Find the largest positive term of the H.P.. Whose first two term are `2/5 and (12)/(23).`

Answer

To find the largest positive term of the Harmonic Progression (H.P.) whose first two terms are \( \frac{2}{5} \) and \( \frac{12}{23} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the first two terms of the H.P.**:
The first two terms are given as \( a_1 = \frac{2}{5} \) and \( a_2 = \frac{12}{23} \).

2. **Convert H.P. to A.P.**:
Since H.P. is the reciprocal of A.P., we can find the corresponding A.P. terms:
\[ b_1 = \frac{5}{2}, \quad b_2 = \frac{23}{12} \]

3. **Find the common difference of the A.P.**:
The common difference \( d \) of the A.P. can be calculated as:
\[ d = b_2 - b_1 = \frac{23}{12} - \frac{5}{2} \]
To perform this subtraction, we need a common denominator. The least common multiple of 12 and 2 is 12.
\[ \frac{5}{2} = \frac{5 \times 6}{2 \times 6} = \frac{30}{12} \]
Now, substituting back:
\[ d = \frac{23}{12} - \frac{30}{12} = \frac{23 - 30}{12} = \frac{-7}{12} \]

4. **Write the general term of the A.P.**:
The \( n \)-th term of the A.P. is given by:
\[ T_n = b_1 + (n-1)d = \frac{5}{2} + (n-1) \left(-\frac{7}{12}\right) \]

5. **Simplify the expression for \( T_n \)**:
\[ T_n = \frac{5}{2} - \frac{7(n-1)}{12} \]
To combine these fractions, we convert \( \frac{5}{2} \) to have a denominator of 12:
\[ \frac{5}{2} = \frac{5 \times 6}{2 \times 6} = \frac{30}{12} \]
Thus,
\[ T_n = \frac{30}{12} - \frac{7(n-1)}{12} = \frac{30 - 7(n-1)}{12} \]

6. **Set the condition for positive terms**:
For \( T_n \) to be positive:
\[ 30 - 7(n-1) > 0 \]
Simplifying this:
\[ 30 - 7n + 7 > 0 \implies 37 > 7n \implies n < \frac{37}{7} \approx 5.2857 \]
Therefore, the largest integer \( n \) is \( n = 5 \).

7. **Calculate the largest positive term**:
Now, substituting \( n = 5 \) back into the formula for \( T_n \):
\[ T_5 = \frac{30 - 7(5-1)}{12} = \frac{30 - 28}{12} = \frac{2}{12} = \frac{1}{6} \]

### Final Answer:
The largest positive term of the H.P. is \( \frac{1}{6} \).