If the sum of the series ` 20 + 19 (3)/(5) + 19 (1)/(5) + 18 (4)/(5) + …..` upto nth tems is 488 and the nth terms is negative then :

Options

A. nth term is -4
B. nth terms `-4 (2)/(5)`
C. n = 60
D. n = 41

Answer

To solve the problem, we need to analyze the series and find the value of \( n \) such that the sum of the series equals 488 and the \( n \)-th term is negative.

### Step-by-step Solution:

1. **Identify the Series**:
The series given is:
\[ 20 + 19 \frac{3}{5} + 19 \frac{1}{5} + 18 \frac{4}{5} + \ldots \]
We can observe that the first term \( a = 20 \) and the common difference \( d \) can be determined from the terms.

2. **Finding the Common Difference**:
The second term is \( 19 \frac{3}{5} = 19.6 \) and the first term is 20. The difference between the first and second term is:
\[ 20 - 19.6 = 0.4 \]
The next term \( 19 \frac{1}{5} = 19.2 \) gives:
\[ 19.6 - 19.2 = 0.4 \]
The next term \( 18 \frac{4}{5} = 18.8 \) gives:
\[ 19.2 - 18.8 = 0.4 \]
Thus, the common difference \( d = -\frac{2}{5} \).

3. **Sum of the Arithmetic Series**:
The sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
Plugging in the values we have:
\[ S_n = \frac{n}{2} \left(2 \cdot 20 + (n-1) \left(-\frac{2}{5}\right)\right) \]
Simplifying this gives:
\[ S_n = \frac{n}{2} \left(40 - \frac{2(n-1)}{5}\right) \]
\[ S_n = \frac{n}{2} \left(40 - \frac{2n - 2}{5}\right) \]
\[ S_n = \frac{n}{2} \left(40 - \frac{2n}{5} + \frac{2}{5}\right) \]
\[ S_n = \frac{n}{2} \left(\frac{200 - 2n + 2}{5}\right) \]
\[ S_n = \frac{n(202 - 2n)}{10} \]

4. **Setting the Sum Equal to 488**:
We set the sum equal to 488:
\[ \frac{n(202 - 2n)}{10} = 488 \]
Multiplying both sides by 10:
\[ n(202 - 2n) = 4880 \]
Rearranging gives:
\[ 2n^2 - 202n + 4880 = 0 \]

5. **Solving the Quadratic Equation**:
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here \( a = 2, b = -202, c = 4880 \):
\[ n = \frac{202 \pm \sqrt{(-202)^2 - 4 \cdot 2 \cdot 4880}}{2 \cdot 2} \]
\[ n = \frac{202 \pm \sqrt{40804 - 39040}}{4} \]
\[ n = \frac{202 \pm \sqrt{1764}}{4} \]
\[ n = \frac{202 \pm 42}{4} \]
This gives us two possible values for \( n \):
\[ n = \frac{244}{4} = 61 \quad \text{and} \quad n = \frac{160}{4} = 40 \]

6. **Finding the n-th Term**:
The n-th term \( a_n \) is given by:
\[ a_n = a + (n-1)d \]
For \( n = 40 \):
\[ a_{40} = 20 + (40-1)\left(-\frac{2}{5}\right) = 20 - \frac{78}{5} = \frac{100 - 78}{5} = \frac{22}{5} = 4.4 \quad (\text{positive}) \]
For \( n = 61 \):
\[ a_{61} = 20 + (61-1)\left(-\frac{2}{5}\right) = 20 - \frac{120}{5} = 20 - 24 = -4 \quad (\text{negative}) \]

7. **Conclusion**:
Since we need the \( n \)-th term to be negative, we take \( n = 61 \) and the \( n \)-th term is \( -4 \).

### Final Answer:
The value of \( n \) is 61 and the \( n \)-th term is \( -4 \).