Let $$a_1, a_2, \ldots, a_n$$ be a given A.P. whose common difference is an integer and $$S_n = a_1 + a_2 + \ldots + a_n$$. If $$a_1 = 1, a_n = 300$$ and $$15 \leq n \leq 50$$, then the ordered pair $$(S_{n-4}, a_{n-4})$$ is equal to:

We are told that $$a_1,a_2,\ldots ,a_n$$ form an arithmetic progression (A.P.) with first term $$a_1=1$$ and last term $$a_n=300$$. Let the common difference be the integer $$d$$. For any A.P. we have the basic formula

$$a_n = a_1 + (n-1)d.$$

Substituting the given values of $$a_1$$ and $$a_n$$, we get

$$300 = 1 + (n-1)d.$$

Rearranging,

$$(n-1)d = 300-1 = 299.$$

Since $$d$$ is an integer and $$299=13 \times 23$$, the integer divisors of $$299$$ are $$1,\,13,\,23,\,299$$. Because $$a_n > a_1$$, the common difference $$d$$ must be positive, and multiplying two positive integers gives $$299$$. Hence one of the factors is $$d$$ and the other is $$n-1$$.

Next, the problem states that $$15 \le n \le 50$$. Therefore $$n-1$$ must lie between $$14$$ and $$49$$ inclusive. Examining the four divisors of $$299$$, only $$23$$ is in that interval. Thus

$$n-1 = 23 \quad\text{and}\quad d = \frac{299}{23}=13.$$

So the progression has

$$n = 24 \quad\text{and}\quad d = 13.$$

Any term of the A.P. can be written as

$$a_k = a_1 + (k-1)d.$$

We need $$a_{\,n-4}$$. Since $$n=24$$,

$$a_{n-4} = a_{20} = 1 + (20-1)\times13 = 1 + 19 \times 13.$$

Evaluating the product, $$19 \times 13 = 247$$, so

$$a_{20} = 1 + 247 = 248.$$

Next we require $$S_{\,n-4} = S_{20}$$, the sum of the first $$20$$ terms. The sum of the first $$m$$ terms of an A.P. is

$$S_m = \frac{m}{2}\bigl[2a_1 + (m-1)d\bigr].$$

Putting $$m = 20,$$ $$a_1 = 1,$$ and $$d = 13,$$ we have

$$S_{20} = \frac{20}{2}\left[2\cdot1 + (20-1)\cdot13\right] = 10\left[2 + 19\cdot13\right].$$

We already found $$19\cdot13 = 247$$, so inside the brackets

$$2 + 247 = 249,$$

and hence

$$S_{20} = 10 \times 249 = 2490.$$

Thus the ordered pair is

$$(S_{\,n-4},\,a_{\,n-4}) = (2490,\,248).$$

Hence, the correct answer is Option 4.