Let the sum of the first n terms of a non-constant AP `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 AP, then the ordered pair `(d, a_(50))` is equal to

Options

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

Explanation

To solve the problem, we need to find the ordered pair \((d, a_{50})\) where \(d\) is the common difference of the arithmetic progression (AP) and \(a_{50}\) is the 50th term of the AP. We start with the given formula for the sum of the first \(n\) terms of the AP:

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

where \(A\) is a constant.

### Step 1: Find the nth term \(a_n\) using the sum formula

The \(n\)th term of an AP can be expressed in terms of the sum of the first \(n\) terms:

\[ a_n = S_n - S_{n-1} \]

First, we need to calculate \(S_{n-1}\):

\[ S_{n-1} = 50(n-1) + \frac{(n-1)(n-8)}{2}A \]

### Step 2: Substitute \(S_n\) and \(S_{n-1}\) into the formula for \(a_n\)

Now, substituting \(S_n\) and \(S_{n-1}\):

\[ a_n = \left(50n + \frac{n(n-7)}{2}A\right) - \left(50(n-1) + \frac{(n-1)(n-8)}{2}A\right) \]

### Step 3: Simplify the expression for \(a_n\)

Expanding \(S_{n-1}\):

\[ S_{n-1} = 50n - 50 + \frac{(n^2 - 8n + 7)}{2}A \]

Now substituting back into \(a_n\):

\[ a_n = 50n + \frac{n(n-7)}{2}A - (50n - 50 + \frac{(n^2 - 8n + 7)}{2}A) \]

Simplifying this gives:

\[ a_n = 50 + \frac{n(n-7) - (n^2 - 8n + 7)}{2}A \]

### Step 4: Further simplify to find a general form for \(a_n\)

Combine the terms in the numerator:

\[ a_n = 50 + \frac{n^2 - 7n - n^2 + 8n - 7}{2}A \]

This simplifies to:

\[ a_n = 50 + \frac{(n + 1)}{2}A \]

### Step 5: Find the common difference \(d\)

The common difference \(d\) can be found by calculating \(a_{n+1} - a_n\):

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

This results in:

\[ d = \frac{A}{2} \]

### Step 6: Calculate \(a_{50}\)

Now, substituting \(n = 50\) into the expression for \(a_n\):

\[ a_{50} = 50 + \frac{(50 + 1)}{2}A = 50 + \frac{51}{2}A \]

### Step 7: Form the ordered pair \((d, a_{50})\)

Now we have:

\[ d = \frac{A}{2}, \quad a_{50} = 50 + \frac{51}{2}A \]

Thus, the ordered pair \((d, a_{50})\) is:

\[ \left(\frac{A}{2}, 50 + \frac{51}{2}A\right) \]

### Final Answer

The ordered pair \((d, a_{50})\) is:

\[ \left(\frac{A}{2}, 50 + \frac{51}{2}A\right) \]