right)\\ & =1-{\displaystyle \frac{1}{2023}}\\ & ={\displaystyle \frac{2022}{2023}}\end{array}\end{array}$.

$\begin{array}{|c|}\hline \begin{array}{rl}\mathbf{\text{Seb}}& \mathbf{\text{agai pengingat kita}}\\ \bullet & 1+2+3+\cdots +n={\displaystyle \frac{n(n+1)}{2}}\\ \bullet & 1+3+5+\cdots +(2n-1)=n^2\\ \bullet & 1^2+2^2+3^2+\cdots +n^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}\\ \bullet & 1^3+2^3+3^3+\cdots +n^3={\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2\\ \bullet & 1+{\displaystyle \frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +n}={\displaystyle \frac{2n}{n+1}}}\\ \bullet & {\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{n(n+1)}={\displaystyle \frac{n}{n+1}}}\\ \bullet & {\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\cdots +\frac{1}{n(n+1)(n+2)}={\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}}}\end{array}\\ \hline\end{array}$.

$\begin{array}{l}\\ 4.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{2023}-\sqrt{2022}}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{1+\sqrt{2}}}& =\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}\\ & ={\displaystyle \frac{1-\sqrt{2}}{1^2-(\sqrt{2}{)}^2}=\frac{1-\sqrt{2}}{1-2}}\\ & ={\displaystyle \frac{1-\sqrt{2}}{-1}=\sqrt{2}-1}\\ {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}}& =\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\\ & ={\displaystyle \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}{)}^2-(\sqrt{3}{)}^2}=\frac{\sqrt{2}-\sqrt{3}}{2-3}}\\ & ={\displaystyle \frac{\sqrt{2}-\sqrt{3}}{-1}=\sqrt{3}-\sqrt{2}}\\ {\displaystyle \frac{1}{\sqrt{3}+\sqrt{4}}}& =\frac{1}{\sqrt{3}+\sqrt{4}}\times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\\ & ={\displaystyle \frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}{)}^2-(\sqrt{4}{)}^2}=\frac{\sqrt{3}-\sqrt{4}}{3-4}}\\ & ={\displaystyle \frac{\sqrt{3}-\sqrt{4}}{-1}=\sqrt{4}-\sqrt{3}}\\ & ⋮\end{array}\\ & \text{Sehingga}\\ & \begin{array}{rl}& =\text{⧸}\sqrt{2}-1+\text{⧸}\sqrt{3}-\text{⧸}\sqrt{2}+\text{⧸}\sqrt{4}-\text{⧸}\sqrt{3}+\cdots +\sqrt{2023}-\text{⧸}\sqrt{2022}\\ & =\sqrt{2023}-1\end{array}\end{array}$

Lanjutan 6 Contoh Soal Barisan dan Deret

$\begin{array}{l}\\ 26.& \text{Syarat untuk deret geometri tak hingga }\\ & \text{dengan suku pertama}a\text{konvergen dengan }\\ & \text{jumlah 2 adalah}.....\\ & \text{A}.-2<a<0\\ & \text{B}.-4<a<0\\ & \text{C}.0<a<2\\ & \text{D}.0<a<4\\ & \text{E}.-4<a<4\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}{S}_{\mathrm{\infty }}=2,\text{dengan}\\ & S_{\mathrm{\infty }}={\displaystyle \frac{a}{1-r}\iff 1-r={\displaystyle \frac{a}{S_{\mathrm{\infty }}}\iff r=1-{\displaystyle \frac{a}{S_{\mathrm{\infty }}}}}}\\ & \iff -1<1-{\displaystyle \frac{a}{S_{\mathrm{\infty }}}<1\iff -2<-{\displaystyle \frac{a}{S_{\mathrm{\infty }}}<0}}\\ & \iff 0<{\displaystyle \frac{a}{S_{\mathrm{\infty }}}<2\iff \iff 0<{\displaystyle \frac{a}{2}<2}}\\ & \iff 0<a<4\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \text{Tiga bilangan membentuk barisan geometri}\\ & \text{dengan jumlah}26.\text{Jika suku tengah ditambah}\\ & \text{4 , maka terbentuklah barisan aritmetika, suku}\\ & \text{suku tengah dari barisan geometri tersebut}.....\\ & \text{A}.2\\ & \text{B}.4\\ & \text{C}.6\\ & \text{D}.10\\ & \text{E}.18\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Barisan Geometri}:U_1+U_2+U_3=26\\ & \bullet U_1+U_3=26-U_2\\ & \bullet U_2^2=U_1.U_3\\ & \text{Barisan Aritmetika}:U_1,U_2+4,U_3\\ & \bullet U_1+U_3=2(U_2+4)=2U_2+8\\ & \text{maka}\\ & 26-U_2=2U_2+8\\ & \iff -2U_2-U_2=8-26\\ & \iff -3U_2=-18\\ & \iff U_2={\displaystyle \frac{-18}{-3}=6}\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Selish suku tengah pada barisan aritmetika}\\ & \text{dengan suku pertama dan terakhir masing-}\\ & \text{masing 1 dan 25 dengan barisan geometri}\\ & \text{yang suku-sukunya positif dengan suku-suku}\\ & \text{pertama dan terakhir juga 1 dan 25 adalah}.....\\ & \text{A}.5\\ & \text{B}.\text{sekitar}7,1\\ & \text{C}.8\\ & \text{D}.13\\ & \text{E}.18\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& U_t=\text{Suku tengah}\\ & \text{Barisan Aritmetika (BA)}:U_{t_{BA}}={\displaystyle \frac{1}{2}(U_1+U_n)}\\ & \iff U_{t_{BA}}={\displaystyle \frac{1}{2}(1+25)=13}\\ & \text{Barisan Geometri (BG)}:U_t^2=U_1.U_n\\ & \iff U_{t_{BG}}=\sqrt{U_1.U_n}=\sqrt{1\times 25}=5\\ & (\text{ambil nilai yang positif})\\ & \text{maka}\\ & U_{t_{BA}}-U_{t_{BG}}=13-5=8\end{array}\end{array}$.

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

$\begin{array}{l}\\ 30.& \mathbf{\text{Soal Mat SNMPTN}}\\ & \text{Agar deret geometri}{\displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}}\\ & \text{jumlahnya memiliki limit, maka nilai}x\\ & \text{harus memenuhi}....\\ & \text{A}.x>0\\ & \text{B}.x<1\\ & \text{C}.0<x<1\\ & \text{D}.x>2\\ & \text{E}.x<0\text{atau}x>2\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Deret Geometri (DG)}:{\displaystyle \frac{x-1}{x},\frac{1}{x},\frac{1}{x(x-1)}}\\ & r={\displaystyle \frac{\frac{1}{x}}{\frac{x-1}{x}}={\displaystyle \frac{1}{x-1}}}\\ & \text{Syarat DG memiliki limit (konvergen)}:\left|r\right|<1\\ & \iff -1<r<1\\ & \iff -1<{\displaystyle \frac{1}{x-1}<1}\end{array}\\ & \begin{array}{rl}& \text{Selesaian 1}\\ & -1<{\displaystyle \frac{1}{x-1}\iff {\displaystyle \frac{1}{x-1}+1>0}}\\ & \iff {\displaystyle \frac{1}{x-1}+\frac{x-1}{x-1}>0\iff \frac{x}{x-1}>0}\\ & \text{Selesaian 2}\\ & {\displaystyle \frac{1}{x-1}<1\iff {\displaystyle \frac{1}{x-1}-1<0}}\\ & \iff {\displaystyle \frac{1}{x-1}-\frac{x-1}{x-1}<0\iff \frac{-x+2}{x-1}<0}\\ & \\ & \text{HP}:\{x<0\text{atau}x>2\}\end{array}\\ & \mathbf{\text{Berikut ilustrasi garis bilangannya}}\\ & \begin{array}{c}\\ (1)& +& +& -& -& -& -& -& +& +& +& +& +\\ & & 0& & & & & 1& & & & \\ \\ (2)& -& -& -& -& -& -& -& +& +& +& +& -\\ & & & & & & & 1& & & & 2& \\ \end{array}\end{array}$.

