Contoh Soal Barisan dan Deret

 $\begin{array}{ll}\\ 1.&(\textbf{EBTANAS 1999})\\ &\textrm{Diketahui jumlah n suku pertama deret}\\ &\textrm{aritmetika dinyatakan sebagai}\: \: S_{n}=n^{2}+2n\\ &\textrm{Beda dari deret tersebut adalah}\: ....\\ &\textrm{a}.\quad 3\: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad -2\\ &\textrm{b}.\quad \color{red}2\qquad\qquad \color{black}\textrm{c}.\quad 1\qquad\qquad \textrm{e}.\quad -3\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: S_{n}=n^{2}+2n,\\ &\textrm{dengan}\: \begin{cases} S_{1} & =U_{1}=a \\ S_{2} & =U_{1}+U_{2} \\ S_{3} & =U_{1}+U_{2}+U_{3} \\ &\vdots \\ S_{n} & =U_{1}+U_{2}+U_{3}+\cdots +U_{n} \end{cases}\\ &\begin{aligned}\textrm{Beda}=b&=U_{2}-U_{1}\\ &=(S_{2}-S_{1})-S_{1}\\ &=S_{2}-2S_{1}\\ &=\left ( 2^{2}+2.(2) \right )-2\left ( 1^{2}+2.(1) \right )\\ &=\left ( 4+4 \right )-2\left ( 1+2 \right )=8-6\\ &=\color{red}2\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&(\textbf{UMPTN 1994})\\ &\textrm{Diketahui jumlah n suku pertama suatu}\\ & \textrm{deret dinyatakan sebagai}\: \: S_{n}=12n-n^{2}.\\ &\textrm{Suku kelima dari deret tersebut adalah}\: ....\\ &\textrm{a}.\quad -1\: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad \color{red}3\\ &\textrm{b}.\quad 1\qquad\qquad \textrm{c}.\quad -3\qquad\qquad \textrm{e}.\quad 0\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: S_{n}=12n-n^{2}\\ &\begin{aligned}U_{5}&=S_{5}-S_{4}\\ &=\left ( 12.(5)-(5)^{2} \right )-\left ( 12.(4)-(4)^{2} \right )\\ &=\left ( 60-25 \right )-\left ( 48-16 \right )\\ &=\color{red}3\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&(\textbf{EBTANAS 2000})\\ &\textrm{Diketahui suku tengah suatu deret }\\ &\textrm{aritmetika adalah 32. Jika jumlah n}\\ &\textrm{suku pertama deret itu adalah 672, }\\ &\textrm{maka banyak suku deret itu adalah}\: ....\\ &\textrm{a}.\quad 17\: \: \: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 23\\ &\textrm{b}.\quad 19\qquad\qquad \textrm{c}.\quad \color{red}21\qquad\qquad \color{black}\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui}\\ &\textrm{Suku tengah}=U_{t}=\displaystyle \frac{U_{1}+U_{n}}{2}=32\: \: \textrm{dan}\\ &\begin{aligned}S_{n}&=\displaystyle \frac{n}{2}\left ( U_{1}+U_{n} \right )\\ &=672\\ n\left ( \displaystyle \frac{U_{1}+U_{2}}{2} \right )&=672\\ 32n&=672\\ n&=\color{red}21 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&(\textbf{UMPTN 1997})\\ &\textrm{Diketahui}\: \: U_{n}\: \: \textrm{adalah suku ke - n }\\ &\textrm{deret aritmetika dengan}\\ &U_{1}+U_{2}+U_{3}=-9\: \: \textrm{dan}\: \: U_{3}+U_{4}+U_{5}=15\\ & \textrm{Maka jumlah lima suku pertama}\\ &\textrm{deret aritmetika tersebut adalah}\: ....\\ &\textrm{a}.\quad 4\: \qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 15\\ &\textrm{b}.\quad \color{red}5\qquad\qquad \color{black}\textrm{c}.\quad 6\qquad\qquad \textrm{e}.\quad 24\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}&U_{1}+U_{2}+U_{3}=-9,\\ &\Rightarrow  a+(a+b)+(a+2b)=3a+3b=-9\\ &U_{3}+U_{4}+U_{5}=15,\\ & \Rightarrow \: \: \qquad(a+2b)+(a+3b)+(a+4b)=3a+9b=15\quad _{-}\\ &  -----------------\\ &\, \,  -6b=-24\Rightarrow b=4\\ &\,  a=-7\\ &\textrm{Maka}\\ &\begin{aligned} S_{5}&=\displaystyle \frac{5}{2}\left ( U_{1}+U_{5} \right )\\ &=\displaystyle \frac{5}{2}\left ( a+a+(5-1)b \right )\\ &=\displaystyle \frac{5}{2}\left ( -7-7+4.4 \right )\\ &=\displaystyle \frac{5}{2}(2)\\ &=\color{red}5\end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&(\textbf{EBTANAS 1999})\\ &\textrm{Nilai dari}\: \: \displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 30.900\: \: \: \: \: \: \quad\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 15.450\\ &\textrm{b}.\quad 30.500\qquad\qquad \textrm{c}.\quad 16.250\qquad\qquad \textrm{e}.\quad \color{red}15.250\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahi}\\ &\begin{aligned}\displaystyle \sum_{k=1}^{100}5k-\sum_{k=1}^{100}(2k-1)&=\sum_{k=1}^{100}(5k-2k+1)\\ &=\displaystyle \sum_{k=1}^{100}(3k+1)\\ &=3\displaystyle \sum_{k=1}^{100}k+1.100\\ &=3\left ( \displaystyle \frac{100}{2}(1+100) \right )+100\\ &=3.(5.050)+100\\ &=15.150+100\\ &=\color{red}15.250\end{aligned} \end{array}$.