If the sum of the series `frac{1}{1. (1 + d)} + frac{1}{(1 + d) (1 + 2d)} + …+ frac{1}{(1 + 9d) (1 + 10 d)}` is equal to 5, then `50d` is equal to:

Options

A. 15
B. 10
C. 5
D. 20

Answer

To solve the problem, we need to find the value of \(50d\) given that the sum of the series

\[ \frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d) \cdot (1 + 2d)} + \ldots + \frac{1}{(1 + 9d) \cdot (1 + 10d)} = 5. \]

### Step 1: Write the general term of the series

The general term of the series can be written as:

\[ T_n = \frac{1}{(1 + (n-1)d)(1 + nd)} \]

for \(n = 1\) to \(10\).

### Step 2: Simplify the general term

We can simplify \(T_n\) using partial fractions:

\[ T_n = \frac{1}{(1 + (n-1)d)(1 + nd)} = \frac{A}{1 + (n-1)d} + \frac{B}{1 + nd} \]

Multiplying through by the denominator \((1 + (n-1)d)(1 + nd)\) gives:

\[ 1 = A(1 + nd) + B(1 + (n-1)d) \]

Expanding and rearranging:

\[ 1 = A + And + B + B(n-1)d \]

Setting up equations for coefficients, we can find \(A\) and \(B\).

### Step 3: Find \(A\) and \(B\)

Setting \(d = 0\) gives \(A + B = 1\).

Setting \(d = 1\) gives \(A + B(n-1) = 0\).

Solving these equations, we find:

\[ A = \frac{1}{d}, \quad B = -\frac{1}{d} \]

Thus,

\[ T_n = \frac{1}{d} \left( \frac{1}{1 + (n-1)d} - \frac{1}{1 + nd} \right) \]

### Step 4: Write the sum of the series

The sum of the series can be expressed as:

\[ S = \frac{1}{d} \left( \left( \frac{1}{1 + 0d} - \frac{1}{1 + 1d} \right) + \left( \frac{1}{1 + 1d} - \frac{1}{1 + 2d} \right) + \ldots + \left( \frac{1}{1 + 9d} - \frac{1}{1 + 10d} \right) \right) \]

This is a telescoping series, which simplifies to:

\[ S = \frac{1}{d} \left( 1 - \frac{1}{1 + 10d} \right) \]

### Step 5: Set the sum equal to 5

We know that:

\[ \frac{1}{d} \left( 1 - \frac{1}{1 + 10d} \right) = 5 \]

Multiplying through by \(d\):

\[ 1 - \frac{1}{1 + 10d} = 5d \]

Rearranging gives:

\[ 1 - 5d = \frac{1}{1 + 10d} \]

### Step 6: Cross-multiply and solve for \(d\)

Cross-multiplying gives:

\[ (1 - 5d)(1 + 10d) = 1 \]

Expanding:

\[ 1 + 10d - 5d - 50d^2 = 1 \]

This simplifies to:

\[ -50d^2 + 5d = 0 \]

Factoring out \(d\):

\[ d(5 - 50d) = 0 \]

Thus, \(d = 0\) or \(d = \frac{1}{10}\).

Since \(d = 0\) is not valid in this context, we have:

\[ d = \frac{1}{10}. \]

### Step 7: Find \(50d\)

Finally, we calculate:

\[ 50d = 50 \times \frac{1}{10} = 5. \]

### Conclusion

Thus, the value of \(50d\) is:

\[ \boxed{5}. \]