First negative term of the series `4,3(5)/(7),3(3)/(7),…` is :

Options

A. 15
B. 16
C. 17
D. None of these

Explanation

To find the first negative term of the series \(4, \frac{3}{7}, \frac{3}{7}, \ldots\), we will follow these steps:

### Step 1: Identify the first term and the common difference
The first term \(a\) of the series is \(4\). To find the common difference \(d\), we will calculate the difference between the second term and the first term.

\[ d = a_2 - a_1 = \frac{3}{7} - 4 \]

To perform this subtraction, we need to express \(4\) in terms of a fraction with a denominator of \(7\):

\[ 4 = \frac{28}{7} \]

Now we can calculate \(d\):

\[ d = \frac{3}{7} - \frac{28}{7} = \frac{3 - 28}{7} = \frac{-25}{7} \]

### Step 2: Write the general term of the arithmetic progression (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 \(d\):

\[ a_n = 4 + (n-1)\left(\frac{-25}{7}\right) \]

### Step 3: Set the general term equal to zero to find when it becomes negative
To find the first negative term, we need to determine when the \(n\)-th term is less than zero:

\[ 4 + (n-1)\left(\frac{-25}{7}\right) < 0 \]

### Step 4: Solve the inequality
First, we set the equation to zero to find the critical point:

\[ 4 + (n-1)\left(\frac{-25}{7}\right) = 0 \]

Rearranging gives:

\[ (n-1)\left(\frac{-25}{7}\right) = -4 \]

Multiplying both sides by \(-1\):

\[ (n-1)\left(\frac{25}{7}\right) = 4 \]

Now, multiply both sides by \(\frac{7}{25}\):

\[ n-1 = \frac{4 \times 7}{25} = \frac{28}{25} \]

Adding \(1\) to both sides gives:

\[ n = 1 + \frac{28}{25} = \frac{25}{25} + \frac{28}{25} = \frac{53}{25} \]

Since \(n\) must be a whole number, we round up to the next whole number, which is \(3\).

### Step 5: Find the first negative term
Now we will check the \(n=3\) term:

\[ a_3 = 4 + (3-1)\left(\frac{-25}{7}\right) = 4 + 2\left(\frac{-25}{7}\right) \]

Calculating this gives:

\[ a_3 = 4 - \frac{50}{7} = \frac{28}{7} - \frac{50}{7} = \frac{-22}{7} \]

Thus, the first negative term occurs at \(n=3\).

### Conclusion
The first negative term of the series is \(\frac{-22}{7}\).

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