$\begin{aligned}15.\quad &\textrm{Di antara bilangan}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\:\: \textrm{terdapat tak hingga banyak}\\ &\textrm{bilangan pecah. Tentukan 999 bilangan pecah di antara}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\\&\textrm{sehingga selisih antara bilangan pecah berikutnya dengan bilangan}\\ &\textrm{pecah sebelumnya konstan}\\ &(\textrm{Maksudnya: jika}\:\: x_{\displaystyle 1},x_{\displaystyle 2},\cdots ,x_{\displaystyle 999}\:\:\textrm{bilangan pecah yang dimaksud}\\ &\textrm{maka},\:\:x_{\displaystyle 2}-x_{\displaystyle 1}=x_{\displaystyle 3}-x_{\displaystyle 2}=\cdots =x_{\displaystyle 999}-x_{\displaystyle 998}) \end{aligned}$
Jawaban: $\begin{aligned}&\textrm{Untuk menyisipkan 999 bilangan pecahan di antara}\:\: \displaystyle \frac{1}{5}\:\: \textrm{dan}\:\: \displaystyle \frac{1}{4}\:\: \textrm{dengan}\\ &\textrm{selisih tetap, maka kita sama saja membentuk sebuah barisan aritmetika}\\&\textrm{dengan} \\ &\circ \quad \text{Suku pertama}:\:\: a=U_{\displaystyle 1}=\displaystyle \frac{1}{5}\\ &\circ \quad \text{Banyak sisipan}:\:\:999\:\: \:\textrm{bilangan pecahan}\\ &\circ \quad \text{Total suku}:k=n+2=999+2=1001\\ &\circ \quad \text{Suku terakhir}:\:\: U_{\displaystyle 1001}=\displaystyle \frac{1}{4}\\ &\circ \quad \text{Beda antar dua suku}:\:\: U_{\displaystyle n}=a+(n-1)b=\displaystyle \frac{1}{4}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow \displaystyle \frac{1}{5}+(1001-1)b=\displaystyle \frac{1}{4}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow 1000b=\displaystyle \frac{1}{4}-\frac{1}{5}=\displaystyle \frac{1}{20}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \Leftrightarrow b=\displaystyle \frac{1}{20.000}\\ &\textrm{Sehingga bilangan pecahan yang dimaksud adalah}:\\&\circ \quad U_{\displaystyle 1}=\displaystyle \frac{1}{5}=\frac{4000}{20000}\\ &\circ \quad U_{\displaystyle 2}=U_{\displaystyle 1}+b=x_{\displaystyle 1}=\frac{4000}{20000}+\frac{1}{20000}=\displaystyle \frac{4001}{20000}\\ &\circ \quad U_{\displaystyle 3}=U_{\displaystyle 2}+b=x_{\displaystyle 2}=\frac{4001}{20000}+\frac{1}{20000}=\displaystyle \frac{4002}{20000}\\ &\qquad\qquad\qquad\vdots\\ &\circ \quad U_{\displaystyle 1000}=U_{\displaystyle 999}+b=x_{\displaystyle 999}=\frac{4998}{20000}+\frac{1}{20000}=\displaystyle \frac{4999}{20000} \end{aligned}$
$\begin{aligned}16.\quad&(\textbf{LM UGM ke-32 Th 2021 Tk.SMA})\\&\textrm{Diberikan barisan}\:(a_{\displaystyle n})\: \textrm{yang memenuhi}\\ &\qquad\qquad\qquad a_{\displaystyle n+1}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{\sqrt{4(a_{\displaystyle n-1})^{\displaystyle 2}-4(a_{\displaystyle n})^{\displaystyle 2}}}\\ &\textrm{Jika}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{\sqrt{2}}\:\: \textrm{dan}\:\: a_{\displaystyle 2}=\displaystyle \frac{1}{2},\: \textrm{nilai dari}\:\: 2^{\displaystyle 2526}\prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}\:\: \textrm{adalah}\:....\\ &\text{a}.\quad \displaystyle \frac{1}{4}\qquad \qquad \qquad\qquad\qquad\:\: \text{d}.\quad \displaystyle 2\\ &\text{b}.\quad \displaystyle \frac{1}{2}\qquad\qquad \text{c}.\quad 1\qquad\qquad \text{e}.\quad 4\end{aligned}$
