How many terms of the series `2+6+18+` . . . . . . . . . . . Must be taken to make the sum equal to 728 ?

Answer

To solve the problem of how many terms of the series \(2 + 6 + 18 + \ldots\) must be taken to make the sum equal to 728, we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \(a\) of the series is 2. To find the common ratio \(r\), we can divide the second term by the first term:
\[ r = \frac{6}{2} = 3 \]
We can also check the ratio between the third term and the second term:
\[ r = \frac{18}{6} = 3 \]
Thus, the common ratio \(r\) is 3.

### Step 2: Write the formula for the sum of the first \(n\) terms of a geometric progression (GP)
Since the common ratio \(r\) is greater than 1, we can use the formula for the sum of the first \(n\) terms of a GP:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Where:
- \(S_n\) is the sum of the first \(n\) terms,
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the number of terms.

### Step 3: Substitute the known values into the formula
We know that \(S_n = 728\), \(a = 2\), and \(r = 3\). Substituting these values into the formula gives:
\[ 728 = \frac{2(3^n - 1)}{3 - 1} \]
This simplifies to:
\[ 728 = \frac{2(3^n - 1)}{2} \]
\[ 728 = 3^n - 1 \]

### Step 4: Solve for \(3^n\)
Now, we can add 1 to both sides to isolate \(3^n\):
\[ 3^n = 728 + 1 \]
\[ 3^n = 729 \]

### Step 5: Express 729 as a power of 3
We can express 729 as a power of 3:
\[ 729 = 3^6 \]

### Step 6: Set the exponents equal to each other
Since the bases are equal, we can set the exponents equal:
\[ n = 6 \]

### Conclusion
Thus, the required number of terms is:
\[ \boxed{6} \]