Find the sum of a geometric series in which ` a=16 , r=(1)/(4) ,l = (1)/(64)` .

Answer

To find the sum of the geometric series given \( a = 16 \), \( r = \frac{1}{4} \), and \( l = \frac{1}{64} \), we can follow these steps:

### Step 1: Identify the formula for the last term
The last term \( l \) of a geometric series can be expressed using the formula:
\[ l = a \cdot r^{n-1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

### Step 2: Substitute the known values into the formula
Substituting the values we have:
\[ \frac{1}{64} = 16 \cdot \left(\frac{1}{4}\right)^{n-1} \]

### Step 3: Simplify the equation
First, we can isolate \( \left(\frac{1}{4}\right)^{n-1} \):
\[ \left(\frac{1}{4}\right)^{n-1} = \frac{1}{64} \cdot \frac{1}{16} \]
Calculating the right side:
\[ \frac{1}{64} = \frac{1}{4^3}, \quad \text{and} \quad \frac{1}{16} = \frac{1}{4^2} \]
Thus,
\[ \frac{1}{64} \cdot \frac{1}{16} = \frac{1}{4^3} \cdot \frac{1}{4^2} = \frac{1}{4^{3+2}} = \frac{1}{4^5} \]
So we have:
\[ \left(\frac{1}{4}\right)^{n-1} = \frac{1}{4^5} \]

### Step 4: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ n - 1 = 5 \]
Solving for \( n \):
\[ n = 6 \]

### Step 5: Calculate the sum of the geometric series
The sum \( S_n \) of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \cdot \frac{r^n - 1}{r - 1} \]
Substituting the known values:
\[ S_6 = 16 \cdot \frac{\left(\frac{1}{4}\right)^6 - 1}{\frac{1}{4} - 1} \]

### Step 6: Calculate \( \left(\frac{1}{4}\right)^6 \)
Calculating \( \left(\frac{1}{4}\right)^6 \):
\[ \left(\frac{1}{4}\right)^6 = \frac{1}{4^6} = \frac{1}{4096} \]
So,
\[ S_6 = 16 \cdot \frac{\frac{1}{4096} - 1}{\frac{1}{4} - 1} \]

### Step 7: Simplify the denominator
Calculating \( \frac{1}{4} - 1 \):
\[ \frac{1}{4} - 1 = \frac{1 - 4}{4} = \frac{-3}{4} \]

### Step 8: Substitute back into the sum formula
Now substituting back:
\[ S_6 = 16 \cdot \frac{\frac{1}{4096} - 1}{\frac{-3}{4}} = 16 \cdot \frac{\frac{1 - 4096}{4096}}{\frac{-3}{4}} = 16 \cdot \frac{-4095/4096}{-3/4} \]

### Step 9: Simplify the expression
This simplifies to:
\[ S_6 = 16 \cdot \frac{4095}{4096} \cdot \frac{4}{3} = \frac{64 \cdot 4095}{4096 \cdot 3} \]

### Step 10: Final calculation
Calculating \( \frac{64 \cdot 4095}{4096 \cdot 3} \):
\[ S_6 = \frac{64 \cdot 4095}{12288} \]
Now simplifying \( \frac{64}{4096} = \frac{1}{64} \):
\[ S_6 = \frac{4095}{64 \cdot 3} = \frac{4095}{192} \]

### Final Answer
Thus, the sum of the geometric series is:
\[ \frac{1365}{64} \]