1. Sequence Analysis & Periodicity

The recurrence relation is given as:


\(a_n=\frac{a_{n-1}}{a_{n-2}}\)


Let \(a_1=x\) and \(a_2=y \). Expressing subsequent terms algebraically:

\(a_3=\frac{a_2}{a_1}=\frac{y}{x} \)
\(a_4=\frac{a_3}{a_2}=\frac{y / x}{y}=\frac{1}{x} \)
\(a_5=\frac{a_4}{a_3}=\frac{1 / x}{y / x}=\frac{1}{y} \)
\(a_6=\frac{a_5}{a_4}=\frac{1 / y}{1 / x}=\frac{x}{y} \)
\(a_7=\frac{a_6}{a_5}=\frac{x / y}{1 / y}=x=a_1 \)
\(a_8=\frac{a_7}{a_6}=\frac{x}{x / y}=y=a_2 \)


This proves the sequence is strictly periodic with a period of 6.

2. Determining the Values

We are given \(a_3=3 \) and \(a_5=\frac{1}{2} \) :

\( a_5=\frac{1}{y}=\frac{1}{2} \Longrightarrow y=2
a_3=\frac{y}{x}=3 \Longrightarrow \frac{2}{x}=3 \Longrightarrow x=\frac{2}{3}\)

Substituting \(x=\frac{2}{3}\) and y=2 into our 6 periodic terms:
\( a_1=\frac{2}{3}
a_2=2 \)


\(a_3=3\)
\( a_4=\frac{3}{2}
a_6=\frac{1}{3}\)


3. Calculating the Product of the First 200 Terms

First, compute the product of one complete period ( \(a_1 \times a_2 \times a_3 \times a_4 \times a_5 \times a_6\) ):


\(P_{\text {period }}=\left(\frac{2}{3}\right) \times 2 \times 3 \times\left(\frac{3}{2}\right) \times\left(\frac{1}{2}\right) \times\left(\frac{1}{3}\right)=1\)


Since \(200=6 \times 33+2\) , the product of the first 200 terms consists of 33 full periods plus the first \(\mathbf{2}\) terms ( \(a_1\) and \(a_2\) ):


\(\begin{gathered}
\operatorname{Product}_{200}=\left(P_{\text {period }}\right)^{33} \times a_1 \times a_2 \\
\operatorname{Product}_{200}=(1)^{33} \times\left(\frac{2}{3}\right) \times 2=\frac{\mathbf{4}}{\mathbf{3}}
\end{gathered}\)



Official Answer
Official Answer: (D) \(\frac{4}{3}\)