The nth term of the series `3, sqrt(3), 1,…` is `(1)/(243)`, then n is

Options

A. 12
B. 13
C. 14
D. 15

Answer

To find the value of \( n \) for which the \( n \)th term of the series \( 3, \sqrt{3}, 1, \ldots \) is \( \frac{1}{243} \), we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the series is:
\[ a = 3 \]
Next, we need to find the common ratio \( r \). The second term is \( \sqrt{3} \), so:
\[ r = \frac{t_2}{t_1} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

### Step 2: Write the formula for the \( n \)th term
The formula for the \( n \)th term of a geometric series is given by:
\[ t_n = a \cdot r^{n-1} \]
Substituting the values of \( a \) and \( r \):
\[ t_n = 3 \cdot \left(\frac{1}{\sqrt{3}}\right)^{n-1} \]

### Step 3: Set the \( n \)th term equal to \( \frac{1}{243} \)
We need to find \( n \) such that:
\[ 3 \cdot \left(\frac{1}{\sqrt{3}}\right)^{n-1} = \frac{1}{243} \]

### Step 4: Simplify the equation
First, divide both sides by 3:
\[ \left(\frac{1}{\sqrt{3}}\right)^{n-1} = \frac{1}{3 \cdot 243} \]
Calculating \( 3 \cdot 243 \):
\[ 3 \cdot 243 = 729 \]
So we have:
\[ \left(\frac{1}{\sqrt{3}}\right)^{n-1} = \frac{1}{729} \]

### Step 5: Express \( \frac{1}{729} \) in terms of powers of \( 3 \)
We know that:
\[ 729 = 3^6 \quad \Rightarrow \quad \frac{1}{729} = 3^{-6} \]
Thus, we can rewrite the equation:
\[ \left(\frac{1}{\sqrt{3}}\right)^{n-1} = 3^{-6} \]

### Step 6: Rewrite \( \frac{1}{\sqrt{3}} \) in terms of powers of \( 3 \)
We know:
\[ \frac{1}{\sqrt{3}} = 3^{-1/2} \]
So we can substitute:
\[ \left(3^{-1/2}\right)^{n-1} = 3^{-6} \]

### Step 7: Set the exponents equal to each other
This gives us:
\[ -\frac{1}{2}(n-1) = -6 \]
Multiplying both sides by -1:
\[ \frac{1}{2}(n-1) = 6 \]
Now, multiply both sides by 2:
\[ n - 1 = 12 \]
Finally, add 1 to both sides:
\[ n = 13 \]

### Final Answer
The value of \( n \) is:
\[ \boxed{13} \]