Animated Solution for Mathematics - Sequence and Series: If the sum of the series $20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+...$ upto $n^{th}$ term is 488 and the $n^{th}$ term is negative, then

Every Arithmetic Progression (AP) is defined by two fundamental parameters: its starting point, $a$, and its common difference, $d$. Our series begins at $a=20$.

To determine the common difference, we examine the second term, $19\frac{3}{5}$, which is equivalent to $\frac{98}{5}$. The common difference $d$ is calculated as the difference between the second and first terms:

The negative sign confirms that the series is on a downward trajectory.

We are given that the sum of the first $n$ terms is $S_n=488$. The standard formula for the sum of an AP is:

Substituting our known values into the formula, we obtain:

By factoring out a $2$ from the bracket, we simplify the expression to:

Finding a common denominator inside the bracket yields:

Expanding the equation above, we arrive at the following quadratic form:

Using the quadratic formula $n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we first calculate the discriminant:

Since the square root of $441$ is $21$, we find two potential solutions for $n$:

This results in two candidates: $n=61$ and $n=40$.

Because the series eventually dips into negative values, the sum increases, reaches a peak, and then decreases as negative terms are added. Both $n=40$ and $n=61$ mathematically satisfy the sum of $488$. However, the problem imposes the constraint that the $n^{th}$ term must be negative.

Testing $n=40$:

Since $4.4>0$, we must reject this solution. Testing $n=61$:

Since $-4<0$, this satisfies all conditions. The number of terms is 61, and the $n^{th}$ term is -4.