If `4 + (4 +d)/( 5) + (4 + 2d)/(5 ^(2)) ……….=1,` then find d.

Answer

To solve the equation \( 4 + \frac{4 + d}{5} + \frac{4 + 2d}{5^2} + \ldots = 1 \), we can follow these steps:

### Step 1: Define the series
Let \( S \) be the sum of the series:
\[ S = 4 + \frac{4 + d}{5} + \frac{4 + 2d}{5^2} + \ldots \]

### Step 2: Multiply the series by the common ratio
The common ratio of the series is \( \frac{1}{5} \). Therefore, we can multiply the entire series \( S \) by \( \frac{1}{5} \):
\[ \frac{S}{5} = \frac{4}{5} + \frac{4 + d}{5^2} + \frac{4 + 2d}{5^3} + \ldots \]

### Step 3: Shift the terms
Now, we can rewrite the right-hand side by shifting the terms:
\[ \frac{S}{5} = \frac{4}{5} + \frac{4}{5^2} + \frac{d}{5^2} + \frac{4}{5^3} + \frac{2d}{5^3} + \ldots \]

### Step 4: Subtract the equations
Now we subtract \( \frac{S}{5} \) from \( S \):
\[ S - \frac{S}{5} = 4 + \left( \frac{4}{5} - \frac{4}{5} \right) + \left( \frac{d}{5^2} + \frac{2d}{5^3} + \ldots \right) \]
This simplifies to:
\[ \frac{4S}{5} = 4 + \left( \frac{d}{5^2} + \frac{2d}{5^3} + \ldots \right) \]

### Step 5: Identify the remaining series
The remaining series \( \frac{d}{5^2} + \frac{2d}{5^3} + \ldots \) can be factored:
\[ \frac{d}{5^2} (1 + \frac{1}{5} + \frac{1}{5^2} + \ldots) \]
This is a geometric series with first term \( 1 \) and common ratio \( \frac{1}{5} \). The sum of this infinite series is:
\[ \frac{1}{1 - \frac{1}{5}} = \frac{5}{4} \]
Thus, we have:
\[ \frac{d}{5^2} \cdot \frac{5}{4} = \frac{d}{4} \]

### Step 6: Substitute back into the equation
Substituting back, we get:
\[ \frac{4S}{5} = 4 + \frac{d}{4} \]

### Step 7: Substitute \( S = 1 \)
Since we know \( S = 1 \):
\[ \frac{4 \cdot 1}{5} = 4 + \frac{d}{4} \]
This simplifies to:
\[ \frac{4}{5} = 4 + \frac{d}{4} \]

### Step 8: Solve for \( d \)
Rearranging gives:
\[ \frac{d}{4} = \frac{4}{5} - 4 \]
Calculating the right-hand side:
\[ \frac{4}{5} - 4 = \frac{4}{5} - \frac{20}{5} = -\frac{16}{5} \]
Thus:
\[ \frac{d}{4} = -\frac{16}{5} \]
Multiplying both sides by 4:
\[ d = -\frac{64}{5} \]

### Final Answer
The value of \( d \) is:
\[ \boxed{-\frac{64}{5}} \]