Lanjutan 5 Contoh Soal Barisan dan Deret

$\begin{array}{l}\\ 21.& \text{Jika jumlah}n\text{suku pertama suatu barisan}\\ & \text{adalah}{S}_n=n^3+2n,\text{maka suku keempat}\\ & \text{adalah}....\\ & \text{A}.33\text{D}.63\\ & \text{B}.39\text{C}.49\text{E}.72\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui jumlah dari suatu barisan bilangan}\\ & \text{adalah}{S}_n=n^3+2n,\text{maka}\\ & \begin{array}{rl}{U}_n& =S_n-S_{n-1}\\ U_4& =(4^3+2(4))-(3^3+2(3))\\ & =(64+8)-(27+6)\\ & =72-33=39\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Dari suatu deret diketahui}{S}_n=3n^2-15n\\ & U_n=0\text{saat}n=....\\ & \text{A}.1\text{D}.4\\ & \text{B}.2\text{C}.3\text{E}.5\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan hal yang diketahui di atas}\\ & \begin{array}{rl}{U}_n& =S_n-S_{n-1}\\ 0& =(3n^2-15n)-(3(n-1{)}^2-15(n-1))\\ 0& =3(n^2-(n-1{)}^2)+15(n-1-n)\\ 0& =3(2n-1)(1)+15(-1)\\ 0& =6n-3-15\\ 0& =6n-18\\ 3& =n\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Diketahui sebuah deret}{U}_n=2an+b+4\\ & \text{dan}{S}_n=3b{n}^2+an,\text{maka nilai}a\text{dan}\\ & b\text{adalah}....\\ & \text{A}.12\text{dan}4\\ & \text{B}.-12\text{dan}4\\ & \text{C}.12\text{dan}-4\\ & \text{D}.-12\text{dan}-4\\ & \text{E}.-4\text{dan}-12\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}{U}_n=2an+b+4\text{dan}\\ & S_n=3b{n}^2+an,\text{maka}\\ & \begin{array}{rl}& U_n=S_n-S_{n-1}\\ & U_2=S_2-S_1\\ & 2a(2)+b+4=(3b{.2}^2+a.2)-(3b{.1}^2+a.1)\\ & \iff 4a+b+4=9b+a\\ & \iff 3a-8b=-4........(1)\end{array}\\ & \text{Dan juga}\\ & \begin{array}{rl}& U_1=S_1\\ & \iff 2a(1)+b+4=3b{.1}^2+a.1\\ & \iff 2a+b+4=3b+a\\ & \iff a-2b=-4\\ & \iff 3a-6b=-12........(2)\end{array}\\ & \begin{array}{rl}& \text{Persamaan (2) disubstitusikan ke (1)}\\ & 3a-8b=-4\\ & \iff 3a-6b-2b=4\\ & \iff (-12)-2b=-4\\ & \iff -2b=-4+12=8\\ & \iff b=-4........(3)\\ & \text{Selanjutnya dikembalikan ke (1), maka}\\ & 3a-8b=-4\\ & \iff 3a-8(-4)=-4\\ & \iff 3a+32=-4\\ & \iff 3a=-4-32=-36\\ & \iff a=-12\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Jumlah}n\text{suku pertam sebuah barisan}\\ & \text{adalah}{S}_n={\displaystyle \frac{1}{6}(4n^3-63n^2-n),\text{suku ke}-n}\\ & \text{akan mempunyai nilai terkecil untuk}n=....\\ & \text{A}.3\text{D}.6\\ & \text{B}.4\text{E}.7\\ & \text{C}.5\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan menggunakan rumus}\\ & U_n=S_n-S_{n-1}\text{dengan}{U}_1=S_1,\text{maka}\\ & \text{akan didapatkan nilai}\\ & U_1=-10,U_2=-27,U_3=-40,U_4=-49\\ & U_5=-54,U_6=-55,U_7=-52\\ & \text{Kesemuanya membentuk barisan aritmetika}\\ & \text{tingkat ke-2}.\text{Berikut ilustrasinya}\\ & \begin{array}{rl}& \underset{+4+4+4+4+4}{\underset{\underbrace{}\underbrace{}\underbrace{}\underbrace{}\underbrace{}}{\underset{-17-13-9-5-1+3}{\underset{\underbrace{}\underbrace{}\underbrace{}\underbrace{}\underbrace{}\underbrace{}}{-10-27-40-49-54-55-52}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Jika suku pertama dan kedua sebuah deret}\\ & \text{geometri masing-masing adalah}{a}^{-4}\text{dan}{a}^x\\ & \text{serta suku kedelapan ialah}{a}^{52},\text{maka nilai}\\ & x\text{adalah}....\\ & \text{A}.-32\text{D}.8\\ & \text{B}.-16\text{C}.12\text{E}.4\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& U_8=a{r}^7=U_1{r}^7=a^{52}\\ & \iff U_8=a^{-4}{r}^7=a^{52}\\ & \iff r^7={\displaystyle \frac{a^{52}}{a^{-4}}=a^{52+4}=a^{56}}\\ & \iff r=a^{{.}^{\frac{56}{7}}}=a^8\\ & \text{Maka nilai}x-\text{nya adalah}\\ & U_2=U_{1}r=a^x\\ & \iff (a^{-4})(a^8)=a^{-4+8}=a^x\\ & \iff x=4\end{array}\end{array}$.

