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: Identify the Series
The given series can be separated into two parts: a constant term and a variable term. We can rewrite it as:
\[ S = 4 + \left( \frac{4 + d}{5} + \frac{4 + 2d}{5^2} + \ldots \right) \]
The series inside the parentheses is a geometric series.

### Step 2: Rewrite the Series
We can express the series in a more manageable form:
\[ S = 4 + \sum_{n=0}^{\infty} \frac{4 + nd}{5^{n+1}} \]
This series can be split into two separate series:
\[ S = 4 + \sum_{n=0}^{\infty} \frac{4}{5^{n+1}} + \sum_{n=0}^{\infty} \frac{nd}{5^{n+1}} \]

### Step 3: Calculate the First Series
The first series is a geometric series:
\[ \sum_{n=0}^{\infty} \frac{4}{5^{n+1}} = 4 \left( \frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + \ldots \right) \]
Using the formula for the sum of an infinite geometric series \( S = \frac{a}{1 - r} \), where \( a = \frac{4}{5} \) and \( r = \frac{1}{5} \):
\[ \sum_{n=0}^{\infty} \frac{4}{5^{n+1}} = \frac{4/5}{1 - 1/5} = \frac{4/5}{4/5} = 1 \]

### Step 4: Calculate the Second Series
The second series involves \( n \):
\[ \sum_{n=0}^{\infty} \frac{nd}{5^{n+1}} = d \sum_{n=0}^{\infty} \frac{n}{5^{n+1}} \]
Using the formula for the sum of \( n \cdot r^n \):
\[ \sum_{n=0}^{\infty} n x^n = \frac{x}{(1 - x)^2} \]
Setting \( x = \frac{1}{5} \):
\[ \sum_{n=0}^{\infty} n \left( \frac{1}{5} \right)^n = \frac{\frac{1}{5}}{\left(1 - \frac{1}{5}\right)^2} = \frac{\frac{1}{5}}{\left(\frac{4}{5}\right)^2} = \frac{1/5}{16/25} = \frac{25}{80} = \frac{5}{16} \]
Thus,
\[ \sum_{n=0}^{\infty} \frac{nd}{5^{n+1}} = d \cdot \frac{5}{16 \cdot 5} = \frac{d}{16} \]

### Step 5: Combine the Results
Now we can combine the results:
\[ S = 4 + 1 + \frac{d}{16} = 5 + \frac{d}{16} \]
Setting this equal to 1:
\[ 5 + \frac{d}{16} = 1 \]

### Step 6: Solve for \( d \)
Subtracting 5 from both sides:
\[ \frac{d}{16} = 1 - 5 = -4 \]
Multiplying both sides by 16:
\[ d = -64 \]

### Final Answer
Thus, the value of \( d \) is:
\[ \boxed{-64} \]