If the sum of first n terms of series `20+19 3/5+19 1/5+18 4/5+....` is 488 and nth term is negative then find n.

Options

A. -4
B. 4
C. 1
D. 6

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the first \( n \) terms of the series \( 20 + 19 \frac{3}{5} + 19 \frac{1}{5} + 18 \frac{4}{5} + \ldots \) equals 488 and the \( n \)-th term is negative.

### Step 1: Identify the first term and the common difference
The first term \( a \) of the series is:
\[ a = 20 \]

To find the common difference \( d \), we can subtract the first term from the second term:
\[ d = 19 \frac{3}{5} - 20 = \frac{98}{5} - \frac{100}{5} = -\frac{2}{5} \]

### Step 2: Write the formula for the sum of the first \( n \) terms
The formula for the sum of the first \( n \) terms \( S_n \) of an arithmetic progression (AP) is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

### Step 3: Substitute known values into the sum formula
We know that \( S_n = 488 \), \( a = 20 \), and \( d = -\frac{2}{5} \). Substituting these values into the sum formula gives:
\[ 488 = \frac{n}{2} \left(2 \times 20 + (n - 1) \left(-\frac{2}{5}\right)\right) \]
\[ 488 = \frac{n}{2} \left(40 - \frac{2(n - 1)}{5}\right) \]

### Step 4: Simplify the equation
Multiplying both sides by 2 to eliminate the fraction:
\[ 976 = n \left(40 - \frac{2(n - 1)}{5}\right) \]
\[ 976 = n \left(40 - \frac{2n - 2}{5}\right) \]
\[ 976 = n \left(40 - \frac{2n}{5} + \frac{2}{5}\right) \]
\[ 976 = n \left(40 + \frac{2}{5} - \frac{2n}{5}\right) \]
\[ 976 = n \left(\frac{200 + 2 - 2n}{5}\right) \]
\[ 976 = \frac{n(202 - 2n)}{5} \]

### Step 5: Multiply through by 5
\[ 4880 = n(202 - 2n) \]
\[ 4880 = 202n - 2n^2 \]
\[ 2n^2 - 202n + 4880 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = -202 \), and \( c = 4880 \):
\[ b^2 - 4ac = (-202)^2 - 4 \times 2 \times 4880 \]
\[ = 40804 - 39040 = 1764 \]
\[ \sqrt{1764} = 42 \]
Now substituting back into the formula:
\[ n = \frac{202 \pm 42}{4} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{244}{4} = 61 \)
2. \( n = \frac{160}{4} = 40 \)

### Step 7: Check which \( n \) gives a negative \( n \)-th term
The \( n \)-th term \( t_n \) is given by:
\[ t_n = a + (n - 1)d \]

**For \( n = 61 \)**:
\[ t_{61} = 20 + (61 - 1) \left(-\frac{2}{5}\right) = 20 - 60 = -40 \quad (\text{Negative}) \]

**For \( n = 40 \)**:
\[ t_{40} = 20 + (40 - 1) \left(-\frac{2}{5}\right) = 20 - 31.6 = -11.6 \quad (\text{Negative}) \]

### Conclusion
Both \( n = 61 \) and \( n = 40 \) yield negative \( n \)-th terms. However, since we need to find \( n \) such that the \( n \)-th term is negative, we can conclude that both values satisfy the condition.

### Final Answer
Thus, the values of \( n \) that satisfy the conditions are \( n = 40 \) and \( n = 61 \).