If the sum of the series $$20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + \ldots$$ up to $$n^{th}$$ term is 488 and the $$n^{th}$$ term is negative, then:

We begin by observing that the sequence

$$20,\;19\dfrac35,\;19\dfrac15,\;18\dfrac45,\ldots$$

is clearly decreasing by equal steps of $$\dfrac25$$. Converting the mixed numbers to fractions in fifths or to decimals helps to see this:

$$20 = 20.0,\qquad19\dfrac35 = 19.6,\qquad19\dfrac15 = 19.2,\qquad18\dfrac45 = 18.8.$$

The common difference is therefore

$$d = 19.6-20 = -0.4 = -\dfrac25.$$

So we are dealing with an arithmetic progression whose first term is

$$a = 20$$

and whose common difference is

$$d = -\dfrac25.$$

We are told that the sum of the first $$n$$ terms is $$488$$. For an AP the sum of the first $$n$$ terms is given by the formula

$$S_n = \dfrac n2\Bigl[2a + (n-1)d\Bigr].$$

Substituting $$a = 20$$, $$d = -\dfrac25$$ and $$S_n = 488$$, we write

$$488 = \dfrac n2\Bigl[2(20) + (n-1)\!\left(-\dfrac25\right)\Bigr].$$

Simplifying inside the bracket first, we have

$$2(20) = 40,$$

so the bracket becomes

$$40 - (n-1)\dfrac25.$$

Hence

$$488 = \dfrac n2\left[40 - \dfrac{2(n-1)}5\right].$$

To clear the denominator 2, multiply both sides by 2:

$$976 = n\left[40 - \dfrac{2(n-1)}5\right].$$

Now expand the right-hand side:

$$976 = 40n - \dfrac{2n(n-1)}5.$$

Multiplying through by 5 to eliminate the denominator gives

$$4880 = 200n - 2n(n-1).$$

Next, expand the quadratic term:

$$2n(n-1) = 2n^2 - 2n,$$

and substitute it back:

$$4880 = 200n - \bigl(2n^2 - 2n\bigr).$$

Removing the brackets carefully, we get

$$4880 = 200n - 2n^2 + 2n.$$

Collecting like terms on the right gives

$$4880 = -2n^2 + 202n.$$

Transposing all terms to one side so that we have zero on the other side, we write

$$-2n^2 + 202n - 4880 = 0.$$

Multiplying by $$-1$$ to make the leading coefficient positive, we obtain

$$2n^2 - 202n + 4880 = 0.$$

Dividing every term by $$2$$ to simplify, we get the quadratic equation

$$n^2 - 101n + 2440 = 0.$$

We solve this using the quadratic formula. For $$ax^2+bx+c=0$$ the roots are

$$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$

Here $$a = 1,\; b = -101,\; c = 2440,$$ so

$$n = \dfrac{101 \pm \sqrt{(-101)^2 - 4(1)(2440)}}{2}.$$

Calculating the discriminant first:

$$(-101)^2 = 10201,\qquad4\cdot1\cdot2440 = 9760,$$

$$\sqrt{10201 - 9760} = \sqrt{441} = 21.$$

Hence

$$n = \dfrac{101 \pm 21}{2}.$$

This gives two possible values:

$$n = \dfrac{101 + 21}{2} = \dfrac{122}{2} = 61,$$

$$n = \dfrac{101 - 21}{2} = \dfrac{80}{2} = 40.$$

We must decide which of these fulfils the extra condition that the $$n^{\text{th}}$$ term itself is negative. The general formula for the $$n^{\text{th}}$$ term of an AP is

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

First, try $$n = 40$$:

$$a_{40} = 20 + (40-1)\!\left(-\dfrac25\right) = 20 - 39\!\left(\dfrac25\right) = 20 - \dfrac{78}{5} = 20 - 15.6 = 4.4,$$

which is positive. Hence $$n = 40$$ is rejected.

Next, try $$n = 61$$:

$$a_{61} = 20 + (61-1)\!\left(-\dfrac25\right) = 20 - 60\!\left(\dfrac25\right) = 20 - 24 = -4.$$

This value is indeed negative, matching the condition. Therefore the required $$n$$ is $$61$$ and the $$n^{\text{th}}$$ term equals $$-4$$.

Among the options provided, the one that states “$$n^{\text{th}}$$ term is $$-4$$” corresponds precisely to our result.

Hence, the correct answer is Option C.