Lanjutan 4 Contoh Soal Barisan dan Deret

 $\begin{array}{l}\\ 16.& \text{Carilah semua barisan bilangan yang berupa }\\ & \text{barisan aritmetika dan sekaligus juga barisan }\\ & \text{geometri}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perhatikanlah bentuk barisan bilangan berikut}:\\ & \underset{\underset{{\displaystyle a}}{\updownarrow }}{\stackrel{\stackrel{{\displaystyle a}}{\updownarrow }}{U_1}},\underset{\underset{{\displaystyle ar}}{\updownarrow }}{\stackrel{\stackrel{{\displaystyle (a+b)}}{\updownarrow }}{U_2}},\underset{\underset{{\displaystyle a{r}^2}}{\updownarrow }}{\stackrel{\stackrel{{\displaystyle (a+2b)}}{\updownarrow }}{U_3}},\underset{\underset{{\displaystyle a{r}^3}}{\updownarrow }}{\stackrel{\stackrel{{\displaystyle (a+3b)}}{\updownarrow }}{U_4}},\underset{\underset{{\displaystyle a{r}^4}}{\updownarrow }}{\stackrel{\stackrel{{\displaystyle (a+4b)}}{\updownarrow }}{U_5}},\cdots ,\underset{\underset{{\displaystyle a{r}^{n-1}}}{\updownarrow }}{\stackrel{\stackrel{{\displaystyle (a+(n-1)b)}}{\updownarrow }}{U_n}}\end{array}\\ & \begin{array}{rl}& \text{Misalkan}\\ & \text{Pada BA berlaku}2U_2=U_1+U_3\\ & \text{yaitu}\{\begin{array}{ll}2(a+b)=a+(a+2b),& \text{atau}\\ 2(ar)=a+(a{r}^2)\end{array}\\ & 2ar=a+a{r}^2,\text{dibagi a masing-masing ruas}\\ & r^2-2r+1=0\\ & (r^2-2r+1)=0\\ & (r-1{)}^2=0\\ & r=1\\ & \text{Sehingga barisannya akan menjadi}\\ & a,a,a,a,\cdots \\ & \text{Pada BG juga berlaku}{U}_2^2=U_1\times U_3\\ & \text{yaitu}\{\begin{array}{ll}(a+b{)}^2=a\times (a+2b),& \text{atau}\\ (ar{)}^2=a\times (a{r}^2)\end{array}\\ & \text{ambil saja}\\ & (a+b{)}^2=a\times (a+2b)\\ & a^2+2ab+b^2=a^2+2ab\\ & b^2=0\\ & b=0\\ & \text{dan barisannya juga sama, yaitu}\\ & a,a,a,a,\cdots \\ & \text{Jadi, semua bilangan memenuhi}a\ne 0\\ & \text{saat}r=1ataub=0\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Pada waktu yang sama Anton mulai }\\ & \text{menabung Rp}10.000.000,00\text{dan Budi }\\ & \text{menabung}\text{Rp}8.000.000,00.\text{Selanjutnya}\\ & \text{Anton menabung Rp}100.000,00\text{tiap}\\ & \text{bulan dan Budi menabung Rp}150.000,00\\ & \text{Setelah berapa bulan tabungan keduanya}\\ & \text{tepat sama}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Soal di atas adalah aplikasi dari deret}\\ & \text{aritmetika}\\ & \bullet \text{barisan pertama}\\ & u_1=a=10.000.000,b=100.000\\ & \bullet \text{barisan kedua}\\ & u_1=a=8.000.000,b=150.000\\ & \text{Selanjutnya adalah}\\ & u_1+(n-1)b=u_1^{{}^{\prime }}+(n-1)b^{\prime }\\ & \iff 10.000.000+(n-1)\times 100.000\\ & =8.000.000+(n-1)\times 150.000\\ & \iff (100.000-150.000)\times (n-1)\\ & =8.000.000-10.000.000\\ & \iff -50.000(n-1)=-2.000.000\\ & \iff (n-1)={\displaystyle \frac{-2.000.000}{-50.000}=40}\\ & n=40+1=41\end{array}\\ & \text{Jadi, tabungan keduanya akan sama setelah}\\ & \text{40 bulan}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Tentukan jumlah semua bilangan asli}\\ & \text{antara 1 dan 150 yang habis dibagi 4}\\ & \text{tetapi tidak habis dibagi 7}?\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Soal di atas adalah aplikasi dari deret}\\ & \text{aritmetika}\\ & \bullet \text{barisan pertama adalah bilangan asli}\\ & \text{yang habis dibagi 4}\\ & u_1=a=4,b=4,U_n=148\\ & U_n=a+(n-1)b=4+(n-1)4=148\\ & \text{didapatkan nilai}n=37\\ & S_n={\displaystyle \frac{n}{2}(a+u_n)=\frac{37}{2}(4+148)=2812}\\ & \bullet \text{barisan kedua adalah bilangan asli}\\ & \text{yang habis dibagi}=4\times 7=28\\ & u_1=a=28,b=28,U_n=140\\ & \text{didapatkan nilai}n=5\\ & S_n={\displaystyle \frac{n}{2}(a+u_n)=\frac{5}{2}(28+140)=420}\end{array}\\ & \text{Jadi, jumlah bilangan yang dimaksud}\\ & \text{adalah}\text{2812}-420=2392\end{array}$.

$\begin{array}{l}\\ 19.& \text{Sebuah tali dibagi menjadi 6 bagian}\\ & \text{dengan panjang membentuk suatu}\\ & \text{barisan geometri. Jika panjang tali}\\ & \text{terpendek adalah 3 cm dan yang}\\ & \text{terpanjang 96 cm, berapakah panjang}\\ & \text{tali sebelum terpotong}?\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan panjang tali yang dimaksud}\\ & a,ar,a{r}^2,a{r}^3,a{r}^4,a{r}^5\\ & \text{Dan diketahui juga bahwa}\\ & \{\begin{array}{ll}a& =\text{3 cm}\\ a{r}^5& =\text{96 cm}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}a{r}^5& =96\\ \iff & 3.r^5=96\\ \iff & r^5=32\\ \iff & r^5=2^5\\ \iff & r=2\\ S_n& ={\displaystyle \frac{a(r^n-1)}{r-1},r\ge 1}\\ & ={\displaystyle \frac{3(2^6-1)}{2-1}=3\times (64-1)}\\ & =3\times 63=189\text{cm}\end{array}\end{array}\\ & \text{Jadi, panjang talinya sebelum dipotong}\\ & \text{adalah}\text{189 cm}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Jumlah penduduk suatu kota setiap }\\ & \text{10 tahun menjadi dua kali lipat}.\\ & \text{Jika menurut perhitungan pada tahun}\\ & \text{2030 nanti akan mencapai 3,2 juta jiwa}\\ & \text{berapakah jumlah penduduk kota tersebut}\\ & \text{pada tahun 1980}?\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari soal diketahui bahwa}\\ & \underset{a}{\begin{array}{c}1980\\ \downarrow \end{array}},\underset{ar}{\begin{array}{c}1990\\ \downarrow \end{array}},\underset{a{r}^2}{\begin{array}{c}2000\\ \downarrow \end{array}},\underset{a{r}^3}{\begin{array}{c}2010\\ \downarrow \end{array}},\underset{a{r}^4}{\begin{array}{c}2020\\ \downarrow \end{array}},\underset{a{r}^5}{\begin{array}{c}2030\\ \downarrow \end{array}}\\ & ?3,2\text{juta}\\ & \text{dengan}\\ & \{\begin{array}{ll}r& =2\\ n& =6\end{array}\\ & \text{maka}\\ & \begin{array}{rlr}{U}_6=a{r}^5& =3.200.000& \\ a{.2}^5& =3.200.000\\ a.32& =3.200.000\\ a& ={\displaystyle \frac{3.200.000}{32}}\\ & =100.000\end{array}\end{array}\\ & \text{Jadi, jumlah penduduk pada tahun 1980}\\ & \text{sejumlah}\text{100.000 jiwa}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti,Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: ESIS.
2. Mauludin, U. 2005. Matematika Program Ilmu Sosial dan Bahasa untuk SMA dan MA Kelas XII. Bandung: SARANA PANCA KARYA NUSA.

