In “P4: “P, = 1: 3,then n is equal to

The correct answer is (a) 5.

The question is asking for the value of $n$ in the following equation:

$$P_n = \frac{1}{3}P_{n-1}$$

We can solve this equation by starting with $P_0 = 1$ and then repeatedly substituting the previous value of $P_n$ into the equation. This gives us the following sequence:

$$P_0 = 1$$
$$P_1 = \frac{1}{3}P_0 = \frac{1}{3}$$
$$P_2 = \frac{1}{3}P_1 = \frac{1}{9}$$
$$P_3 = \frac{1}{3}P_2 = \frac{1}{27}$$
$$P_4 = \frac{1}{3}P_3 = \frac{1}{81}$$

As you can see, the sequence is decreasing by a factor of $\frac{1}{3}$ each time. This means that the value of $n$ must be a multiple of

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