How many terms of the geometric progression `1+4+16+64+ . . . . .. . .` must be added to get sum equal to 5461 ?

Answer

To solve the problem of how many terms of the geometric progression \(1 + 4 + 16 + 64 + \ldots\) must be added to get a sum equal to 5461, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the first term and common ratio:**
- The first term \(a = 1\).
- The common ratio \(r\) can be found by dividing the second term by the first term:
\[ r = \frac{4}{1} = 4. \]

2. **Use the formula for the sum of the first \(n\) terms of a geometric progression:**
- The formula for the sum of the first \(n\) terms \(S_n\) of a geometric progression is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1}. \]
- Here, we need to find \(n\) such that \(S_n = 5461\).

3. **Substitute the known values into the formula:**
- Substitute \(a = 1\), \(r = 4\), and \(S_n = 5461\) into the formula:
\[ 5461 = \frac{1(4^n - 1)}{4 - 1}. \]
- This simplifies to:
\[ 5461 = \frac{4^n - 1}{3}. \]

4. **Multiply both sides by 3 to eliminate the fraction:**
\[ 5461 \times 3 = 4^n - 1. \]
- Calculate \(5461 \times 3\):
\[ 16383 = 4^n - 1. \]

5. **Add 1 to both sides:**
\[ 16384 = 4^n. \]

6. **Express \(16384\) as a power of \(4\):**
- We can rewrite \(16384\) as \(4^7\) since \(4^7 = (2^2)^7 = 2^{14} = 16384\).

7. **Equate the exponents:**
\[ n = 7. \]

### Conclusion:
Thus, **7 terms** of the geometric progression must be added to get a sum equal to 5461.