Jawaban: $\begin{aligned}&\textbf{Jawab: d}\\&\textrm{Perhatikan bahwa}\\ &a_{\displaystyle n+1}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{\sqrt{4(a_{\displaystyle n-1})^{\displaystyle 2}-4(a_{\displaystyle n})^{\displaystyle 2}}}=\displaystyle \frac{a_{\displaystyle n}a_{\displaystyle n-1}}{2\sqrt{a_{\displaystyle n-1}^{\displaystyle 2}-a_{\displaystyle n}^{\displaystyle 2}}}\\ &\Leftrightarrow a_{\displaystyle n+1}^{\displaystyle 2}=\displaystyle \frac{a_{\displaystyle n}^{\displaystyle 2}a_{\displaystyle n-1}^{\displaystyle 2}}{\displaystyle 4\left( a_{\displaystyle n-1}^{\displaystyle 2}-a_{\displaystyle n}^{\displaystyle 2} \right)}\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle n+1}^{\displaystyle 2}}=\displaystyle \frac{4}{a_{\displaystyle n}^{\displaystyle 2}}-\frac{4}{a_{\displaystyle n-1}^{\displaystyle 2}}\\ &\textrm{dengan}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{\sqrt{2}}\:,\: a_{\displaystyle 2}=\displaystyle \frac{1}{2},\: \textrm{maka}\:\: a_{\displaystyle 1}^{\displaystyle 2}=\displaystyle \frac{1}{2}\:\: \textrm{dan}\:\: a_{\displaystyle 2}^{\displaystyle 2}=\displaystyle \frac{1}{4}\\ &\textrm{Misalkan saja}\:\: b_{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}},\: \textrm{maka}\:\: b_{\displaystyle n+1}=4\left( b_{\displaystyle n}-b_{\displaystyle n-1} \right)\\ &\textrm{dengan}\:\: b_{\displaystyle 1}=2,\: b_{\displaystyle 2}=4,\: \textrm{maka akan diperoleh}:\\ &\circ \quad b_{\displaystyle 3}=4(4-2)=4.2=8\\ &\circ \quad b_{\displaystyle 4}=4(8-4)=4.4=16\\ &\textrm{Tampak bahwa}\:\: b_{\displaystyle n}=2^{\displaystyle n}\: \textrm{dan}\: b_{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}},\\ &\textrm{maka}\:\: 2^{\displaystyle n}=\displaystyle \frac{1}{a_{\displaystyle n}^{\displaystyle 2}}\Leftrightarrow a_{\displaystyle n}=\displaystyle 2^{\displaystyle -\frac{n}{2}}.\\ &\textrm{Karena}\:\: a_{\displaystyle n}\gt 0,\:\: \textrm{maka}\:\: \prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}=\prod_{\displaystyle n=1}^{\displaystyle 100}2^{\displaystyle -\frac{1}{2}n}\\ &=2^{\displaystyle -\frac{1}{2}(1+2+3+4+\cdots +99+100)}=2^{\displaystyle -\frac{1}{2}\left( \frac{100\times 101}{2} \right)}\\ &=2^{\displaystyle -2525}\\ &\textrm{Akibatnya adalah}:\\ &2^{\displaystyle 2526}\prod_{\displaystyle n=1}^{\displaystyle 100}a_{\displaystyle n}=2^{\displaystyle 2526}\times 2^{\displaystyle -2525}=2^{\displaystyle 1}=2 \end{aligned}$