Lanjutan 3 Contoh Soal Barisan dan Deret

 $\begin{array}{l}\\ 11.& (\mathbf{\text{UN 2014}})\\ & \text{Diketahui seutas kawat dipotong menjadi 5 }\\ & \text{bagian dan hasil potongannya membentuk }\\ & \text{deret geometri. Jika panjang kawat terpendek}\\ & \text{16 cm dan terpanjang 81 cm, maka panjang }\\ & \text{kawat semula adalah}....\text{cm}\\ & \text{A}.121\text{D}.211\\ & \text{B}.130\text{C}.133\text{E}.242\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketahui}& \{\begin{array}{ll}a& =U_1=16\\ U_n& =a.r^{n-1}\\ S_n& ={\displaystyle \frac{a(r^n-1)}{r-1}}\end{array}\end{array}\\ & \begin{array}{rl}{\displaystyle \frac{U_5}{U_1}=\frac{a{r}^4}{a}}& ={\displaystyle \frac{81}{16}}\\ r^4& ={\displaystyle \frac{3^4}{2^4}}\\ r& ={\displaystyle \frac{3}{2},(r>1)}\\ \text{Sehingga},& \\ S_5& ={\displaystyle \frac{16({\left({\displaystyle \frac{3}{2}}\right)}^5-1)}{{\displaystyle \frac{3}{2}-1}}}\\ & ={\displaystyle \frac{16\left(\frac{3^5-2^5}{2^5}\right)}{{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle 32\left(\frac{243-32}{32}\right)}\\ & =211\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& (\mathbf{\text{UMPTN 2001}})\\ & \text{Diketahui sepotong kawat dengan panjang }\\ & \text{124 cm akan dipotong menjadi 5 bagian }\\ & \text{dan hasil potongan kawatnya membentuk }\\ & \text{barisan geometri. Jika pajang potongan }\\ & \text{kawat yang terpendek adalah 4 cm, maka}\\ & \text{potongan kawat yang terpanjang adalah}....\text{cm}\\ & \text{A}.60\text{D}.72\\ & \text{B}.64\text{C}.68\text{E}.76\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan hal yang diketahui di atas}\\ & \begin{array}{rl}{S}_5& =U_1+U_2+U_3+U_4+U_5\\ S_5& =a+ar+a{r}^2+a{r}^3+a{r}^4\\ 124& =4+4r+4r^2+4r^3+4r^4\\ 31& =1+r+r^2+r^3+r^4\\ 30& =r+r^2+r^3+r^4\\ & r^4+r^3+r^2+r-30=0\end{array}\\ & \begin{array}{rl}& \text{Perhatikanlah bahwa pada polinom}\\ & r^4+r^3+r^2+r-30=0\\ & \text{faktor prima dari 30 adalah 2, 3, dan 5.}\\ & \text{Dan faktor yang memenuhi adalah r = 2}\\ & \text{Sehingga},U_5\text{sebagai potongan kawat terpanjang;}\\ & U_5=a{r}^4={4.2}^4=4.16=64\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& (\mathbf{\text{UMPTN 2001}})\\ & \text{Diketahui rasio sebuah deret geometri tak hingga }\\ & \text{adalah}{}^3\log (2x-1).\text{Jika deret tersebut memiliki }\\ & \text{jumlah (konvergen), maka nilai}x\text{yang memenuhi }\\ & \text{adalah}.....\\ & \text{A}.{\displaystyle \frac{1}{2}<x<\frac{2}{3}}\\ \\ & \text{B}.{\displaystyle \frac{1}{2}<x<2}\\ \\ & \text{C}.{\displaystyle \frac{2}{3}<x<2}\\ \\ & \text{D}.{\displaystyle \frac{2}{3}\le x\le 2}\\ \\ & \text{E}.{\displaystyle \frac{1}{2}\le x\le \frac{3}{2}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Syarat}\mathbf{\text{konvergen adalah}}\left|r\right|<1\\ & \text{atau}:-1<r<1\\ & \text{maka}\\ & -1{<}^3\log (2x-1)<1\\ & \iff (-1).({}^3\log 3)<{}^3\log (2x-1)<1.({}^3\log 3)\\ & \iff {}^3\log 3^{-1}<{}^3\log (2x-1)<{}^3\log 3^1\\ & \iff 3^{-1}<(2x-1)<3\\ & \iff {\displaystyle \frac{1}{3}<2x-1<3}\\ & \iff {\displaystyle \frac{1}{3}+1<2x-1+1<3+1}\\ & \iff {\displaystyle \frac{4}{3}<2x<4}\\ & \iff {\displaystyle \frac{2}{3}<x<2}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& (\mathbf{\text{UN 2010}})\\ & \text{Tiga buah bilangan membentuk barisan aritmetika }\\ & \text{dengan beda 3. Jika suku kedua dikurangi 1, maka }\\ & \text{terbentuklah deret geometri dengan jumlah 14.}\\ & \text{Rasio barisan tersebut adalah ... .}\\ & \text{A}.4\text{D}.{\displaystyle -\frac{1}{2}}\\ & \text{B}.2\text{C}.{\displaystyle \frac{1}{2}\text{E}.-2}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Barisan Aritmetika (BA)}\{\begin{array}{ll}{U}_1& =a\\ U_2& =a+3\\ U_3& =a+6\end{array}\\ & \text{Barisan Geometri (BG)}\{\begin{array}{ll}{U}_1& =a\\ U_2& =a+3-1=a+2\\ U_3& =a+6\end{array}\end{array}\\ & \begin{array}{rl}& \text{dan untuk deret geometri (DG)}\\ & U_1+U_2+U_3=14\\ & a+(a+2)+(a+6)=14\\ & 3a+8=14\\ & 3a=6\\ & a=2\\ & \text{sehingga},\\ & r={\displaystyle \frac{U_2}{U_1}=\frac{a+2}{a}=\frac{4}{2}=2}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& (\mathbf{\text{UN 2009}})\\ & \text{Jumlah tiga bilangan barisan aritmetika adalah}\\ & \text{45. Jika suku kedua dikurangi 1 dan suku ketiga}\\ & \text{ditambah 5, maka barisan tersebut menjadi }\\ & \text{barisan geometri. Rasio barisan geometri tersebut}\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{2}\text{D}.2}\\ \\ & \text{B}.{\displaystyle \frac{3}{4}\text{C}.1{\displaystyle \frac{1}{2}\text{E}.3}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{Ada 2 pilihan, yaitu A dan D}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \{\begin{array}{l}{U}_1+U_2+U_3=45(DA)\\ \iff a+(a+b)+(a+2b)=45\\ \iff a+b=15\Rightarrow a=15-b\\ U_1,(U_2-1),(U_3+5)(BG)\\ \Rightarrow a,(a+b-1),(a+2b+5)\\ \Rightarrow a,(14),(20+b)\end{array}\end{array}\\ & \begin{array}{rl}& \text{Pada BG berlaku}\\ & {14}^2=a.(20+b)\\ & 196=(15-b)(20+b)\\ & 196=300-5b-b^2\\ & b^2+5b-104=0\\ & (b+13)(b-8)=0\\ & b=-13\text{atau}b=8\\ & \text{untuk}b=-13\Rightarrow a=15-(-13)=28\\ & \{\begin{array}{ll}B{A}_1:& 28,15,2\\ B{G}_1:& 28,14,7\end{array}\\ & \text{untuk}b=8\Rightarrow a=15-(8)=7\\ & \{\begin{array}{ll}B{A}_2:& 7,15,23\\ B{G}_2:& 7,14,28\end{array}\\ & \text{Jadi, rasio dari barisan geometrianya ada 2 yaitu}:\\ & r_1={\displaystyle \frac{1}{2},\text{dan}{r}_2=2}\end{array}\end{array}$.

