Which term of the progression `sqrt(3),3,3sqrt(3)...` is 729 ?

Explanation

To find which term of the progression \( \sqrt{3}, 3, 3\sqrt{3}, \ldots \) is equal to 729, we first need to identify the type of progression and the parameters involved.

### Step 1: Identify the first term and common ratio
The first term \( a \) of the progression is:
\[ a = \sqrt{3} \]
The second term is \( 3 \), and the common ratio \( r \) can be found by dividing the second term by the first term:
\[ r = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Step 2: Write the general term of the progression
The \( n \)-th term of a geometric progression (G.P.) is given by the formula:
\[ a_n = a \cdot r^{n-1} \]
Substituting the values of \( a \) and \( r \):
\[ a_n = \sqrt{3} \cdot (\sqrt{3})^{n-1} \]
This simplifies to:
\[ a_n = \sqrt{3} \cdot 3^{\frac{n-1}{2}} = 3^{\frac{1}{2}} \cdot 3^{\frac{n-1}{2}} = 3^{\frac{n}{2}} \]

### Step 3: Set the general term equal to 729
We need to find \( n \) such that:
\[ 3^{\frac{n}{2}} = 729 \]

### Step 4: Express 729 as a power of 3
We know that:
\[ 729 = 3^6 \]
Thus, we can rewrite the equation:
\[ 3^{\frac{n}{2}} = 3^6 \]

### Step 5: Equate the exponents
Since the bases are the same, we can set the exponents equal to each other:
\[ \frac{n}{2} = 6 \]

### Step 6: Solve for \( n \)
To find \( n \), multiply both sides by 2:
\[ n = 12 \]

### Conclusion
Thus, the term of the progression that is equal to 729 is the 12th term.