The least positive term of an arithmetic progression whose first two term are `(5)/(2)` and `(23)/(12)` is

Options

A. 6
B. 5
C. `(1)/(6)`
D. `(37)/(7)`

Answer

To find the least positive term of the arithmetic progression (AP) whose first two terms are \( \frac{5}{2} \) and \( \frac{23}{12} \), we will follow these steps:

### Step 1: Identify the first term and the second term
Let the first term \( a = \frac{5}{2} \) and the second term \( a + d = \frac{23}{12} \).

### Step 2: Calculate the common difference \( d \)
To find \( d \), we can rearrange the equation:
\[ d = a + d - a = \frac{23}{12} - \frac{5}{2} \]
Now, we need to find a common denominator to perform the subtraction. The least common multiple of 12 and 2 is 12.

Convert \( \frac{5}{2} \) to have a denominator of 12:
\[ \frac{5}{2} = \frac{5 \times 6}{2 \times 6} = \frac{30}{12} \]
Now substitute back into the equation:
\[ d = \frac{23}{12} - \frac{30}{12} = \frac{23 - 30}{12} = \frac{-7}{12} \]

### Step 3: Write the general term of the AP
The \( n \)-th term of an AP can be expressed as:
\[ T_n = a + (n - 1)d \]
Substituting the values of \( a \) and \( d \):
\[ T_n = \frac{5}{2} + (n - 1) \left(-\frac{7}{12}\right) \]
This simplifies to:
\[ T_n = \frac{5}{2} - \frac{7(n - 1)}{12} \]

### Step 4: Set the general term greater than zero
To find the least positive term, we need to solve for \( n \) such that \( T_n > 0 \):
\[ \frac{5}{2} - \frac{7(n - 1)}{12} > 0 \]
Multiply through by 12 to eliminate the fraction:
\[ 12 \cdot \frac{5}{2} - 7(n - 1) > 0 \]
This simplifies to:
\[ 30 - 7(n - 1) > 0 \]
Expanding gives:
\[ 30 - 7n + 7 > 0 \quad \Rightarrow \quad 37 - 7n > 0 \]
Rearranging leads to:
\[ 37 > 7n \quad \Rightarrow \quad n < \frac{37}{7} \approx 5.2857 \]

### Step 5: Find the largest integer \( n \)
The largest integer \( n \) that satisfies this inequality is \( n = 5 \).

### Step 6: Calculate the 5th term
Now substitute \( n = 5 \) back into the formula for \( T_n \):
\[ T_5 = \frac{5}{2} + (5 - 1) \left(-\frac{7}{12}\right) \]
This simplifies to:
\[ T_5 = \frac{5}{2} - \frac{28}{12} \]
Convert \( \frac{5}{2} \) to have a denominator of 12:
\[ \frac{5}{2} = \frac{30}{12} \]
Now substitute:
\[ T_5 = \frac{30}{12} - \frac{28}{12} = \frac{2}{12} = \frac{1}{6} \]

### Conclusion
The least positive term of the arithmetic progression is:
\[ \boxed{\frac{1}{6}} \]