Lanjutan 2 Contoh Soal Barisan dan Deret

$\begin{array}{l}\\ 6.& (\mathbf{\text{EBTANAS 2000}})\\ & \text{Diketahui}{\displaystyle \sum _{k=5}^{25}(2-pk)=0,\text{maka nilai}}\\ & \sum _{k=5}^{25}pk=....\\ & \text{A}.20\text{D}.42\\ & \text{B}.28\text{C}.30\text{E}.112\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\displaystyle \sum _{k=5}^{25}(2-pk)={\displaystyle \sum _{k=5}^{25}2-\sum _{k=5}^{25}pk}}& =0\\ {\displaystyle \sum _{k=5-4}^{25-4}2-\sum _{k=5}^{25}pk}& =0\\ {\displaystyle \sum _{k=5}^{25}pk}& ={\displaystyle \sum _{k=1}^{21}2}\\ & =21.2\\ & =42\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& (\mathbf{\text{EBTANAS 2000}})\\ & \text{Nilai dari}{\displaystyle \sum _{k=1}^7{\left({\displaystyle \frac{1}{2}}\right)}^{k+1}=....}\\ & \text{A}.{\displaystyle \frac{127}{1024}\text{D}.{\displaystyle \frac{127}{128}}}\\ \\ & \text{B}.{\displaystyle \frac{127}{256}\text{C}.{\displaystyle \frac{255}{512}\text{E}.{\displaystyle \frac{255}{256}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \sum _{k=1}^7{\left({\displaystyle \frac{1}{2}}\right)}^{k+1}}\\ & ={\left({\displaystyle \frac{1}{2}}\right)}^2+{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^4+{\left(\frac{1}{2}\right)}^5+{\left(\frac{1}{2}\right)}^6+{\left(\frac{1}{2}\right)}^7+{\left({\displaystyle \frac{1}{2}}\right)}^8\\ & ={\displaystyle \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}}\\ & ={\displaystyle \frac{64+32+16+8+4+2+1}{256}}\\ & ={\displaystyle \frac{127}{256}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& (\mathbf{\text{UN 2013}})\\ & \text{Diketahui suatu barisan aritmetika dengan suku }\\ & \text{ketiga adalah 4 dan suku ketujuhnya adalah 16}.\\ & \text{Jumlah 10 suku pertama dari deret tersebut }\\ & \text{adalah}....\\ & \text{A}.115\text{D}.135\\ & \text{B}.125\text{C}.130\text{E}.{\displaystyle 140}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dari soal diketahui bahwa}\\ & \begin{array}{rl}{U}_3=a+2b& =4\\ U_7=a+6b& =16{}_{-}\\ ------& --\\ -4b& =-12\\ b& =3\\ \text{Sehingga}& \text{didapatkan}\\ \text{nilai}a=& 4-2b\\ =& 4-2.3\\ =& -2\end{array}\\ & \begin{array}{rl}\text{Maka}& \text{jumlah 10 suku pertama deret tersebut adalah}\\ S_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)}\\ S_{10}& ={\displaystyle \frac{10}{2}(2.(-2)+(10-1).3)}\\ & =5(-4+27)\\ & =5(23)\\ & =115\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& (\mathbf{\text{UN 2014}})\\ & \text{Diketahui tempat duduk gedung pertunjukan}\\ & \text{film diatur mulai dari baris depan ke belakang }\\ & \text{dengan banyak banyak baris dibelakang lebih }\\ & \text{4 kursi dari baris di depannya. Jika dalam }\\ & \text{gedung pertunjukan terdapat 15 baris kursi dan}\\ & \text{baris terdepan ada 20 kursi, maka kapasitas }\\ & \text{gedung pertunjukan tersebut adalah}....\text{kursi}\\ & \text{A}.1200\text{D}.600\\ & \text{B}.800\text{C}.720\text{E}.300\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketahui}& \{\begin{array}{ll}a& =U_1=20\\ b& =4\\ n& =15\\ S_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)}\end{array}\end{array}\\ & \begin{array}{rl}{S}_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)}\\ & ={\displaystyle \frac{15}{2}(2(20)+(15-1).4)}\\ & =15(20+28)\\ & =15(48)\\ & =750\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& (\mathbf{\text{UN 2015}})\\ & \text{Diketahui suatu barisan aritmetika dengan suku}\\ & \text{ke-3 adalah 2 dan suku ke-8 adalah -13. Jumlah}\\ & \text{20 suku pertama dari deret tersebut adalah}....\\ & \text{A}.-580\text{D}.-410\\ & \text{B}.-490\text{C}.-440\text{E}.-380\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dari soal di atas diketahui bahwa}\\ & \begin{array}{rl}{U}_3=a+2b& =2\\ U_8=a+7b& =-13{}_{-}\\ ------& ---\\ -5b& =15\\ b& =-3\\ \text{Sehingga}& \text{didapatkan}\\ \text{nilai}a=& 2-2b\\ =& 2-2.(-3)\\ =& 8\end{array}\\ & \begin{array}{rl}\text{Maka}& \text{jumlah 20 suku pertama deret tersebut adalah}\\ S_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)}\\ S_{20}& ={\displaystyle \frac{20}{2}(2.(8)+(20-1).(-3))}\\ & =10(16-57)\\ & =10(-41)\\ & =-410\end{array}\end{array}$.

