Let the sum of the first n terms of a non-constant A.P., `a_(1), a_(2), a_(3),... " be " 50n + (n (n -7))/(2)A`, where A is a constant. If d is the common difference of this A.P., then the ordered pair `(d, a_(50))` is equal to

Options

A. (50, 50+46A)
B. (A, 50 + 45 A)
C. (A, 50 + 46A)
D. (50, 50 + 45A)

Explanation

To solve the problem, we need to find the ordered pair \((d, a_{50})\) for the given arithmetic progression (A.P.) where the sum of the first \(n\) terms is given as:

\[ S_n = 50n + \frac{n(n-7)}{2}A \]

where \(A\) is a constant.

### Step 1: Express the sum of the first \(n\) terms of an A.P.

The sum of the first \(n\) terms of an A.P. can be expressed as:

\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

where \(a\) is the first term and \(d\) is the common difference.

### Step 2: Set the two expressions for \(S_n\) equal to each other.

Equating the two expressions for \(S_n\):

\[ \frac{n}{2} \left(2a + (n-1)d\right) = 50n + \frac{n(n-7)}{2}A \]

### Step 3: Multiply through by 2 to eliminate the fraction.

\[ n(2a + (n-1)d) = 100n + n(n-7)A \]

### Step 4: Simplify the equation.

Dividing both sides by \(n\) (assuming \(n \neq 0\)) gives:

\[ 2a + (n-1)d = 100 + (n-7)A \]

### Step 5: Rearrange the equation.

Rearranging gives:

\[ (n-1)d = 100 + (n-7)A - 2a \]

### Step 6: Analyze the equation for different values of \(n\).

1. For \(n = 1\):
\[ 0 = 100 + (1-7)A - 2a \implies 0 = 100 - 6A - 2a \]
Thus, we have:
\[ 2a + 6A = 100 \quad \text{(Equation 1)} \]

2. For \(n = 2\):
\[ d = 100 + (2-7)A - 2a \implies d = 100 - 5A - 2a \quad \text{(Equation 2)} \]

### Step 7: Solve the equations.

From Equation 1, we can express \(a\) in terms of \(A\):

\[ 2a = 100 - 6A \implies a = 50 - 3A \]

Substituting \(a\) into Equation 2:

\[ d = 100 - 5A - 2(50 - 3A) \]
\[ d = 100 - 5A - 100 + 6A = A \]

### Step 8: Find \(a_{50}\).

The \(n\)th term of an A.P. is given by:

\[ a_n = a + (n-1)d \]

For \(n = 50\):

\[ a_{50} = a + 49d \]
Substituting \(a\) and \(d\):

\[ a_{50} = (50 - 3A) + 49A = 50 - 3A + 49A = 50 + 46A \]

### Step 9: Form the ordered pair \((d, a_{50})\).

Since \(d = A\), we have:

\[ (d, a_{50}) = (A, 50 + 46A) \]

### Final Result:

The ordered pair is:

\[ \boxed{(A, 50 + 46A)} \]