Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant.
  • The $k^{th}$ term: $a_k = a_1 + (k-1)d$, where $a_1$ is the first term and $d$ is the common difference.
  • The sum of the first $k$ terms: $S_k = \frac{k}{2}(a_1 + a_k)$ or $S_k = \frac{k}{2}(2a_1 + (k-1)d)$.
• Integer Properties: Understanding integer factors and their constraints is crucial.

Step-by-Step Solution

Step 1: Determine the Number of Terms ($n$) and Common Difference ($d$)

We are given $a_1 = 1$, $a_n = 300$, $d$ is an integer, and $15 \le n \le 50$. Using the formula for the $n^{th}$ term: $a_n = a_1 + (n-1)d$ Substituting the given values: $300 = 1 + (n-1)d$ Simplifying the equation: $(n-1)d = 300 - 1$ $(n-1)d = 299$ We need to find integer factors of 299. Let's find the prime factorization of 299. We test small prime numbers: 299 is not divisible by 2, 3, 5, 7, 11. Testing 13: $299 \div 13 = 23$. So, $299 = 13 \times 23$. Both 13 and 23 are prime numbers. The possible integer factor pairs for $(n-1, d)$ are $(1, 299), (299, 1), (13, 23), (23, 13)$. Since $a_1 < a_n$ and $n \ge 15 > 1$, both $n-1$ and $d$ must be positive. We are also given the constraint $15 \le n \le 50$. This implies $14 \le n-1 \le 49$. Let's examine the factor pairs for $(n-1, d)$:

• If $n-1 = 1$, then $n=2$. This is not in the range $[15, 50]$.
• If $n-1 = 299$, then $n=300$. This is not in the range $[15, 50]$.
• If $n-1 = 13$, then $n=14$. This is not in the range $[15, 50]$ as $n$ must be greater than or equal to 15.
• If $n-1 = 23$, then $n=24$. This value of $n=24$ is within the range $[15, 50]$. If $n-1 = 23$, then $d = 13$. Since $d=13$ is an integer, this is the correct pair.

Thus, we have $n=24$ and $d=13$.

Step 2: Calculate $a_{n-4}$

We need to find $a_{n-4}$. Since $n=24$, we need to find $a_{24-4} = a_{20}$. Using the formula $a_k = a_1 + (k-1)d$: $a_{20} = a_1 + (20-1)d$ Substitute $a_1=1$ and $d=13$: $a_{20} = 1 + (19)(13)$ $a_{20} = 1 + 247$ $a_{20} = 248$ So, $a_{n-4} = 248$.

Step 3: Calculate $S_{n-4}$

We need to find $S_{n-4}$. Since $n=24$, we need to find $S_{24-4} = S_{20}$. Using the sum formula $S_k = \frac{k}{2}(a_1 + a_k)$: $S_{20} = \frac{20}{2}(a_1 + a_{20})$ Substitute $a_1=1$ and $a_{20}=248$: $S_{20} = 10(1 + 248)$ $S_{20} = 10(249)$ $S_{20} = 2490$ So, $S_{n-4} = 2490$.

Step 4: Form the Ordered Pair

The ordered pair $(S_{n-4}, a_{n-4})$ is $(2490, 248)$.

Comparing this with the given options: (A) (2480, 249) (B) (2480, 248) (C) (2490, 248) (D) (2490, 249)

Our calculated pair matches option (C).

Common Mistakes & Tips

• Constraint Check: Always verify that the determined values of $n$ and $d$ satisfy all given constraints, especially the range for $n$.
• Factorization Precision: When solving $(n-1)d = C$, systematically listing all factor pairs of $C$ and then applying the constraints on $n$ (and consequently $n-1$) is crucial.
• Index Calculation: Be careful when calculating the required term $a_{n-4}$ and sum $S_{n-4}$. Ensure you are using the correct index value ($n-4$).

Summary

By utilizing the formulas for the $n^{th}$ term and the sum of an arithmetic progression, along with the given constraints on the number of terms and the integer nature of the common difference, we uniquely determined $n=24$ and $d=13$. Subsequently, we calculated the required term $a_{n-4} = a_{20} = 248$ and the sum $S_{n-4} = S_{20} = 2490$. This leads to the ordered pair $(2490, 248)$.

The final answer is $\boxed{(2490, 248)}$, which corresponds to option (C).