Contoh Soal Barisan dan Deret

 $\begin{array}{l}\\ 1.& (\mathbf{\text{EBTANAS 1999}})\\ & \text{Diketahui jumlah n suku pertama deret}\\ & \text{aritmetika dinyatakan sebagai}{S}_n=n^2+2n\\ & \text{Beda dari deret tersebut adalah}....\\ & \text{a}.3\text{d}.-2\\ & \text{b}.2\text{c}.1\text{e}.-3\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}{S}_n=n^2+2n,\\ & \text{dengan}\{\begin{array}{ll}{S}_1& =U_1=a\\ S_2& =U_1+U_2\\ S_3& =U_1+U_2+U_3\\ & ⋮\\ S_n& =U_1+U_2+U_3+\cdots +U_n\end{array}\\ & \begin{array}{rl}\text{Beda}=b& =U_2-U_1\\ & =(S_2-S_1)-S_1\\ & =S_2-2S_1\\ & =(2^2+2.(2))-2(1^2+2.(1))\\ & =(4+4)-2(1+2)=8-6\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& (\mathbf{\text{UMPTN 1994}})\\ & \text{Diketahui jumlah n suku pertama suatu}\\ & \text{deret dinyatakan sebagai}{S}_n=12n-n^2.\\ & \text{Suku kelima dari deret tersebut adalah}....\\ & \text{a}.-1\text{d}.3\\ & \text{b}.1\text{c}.-3\text{e}.0\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}{S}_n=12n-n^2\\ & \begin{array}{rl}{U}_5& =S_5-S_4\\ & =(12.(5)-(5{)}^2)-(12.(4)-(4{)}^2)\\ & =(60-25)-(48-16)\\ & =3\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& (\mathbf{\text{EBTANAS 2000}})\\ & \text{Diketahui suku tengah suatu deret }\\ & \text{aritmetika adalah 32. Jika jumlah n}\\ & \text{suku pertama deret itu adalah 672, }\\ & \text{maka banyak suku deret itu adalah}....\\ & \text{a}.17\text{d}.23\\ & \text{b}.19\text{c}.21\text{e}.25\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui}\\ & \text{Suku tengah}=U_t={\displaystyle \frac{U_1+U_n}{2}=32\text{dan}}\\ & \begin{array}{rl}{S}_n& ={\displaystyle \frac{n}{2}(U_1+U_n)}\\ & =672\\ n\left({\displaystyle \frac{U_1+U_2}{2}}\right)& =672\\ 32n& =672\\ n& =21\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& (\mathbf{\text{UMPTN 1997}})\\ & \text{Diketahui}{U}_n\text{adalah suku ke - n }\\ & \text{deret aritmetika dengan}\\ & U_1+U_2+U_3=-9\text{dan}{U}_3+U_4+U_5=15\\ & \text{Maka jumlah lima suku pertama}\\ & \text{deret aritmetika tersebut adalah}....\\ & \text{a}.4\text{d}.15\\ & \text{b}.5\text{c}.6\text{e}.24\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}& U_1+U_2+U_3=-9,\\ & \Rightarrow a+(a+b)+(a+2b)=3a+3b=-9\\ & U_3+U_4+U_5=15,\\ & \Rightarrow (a+2b)+(a+3b)+(a+4b)=3a+9b=15{}_{-}\\ & -----------------\\ & -6b=-24\Rightarrow b=4\\ & a=-7\\ & \text{Maka}\\ & \begin{array}{rl}{S}_5& ={\displaystyle \frac{5}{2}(U_1+U_5)}\\ & ={\displaystyle \frac{5}{2}(a+a+(5-1)b)}\\ & ={\displaystyle \frac{5}{2}(-7-7+4.4)}\\ & ={\displaystyle \frac{5}{2}(2)}\\ & =5\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& (\mathbf{\text{EBTANAS 1999}})\\ & \text{Nilai dari}{\displaystyle \sum _{k=1}^{100}5k-\sum _{k=1}^{100}(2k-1)\text{adalah}....}\\ & \text{a}.30.900\text{d}.15.450\\ & \text{b}.30.500\text{c}.16.250\text{e}.15.250\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahi}\\ & \begin{array}{rl}{\displaystyle \sum _{k=1}^{100}5k-\sum _{k=1}^{100}(2k-1)}& =\sum _{k=1}^{100}(5k-2k+1)\\ & ={\displaystyle \sum _{k=1}^{100}(3k+1)}\\ & =3{\displaystyle \sum _{k=1}^{100}k+1.100}\\ & =3\left({\displaystyle \frac{100}{2}(1+100)}\right)+100\\ & =3.(5.050)+100\\ & =15.150+100\\ & =15.250\end{array}\end{array}$.

Lanjutan Materi Barisan dan Deret

 E. Barisan Geometri

Perhatikan susunan bilangan-bilangan berikut 

$\begin{array}{r}1,{\displaystyle \frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\cdots }\end{array}$.

dengan rincian

$\begin{array}{r}\underset{\begin{array}{c}\downarrow \\ u_1\end{array}}{1},\underset{\begin{array}{c}\downarrow \\ u_2\end{array}}{{\displaystyle \frac{1}{2}}},\underset{\begin{array}{c}\downarrow \\ u_3\end{array}}{{\displaystyle \frac{1}{4}}},\underset{\begin{array}{c}\downarrow \\ u_4\end{array}}{{\displaystyle \frac{1}{8}}},\underset{\begin{array}{c}\downarrow \\ u_5\end{array}}{{\displaystyle \frac{1}{16}}},\cdots \end{array}$.

Dari pola di atas kita dapat tuliskan menjadi

$\begin{array}{r}1,{\displaystyle \frac{1}{2},\frac{1}{2}\times \frac{1}{2},\frac{1}{2}\times \frac{1}{4},\frac{1}{2}\times \frac{1}{8},\cdots }\end{array}$.

Dari pola yang tersusun di atas terdapat hal yang menarik yaitu:

$\begin{array}{rl}& {\displaystyle \frac{u_2}{u_1}=\frac{u_3}{u_2}=\frac{u_4}{u_3}=\cdots =\frac{u_n}{u_{n-1}}=\frac{1}{2}}\end{array}$.

Selanjutnya perhatikan

$\begin{array}{rl}& u_1=a=1\\ & u_2=u_1\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{2}\iff u_2=a.r}}\\ & u_3=u_2\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{2}\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{4}\iff u_3=a.r^2}}}}\\ & u_4=u_3\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{2}\times {\displaystyle \frac{1}{2}\times {\displaystyle \frac{1}{2}=1\times {\displaystyle \frac{1}{8}\iff u_4=a.r^3}}}}}\\ & ...=...\\ & u_n=a.r^{n-1}\end{array}$.

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

$S_n=a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots +a{r}^{n-1}$.

Untuk mencari besar $S_n$ adalah dengan mensiasatinya yaitu mengalikan  $r$  ke  $S_n$ sehingga menjadi bentuk

$r{S}_n=ar+a{r}^2+a{r}^3+a{r}^4+a{r}^5+\cdots +a{r}^n$.

Selanjutnya kita kondisikan sebagai berikut

$\begin{array}{rl}& S_n-r{S}_n\\ & =(a+ar+a{r}^2+a{r}^3+a{r}^4+...+a{r}^{n-2}+a{r}^{n-1})\\ & -(ar+a{r}^2+a{r}^3+a{r}^4+a{r}^5+...+a{r}^{n-1}+a{r}^n)\\ & (1-r)S_n=a-a{r}^n=a(1-r^n)\\ & S_n=\frac{a(1-r^n)}{1-r}\end{array}$.

Sebagai rangkuman dari materi barisan dan deret geometri ini, perhatikan tabel berikut

$\begin{array}{|cccc|}\hline \text{No}& \text{Barisan Geometri}& \text{Deret Geometri (Ukur)}& \text{Syarat}\\ 1& \begin{array}{rl}& U_1,U_2,U_3,U_4,...\\ & \\ & \text{Selanjutnya}\\ & U_1,U_2,U_3,\cdots \\ & \text{disebut suku-suku}\\ & \text{dan}{U}_1=a=\\ & \text{suku pertama}\end{array}& \begin{array}{rl}& U_1+U_2+U_3+U_4+...\\ & \\ & \text{Selanjutnya}\\ & U_1,U_2,U_3,\cdots \\ & \text{disebut suku-suku}\\ & \text{dan}{U}_1=a=\\ & \text{suku pertama}\end{array}& \begin{array}{rl}\text{Rasio}=r& ={\displaystyle \frac{U_2}{U_1}}\\ & ={\displaystyle \frac{U_3}{U_2}}\\ & ={\displaystyle \frac{U_4}{U_3}}\\ & =\cdots \\ & ={\displaystyle \frac{U_n}{U_{(n-1)}}}\end{array}\\ 2& U_n=a.r^{(n-1)}& U_n=a.r^{(n-1)}& \begin{array}{rl}{U}_t& ={\displaystyle \sqrt{a.U_n}}\\ & =\text{Suku tengah}\end{array}\\ \hline\end{array}$.

$\begin{array}{|cccc|}\hline \text{No}& \begin{array}{rl}& \text{Barisan}\\ & \text{Geometri}\end{array}& \begin{array}{rl}& \text{Deret }\\ & \text{Geometri (Ukur)}\end{array}& \text{Syarat}\\ 3& & \begin{array}{rl}{S}_n& ={\displaystyle \frac{a(r^n-1)}{r-1}}\\ & \mathbf{\text{atau}}\\ S_n& ={\displaystyle \frac{a(1-r^n)}{1-r}}\end{array}& \begin{array}{rl}& \text{sisipan}k\text{bilangan}\\ & \text{misal}\\ & U_1\cdots \cdots \cdots U_m\\ & \text{ingin disisipkan}k\text{bilangan}\\ & \text{Rasio baru}=r^{\prime }={\displaystyle \sqrt[k+1]{{\displaystyle \frac{U_m}{U_1}}}}\end{array}\\ \hline\end{array}$.

