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 will follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the progression is:
\[ a = \sqrt{3} \]
To find the common ratio \( r \), we can take the ratio of the second term to the first term:
\[ r = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Step 2: Write the formula for the nth term of a geometric progression
The nth term \( a_n \) of a geometric progression can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
Substituting the values of \( a \) and \( r \):
\[ a_n = \sqrt{3} \cdot (\sqrt{3})^{n-1} \]

### Step 3: Simplify the nth term expression
Using the property of exponents, we can combine the terms:
\[ a_n = \sqrt{3} \cdot \sqrt{3}^{n-1} = \sqrt{3}^{1 + (n-1)} = \sqrt{3}^n \]

### Step 4: Set the nth term equal to 729
We know that \( a_n = 729 \), so we can set up the equation:
\[ \sqrt{3}^n = 729 \]

### Step 5: Express 729 in terms of powers of 3
We can express 729 as a power of 3:
\[ 729 = 3^6 \]
Also, we can express \( \sqrt{3} \) as:
\[ \sqrt{3} = 3^{1/2} \]
Thus, we can rewrite the equation:
\[ (3^{1/2})^n = 3^6 \]

### Step 6: Simplify the left side
Using the property of exponents, we simplify the left side:
\[ 3^{n/2} = 3^6 \]

### Step 7: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal:
\[ \frac{n}{2} = 6 \]

### Step 8: Solve for \( n \)
Multiplying both sides by 2 gives:
\[ n = 12 \]

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

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