Key Concepts and Formulas

• The sum of the first $n$ terms of an arithmetic progression (A.P.) is denoted by $S_n$.
• The $n$-th term of an A.P., $a_n$, can be found using the relation $a_n = S_n - S_{n-1}$ for $n > 1$, and $a_1 = S_1$.
• For an A.P., the $n$-th term $a_n$ is a linear function of $n$, i.e., $a_n = dn + c$, where $d$ is the common difference.
• The sum of the first $n$ terms of an A.P. can be expressed as a quadratic in $n$: $S_n = Pn^2 + Qn$. In this form, the common difference $d = 2P$.

Step-by-Step Solution

Step 1: Derive the general formula for the $n$-th term ($a_n$)

• Why: To find the common difference and any specific term like $a_{50}$, we first need to express the $n$-th term of the A.P. in a general form.
• Math: We are given $S_n = 50n + \frac{n(n-7)}{2}A$. We use the relation $a_n = S_n - S_{n-1}$ for $n > 1$. First, let's find $S_{n-1}$ by substituting $(n-1)$ for $n$ in the expression for $S_n$: $S_{n-1} = 50(n-1) + \frac{(n-1)((n-1)-7)}{2}A$ $S_{n-1} = 50(n-1) + \frac{(n-1)(n-8)}{2}A$ Now, we compute $a_n$: $a_n = S_n - S_{n-1} = \left(50n + \frac{n(n-7)}{2}A\right) - \left(50(n-1) + \frac{(n-1)(n-8)}{2}A\right)$ Group terms not involving $A$ and terms involving $A$: $a_n = (50n - 50(n-1)) + \left(\frac{n(n-7)}{2}A - \frac{(n-1)(n-8)}{2}A\right)$ Simplify the first part: $50n - 50(n-1) = 50n - 50n + 50 = 50$ Simplify the second part: $\frac{A}{2} [n(n-7) - (n-1)(n-8)]$ Expand the terms inside the bracket: $n(n-7) = n^2 - 7n$ $(n-1)(n-8) = n^2 - 8n - n + 8 = n^2 - 9n + 8$ Substitute these back: $\frac{A}{2} [(n^2 - 7n) - (n^2 - 9n + 8)] = \frac{A}{2} [n^2 - 7n - n^2 + 9n - 8]$ $= \frac{A}{2} [2n - 8] = A(n - 4)$ Combining the simplified parts, we get the general term: $a_n = 50 + A(n-4)$
• Reasoning: This formula allows us to find any term of the A.P. and also extract information about the common difference.

Step 2: Determine the common difference ($d$)

• Why: The common difference is a fundamental property of an A.P. and is directly related to the coefficient of $n$ in the general term $a_n$.
• Math: We can rewrite the general term $a_n$ as: $a_n = 50 + An - 4A$ $a_n = An + (50 - 4A)$ This is in the form $a_n = dn + c$, where $d$ is the common difference. By comparing the coefficients, we see that $d = A$. Alternatively, we can calculate $a_{n-1}$: $a_{n-1} = 50 + A((n-1)-4) = 50 + A(n-5)$ The common difference $d = a_n - a_{n-1}$: $d = (50 + A(n-4)) - (50 + A(n-5))$ $d = 50 + An - 4A - 50 - An + 5A$ $d = A$
• Reasoning: The common difference is the constant value added to each term to get the next term in an A.P. Our calculation confirms it is $A$.

Step 3: Calculate the 50th term ($a_{50}$)

• Why: The problem explicitly asks for the value of the 50th term of the A.P.
• Math: Substitute $n=50$ into the general formula for $a_n$: $a_n = 50 + A(n-4)$ $a_{50} = 50 + A(50-4)$ $a_{50} = 50 + A(46)$ $a_{50} = 50 + 46A$
• Reasoning: By plugging in the specific value of $n=50$ into the derived general formula for $a_n$, we obtain the value of the 50th term.

Step 4: Form the ordered pair ($d, a_{50}$)

• Why: The question requires the answer in the form of an ordered pair $(d, a_{50})$.
• Math: From Step 2, we found $d = A$. From Step 3, we found $a_{50} = 50 + 46A$. Therefore, the ordered pair is: $(d, a_{50}) = (A, 50 + 46A)$
• Reasoning: We combine the computed values of $d$ and $a_{50}$ to form the final ordered pair.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when expanding brackets and simplifying expressions, especially when dealing with subtractions of terms involving $n-1$ or $n-8$.
• Using $a_n = S_n - S_{n-1}$ for $n=1$: Remember that $a_1 = S_1$. While the derived formula for $a_n$ often holds for $n=1$ as well, it's good practice to verify or be aware of this distinction.
• Quadratic $S_n$ Shortcut: For an A.P., if $S_n = Pn^2 + Qn$, then $d=2P$. Our given $S_n = 50n + \frac{n(n-7)}{2}A = \frac{A}{2}n^2 + (50 - \frac{7A}{2})n$. Here, $P = \frac{A}{2}$, so $d = 2 \times \frac{A}{2} = A$. This provides a quick check for the common difference.

Summary

We determined the general $n$-th term of the arithmetic progression by utilizing the relationship $a_n = S_n - S_{n-1}$. This allowed us to identify the common difference $d$ as $A$. Subsequently, we calculated the 50th term, $a_{50}$, by substituting $n=50$ into the general term formula, yielding $50 + 46A$. The ordered pair $(d, a_{50})$ is thus $(A, 50 + 46A)$.

The final answer is $\boxed{(A, 50+46A)}$ which corresponds to option (A).