$\begin{array}{|cccl|}\hline \text{No}& \begin{array}{rl}& \text{Barisan}\\ & \text{Geometri}\end{array}& \begin{array}{rl}& \text{Deret }\\ & \text{Geometri (Ukur)}\end{array}& \text{Syarat}\\ & \text{Deret tak Hingga}& \text{Deret tak Hingga}& \text{Hubungan}\\ 4& \text{Konvergen}& \text{Divergen}& \text{suku dan jumlah}\\ & S_{\mathrm{\infty }}={\displaystyle \frac{a}{1-r},\left|r\right|<1}& r\le -1\text{atau}r\ge 1& \begin{array}{rl}{U}_1& =S_1=a\\ U_n& =S_n-S_{(n-1)}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan suku ke}-12\text{dari barisan}\\ & \text{berikut}\\ & 4,1,{\displaystyle \frac{1}{4},\frac{1}{16},\cdots }\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \{\begin{array}{ll}{u}_1& =a=4\\ r& ={\displaystyle \frac{u_2}{u_1}=\frac{u_3}{u_2}=\cdots =\frac{1}{4}}\end{array}\\ & \text{Untuk mencari suku ke}-12,\text{maka}\\ & \begin{array}{rl}{u}_{12}& =a.r^{(12-1)}=a{r}^{11}\\ & =4.{\left({\displaystyle \frac{1}{4}}\right)}^{11}\\ & =4^1{.4}^{-11}=4^{1-11}=4^{-10}\\ & ={\displaystyle \frac{1}{4^{10}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan jumlah 12 suku pertama}\\ & \text{dari}\\ & 4+1+{\displaystyle \frac{1}{4}+\frac{1}{16}+\cdots +\frac{1}{4^{10}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}{S}_n& ={\displaystyle \frac{a(1-r^n)}{1-r}}\\ S_{12}& ={\displaystyle \frac{4(1-{\left({\displaystyle \frac{1}{4}}\right)}^{12})}{1-{\displaystyle \frac{1}{4}}}}\\ & ={\displaystyle \frac{4(1-{\left({\displaystyle \frac{1}{4}}\right)}^{12})}{{\displaystyle \frac{3}{4}}}}\\ & ={\displaystyle \frac{16}{3}(1-{\left({\displaystyle \frac{1}{4}}\right)}^{12})}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Suatu deret geometri dengan jumlah}\\ & S_n={3.2}^n-1,\text{maka suku ke}-2022\\ & \text{dari deret tersebut adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}{S}_n={3.2}^n-1,\text{maka}\\ & u_{2022}=S_{2022}-S_{2021}\\ & \text{Sehingga}\\ & \begin{array}{rl}{u}_{2022}& =S_{2022}-S_{2021}\\ & =({3.2}^{2022}-1)-({3.2}^{2021}-1)\\ & ={3.2}^{2022}-{3.2}^{2021}={3.2}^{2021}(2-1)\\ & ={2.3}^{2021}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui deret geometri dengan}{\displaystyle \frac{u_4}{u_6}=k}\\ & \text{dan}{u}_2\times u_8={\displaystyle \frac{1}{k},\text{maka suku pertama}}\\ & \text{deret geometri ini adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \{\begin{array}{ll}& {\displaystyle \frac{u_4}{u_6}=k}\\ & u_2\times u_8={\displaystyle \frac{1}{k}}\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}& {\displaystyle \frac{u_6}{u_4}=\frac{a{r}^5}{a{r}^3}=r^2=\frac{1}{k}\text{dan}}\\ & u_2\times u_8\\ & =ar\times a{r}^7=a^2{r}^8={\left(a{r}^4\right)}^2=\frac{1}{k}\\ & \iff {\left(u_5\right)}^2={\displaystyle \frac{1}{k}\iff u_5=\sqrt{\frac{1}{k}}}\\ & \text{sehingga}\\ & u_5=a{r}^4=a{\left(r^2\right)}^2\iff \sqrt{\frac{1}{k}}=a{\left(\frac{1}{k}\right)}^2\\ & \begin{array}{rl}a& =k^2\times \sqrt{{\displaystyle \frac{1}{k}}}\\ & =k\times k\times k^{-\frac{1}{2}}\\ & =k\times k^{{}^{{.}^{\frac{1}{2}}}}=k\sqrt{k}\end{array}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Waji, J., Linggih, S., Syahrudin,Y.R. 1981. Ringkasan Materi IPA. Bandung: GANECA EXACT.

Barisan dan Deret

 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 

Jika kita cermati susunan susunan kue donat dalam kotak terkecil ke terbesar atau begitu seterusnya pada tiap-tiap terisi sejumlah : 1, 4, 9, 16, 25. Sehingga saat kita rinci 

Dapatkan Anda menentukan kelompok kotak berikutnya setelah kotak ke-5, misalnya ketak ke-6, 7, 8, dan seterusnya

Jika kue donat dalam kotak kita tabelkan akan berupa ilustrasi berikut

$\begin{array}{|ccc|}\hline \text{Kelompok}& \text{Kue Donat dalam Kotak}& \text{Pola}\\ K_1& 1& 1=1\times 1\\ K_2& 4& 4=2\times 2\\ K_3& 9& 9=3\times 3\\ K_4& 16& 16=4\times 4\\ K_5& 25& 25=5\times 5\\ ⋮& ⋮& ⋮\\ K_{\text{n}}& ?& ?=\text{n}\times \text{n}\\ \hline\end{array}$.

Dengan memperhatikan pola yang ada di atas, maka akan dengan mudah kita menentukan isi kotak ke-6, yaitu berisi 6x6 = 36 buah kue donat dean demikian seterusnya.

B. Menemukan Pola Barisan dan Deret suatu Bilangan

Misalkan diberikan susunan bilangan berikut

$\begin{array}{rl}& {\displaystyle \frac{1}{2},\frac{1}{6},\frac{1}{12},\frac{1}{20},\frac{1}{30},\cdots ,\frac{1}{9900}}\end{array}$.

Andai kita tabelkan akan berupa

$\begin{array}{|ccc|}\hline \text{Suku ke-}& \text{Nilai}& \text{Pola}\\ U_1& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}=\frac{1}{1\times 2}}\\ U_2& {\displaystyle \frac{1}{6}}& {\displaystyle \frac{1}{6}=\frac{1}{2\times 3}}\\ U_3& {\displaystyle \frac{1}{12}}& {\displaystyle \frac{1}{12}={\displaystyle \frac{1}{3\times 4}}}\\ U_4& {\displaystyle \frac{1}{20}}& {\displaystyle \frac{1}{20}=\frac{1}{4\times 5}}\\ U_5& {\displaystyle \frac{1}{30}}& {\displaystyle \frac{1}{30}=\frac{1}{5\times 6}}\\ ⋮& ⋮& ⋮\\ U_{\text{99}}& {\displaystyle \frac{1}{9900}}& {\displaystyle \frac{1}{9900}={\displaystyle \frac{1}{99\times 100}}}\\ \hline\end{array}$.

