Which term of the progression `19, 18 1/5, 17 2/5` ………. Is the first negative term?

Answer

To find the first negative term of the arithmetic progression (AP) given by the sequence \(19, 18 \frac{1}{5}, 17 \frac{2}{5}, \ldots\), we will follow these steps:

### Step 1: Identify the first term and the common difference
The first term \(a\) of the sequence is:
\[ a = 19 \]

Next, we need to find the common difference \(d\). We can calculate it by subtracting the second term from the first term:
\[ d = A_2 - A_1 = 18 \frac{1}{5} - 19 = \frac{91}{5} - \frac{95}{5} = -\frac{4}{5} \]

### Step 2: Write the general term of the AP
The \(n\)-th term of an arithmetic progression can be expressed as:
\[ A_n = a + (n-1)d \]
Substituting the values of \(a\) and \(d\):
\[ A_n = 19 + (n-1)\left(-\frac{4}{5}\right) \]

### Step 3: Set up the inequality for the first negative term
We need to find the smallest \(n\) for which \(A_n < 0\):
\[ 19 + (n-1)\left(-\frac{4}{5}\right) < 0 \]

### Step 4: Solve the inequality
Rearranging the inequality:
\[ 19 - \frac{4(n-1)}{5} < 0 \]
Multiplying through by 5 to eliminate the fraction:
\[ 95 - 4(n-1) < 0 \]
Expanding and simplifying:
\[ 95 - 4n + 4 < 0 \implies 99 - 4n < 0 \]
This leads to:
\[ 4n > 99 \implies n > \frac{99}{4} \implies n > 24.75 \]

Since \(n\) must be a whole number, the smallest integer \(n\) satisfying this condition is:
\[ n = 25 \]

### Step 5: Verify the 25th term
Now, we will calculate the 25th term to ensure it is negative:
\[ A_{25} = 19 + (25-1)\left(-\frac{4}{5}\right) = 19 + 24\left(-\frac{4}{5}\right) \]
Calculating:
\[ A_{25} = 19 - \frac{96}{5} = \frac{95}{5} - \frac{96}{5} = -\frac{1}{5} \]

Thus, the first negative term occurs at \(n = 25\).

### Conclusion
The first negative term of the progression is at \(n = 25\).

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