Sehingga dari pola bilangan di atas kita dengan mudah menentukan urutan suku ke-n atau  $U_n$  yaitu ${\displaystyle \frac{1}{n\times (n+1)}}$ dan andai kita diminta menentukan besar suku ke-2022 adalah ${\displaystyle \frac{1}{2022\times 2023}=\frac{1}{4090506}}$.

C. Barisan Aritmetika

Secara definisi barisan aritmetika adalah barisan bilangan di mana beda setiap suku dengan tepat suku setelahnya memiliki selisih tetap.

Sebagai ilustrasinya misalkan $u_1=a$, $u_2=a+b$, dan untuk suku ke-3 adalah $u_3=a+b+b=a+2b$, demikian seterusnya akan selalu ditambahkan $b$ dan selanjutnya nilai $b$=$u_2-u_1=u_3-u_2=u_4-u_3=\cdots$.

Perhatikan ilustrasi berikut

$\begin{array}{rl}& u_1=a\\ & u_2=u_1+b=u_1+b\\ & u_3=u_2+b=u_1+2b\\ & u_4=u_3+b=u_1+3b\\ & u_5=u_4+b=u_1+4b\\ & ⋮⋮\\ & u_n=u_{(n-1)}+b=u_1+(n-1)b\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan suku ke-50 dari barisan berikut}\\ & 5,-2,-9,-16,\cdots \\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}& \underset{\begin{array}{c}\downarrow \\ u_1\end{array}}{5},\underset{\begin{array}{c}\downarrow \\ u_2\end{array}}{-2},\underset{\begin{array}{c}\downarrow \\ u_3\end{array}}{-9},\underset{\begin{array}{c}\downarrow \\ u_4\end{array}}{-16},\cdots ,\underset{\begin{array}{c}\downarrow \\ u_n\end{array}}{u_1+(n-1)b}\end{array}\\ & \text{Jelas bahwa}\\ & \begin{array}{rl}b=& u_2-u_1=-2-5=-7\\ \text{ma}& \text{ka}\\ U_{50}& =u_1+(50-1).(-7)\\ & =5+49.(-7)=5-343\\ & =-338\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika diketahui barisan aritmetika dengan}\\ & \text{suku ke}-3=-4{\displaystyle \frac{1}{2}\text{dan suku ke}-8=-2}\\ & \text{Tentukan suku pertama, beda serta rumus}\\ & \text{suku ke}-n\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel berikut}\\ & \begin{array}{|cc|}\hline \begin{array}{lll}& u_3=-4{\displaystyle \frac{1}{2}=a+2b}& \\ & u_8=-2=a+7b& -\\ & -4{\displaystyle \frac{1}{2}-(-2)=-5b}\\ & -2{\displaystyle \frac{1}{2}=-5b}\\ & 5b={\displaystyle \frac{5}{2}}\\ & b={\displaystyle \frac{1}{2}}\end{array}& \begin{array}{rl}{u}_3& =a+2b=-4{\displaystyle \frac{1}{2}}\\ & a+2\left({\displaystyle \frac{1}{2}}\right)=-4{\displaystyle \frac{1}{2}}\\ & a+1=-4{\displaystyle \frac{1}{2}}\\ & a=-4{\displaystyle \frac{1}{2}-1}\\ & a=-5{\displaystyle \frac{1}{2}}\end{array}\\ \hline\end{array}\\ & \text{maka}{u}_n=a+(n-1)b\\ & \begin{array}{rl}{u}_n& =-5{\displaystyle \frac{1}{2}+(n-1).{\displaystyle \frac{1}{2}}}\\ & =-5{\displaystyle \frac{1}{2}+{\displaystyle \frac{1}{2}n-\frac{1}{2}={\displaystyle \frac{1}{2}n-6}}}\end{array}\end{array}$.

D. Deret Aritmetika

Jika pada barisan aritmetika di atas dijumlahkan semua sukunya, maka akan terbentuklah sebuah deret hitung yang selanjutnya adalah nama lain dari deret aritmetika

$\begin{array}{rl}{S}_n& =a+(a+b)+(a+2b)+(a+3b)+\cdots \end{array}$.

dan

$\begin{array}{ll}{S}_n& =a+(a+b)+(a+2b)+\cdots +a+(n-1)b\\ S_n& =a+(n-1)b+\cdots +(a+2b)+(a+b)+a& +\\ 2S_n& =2a+(n-1)b+\cdots +2a+(n-1)b\\ 2S_n& =n(2a+(n-1)b)\\ S_n& ={\displaystyle \frac{n}{2}(2a+(n-1)b)\mathbf{\text{atau}}}\\ S_n& ={\displaystyle \frac{n}{2}(a+a+(n-1)b)=\frac{n}{2}(u_1+u_n)}\end{array}$.

$\text{CONTOH SOAL}$.

Pada contoh soal no.1 di atas tentukanlah jumlah 50 suku pertema deret aritmetika tersebut

Jawab:

$\begin{array}{rl}\text{Dik}& \text{etahui}\\ S_{50}& =5+(-2)+(-9)+(-16)+\cdots +(-338)\\ \text{De}& \text{ngan}{S}_n={\displaystyle \frac{1}{2}n(u_1+u_n)}\\ S_{50}& ={\displaystyle \frac{1}{2}.50.(5+(-338))}\\ & =25.(-333)\\ & =-8325\end{array}$.

Sebagai rangkumannya perhatikan tabel berikut terkait barisan dan deret aritmetika

$\begin{array}{|clll|}\hline \text{No}& \text{Barisan Aritmetika}& \text{Deret Aritmetika (Hitung)}& \text{Syarat}\\ 1& \begin{array}{rl}& U_1,U_2,U_3,U_4,...\\ & \\ & \text{Selanjutnya}\\ & U_1,U_2,U_3,\cdots \\ & \text{disebut suku-suku}\\ & \text{dan}{U}_1=a=\\ & \text{suku pertama}\end{array}& \begin{array}{rl}& U_1+U_2+U_3+U_4+...\\ & \\ & \text{Selanjutnya}\\ & U_1,U_2,U_3,\cdots \\ & \text{disebut suku-suku}\\ & \text{dan}{U}_1=a=\\ & \text{suku pertama}\end{array}& \begin{array}{rl}\text{Beda}=b& =U_2-U_1\\ & =U_3-U_2\\ & =U_4-U_3\\ & =\cdots \\ & =U_n-U_{(n-1)}\\ & \\ & \end{array}\\ 2& U_n=a+(n-1)b& U_n=a+(n-1)b& \begin{array}{rl}{U}_t& ={\displaystyle \frac{U_1+U_n}{2}}\\ & =\text{Suku tengah}\end{array}\\ 3& & S_n={\displaystyle \frac{1}{2}n(a+U_n)}& \begin{array}{rl}& \text{sisipan}k\text{bilangan}\\ & \text{misal},\\ & U_1\cdots \cdots \cdots U_m\\ & \text{ingin disisipkan}k\text{bilangan}\\ & \text{beda baru}=b^{\prime }={\displaystyle \frac{U_m-U_1}{k+1}}\end{array}\\ \hline\end{array}$

DAFTAR PUSTAKA

1. Kementerian Pendidikan dan Kebudayaan Republik Indonesia. 2013. Matematika Kelas X. Jakarta: Kementerian Pendidikan dan Kebudayaan
2. Susanto, D., dkk. 2021. Matematika untuk SMA/SMK Kelas X. Klaten: MACANANJAYA CEMERLANG.