What is the greatest positive term of the H.P. whose first two terms are $\frac{2}{5}$ and $\frac{12}{23}$?

Key Concepts Needed

1. Harmonic Progression (HP)
   An HP is a sequence $a_1, a_2, a_3, \dots$ where the reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \dots$ form an Arithmetic Progression (AP).

2. Arithmetic Progression (AP)
   An AP is a sequence with a constant difference $d$:
   $A,\; A+d,\; A+2d,\; A+3d,\; \dots$
   Here $A$ is the first term (often denoted $a$) and $d$ is the common difference.

3. Link Between HP and AP
   If $a_n$ is the $n$-th term of the HP, then its reciprocal forms the AP:
   $\frac{1}{a_n}=A + (n-1)d$
   which implies
   $a_n = \frac{1}{\,A + (n-1)d\,}$

Applying the Theory to the Given Problem

1. Identify the first two HP terms
   $a_1 = \frac{2}{5}, \qquad a_2 = \frac{12}{23}$

2. Form their reciprocals (to get the AP)
   $\frac{1}{a_1} = \frac{5}{2}, \qquad \frac{1}{a_2} = \frac{23}{12}$

3. Find the common difference $d$ of the AP
   $d = \frac{23}{12} - \frac{5}{2} = \frac{23}{12} - \frac{30}{12} = -\frac{7}{12}$
   (Negative $d$ tells us the AP terms decrease.)

4. General $n$-th reciprocal term
   $\frac{1}{a_n} = A + (n-1)d = \frac{5}{2} + (n-1)\Bigl(-\frac{7}{12}\Bigr)$

5. Condition for positivity of HP terms
   We need the denominator (the AP term) to stay strictly positive:
   $\frac{5}{2} + (n-1)\Bigl(-\frac{7}{12}\Bigr) > 0$

6. Solve the inequality
   Convert everything to twelfths to ease calculation:
   $\frac{5}{2} = \frac{30}{12}$
   So,

   $30 - 7(n-1) > 0$
   $30 > 7(n-1)$
   $n - 1 < \frac{30}{7} \approx 4.2857$
   Since $n$ is an integer, the largest permissible $n$ is 5.

7. Compute the 5th HP term

   = \frac{5}{2} - \frac{28}{12} = \frac{5}{2} - \frac{7}{3}$$ Put on a common denominator (6): $$\frac{5}{2} = \frac{15}{6}, \qquad \frac{7}{3} = \frac{14}{6}$$ $$\frac{1}{a_5} = \frac{15}{6} - \frac{14}{6} = \frac{1}{6}$$ Hence $$a_5 = \frac{1}{1/6} = 6$$

8. Conclusion
   The largest positive term of the HP is 6.

Imagine a Line of Friends Sharing Chocolates

1. You have 2 friends. The first gets 2 chocolates out of 5 (that is $\frac{2}{5}$) and the second gets 12 chocolates out of 23 (that is $\frac{12}{23}$).
2. These shares form something called a Harmonic Progression (HP).
3. In an HP, if you flip the fractions (find their reciprocals), those new numbers make a straight-line pattern called an Arithmetic Progression (AP).
4. We flip $\tfrac{2}{5}$ and $\tfrac{12}{23}$ to get $\tfrac{5}{2}$ and $\tfrac{23}{12}$.
5. In an AP, we move by the same step each time. Here that step (difference) is negative, so every next flipped number gets smaller.
6. As long as those flipped numbers stay above zero, our original HP terms stay positive.
7. The moment the flipped number would turn zero or negative, we must stop, because dividing by 0 or a negative makes our HP term explode or turn negative.
8. The last positive term we can safely take is when the flipped number is just bigger than zero. That gives the biggest original fraction.
9. Do the quick maths and you find that the last safe flipped number is $\tfrac{1}{6}$, so the biggest original fraction is $\tfrac{1}{1/6} = 6$.
   So, the greatest positive term is 6.

Step-by-Step Solution

1. Given first two HP terms
   $a_1 = \frac{2}{5},\quad a_2 = \frac{12}{23}$

2. Convert to AP by taking reciprocals
   $A = \frac{1}{a_1} = \frac{5}{2}$
   $A + d = \frac{1}{a_2} = \frac{23}{12}$

3. Find common difference $d$
   $d = \frac{23}{12} - \frac{5}{2} = \frac{23}{12} - \frac{30}{12} = -\frac{7}{12}$

4. General reciprocal term
   $\frac{1}{a_n} = A + (n-1)d = \frac{5}{2} + (n-1)\Bigl(-\frac{7}{12}\Bigr)$

5. Ensure positivity
   $\frac{5}{2} - (n-1)\frac{7}{12} > 0$
   Multiply by 12:
   $30 - 7(n-1) > 0$
   $30 > 7(n-1)$
   $n-1 < \frac{30}{7} \Rightarrow n \le 5$

6. Compute $a_5$

   = \frac{5}{2} - \frac{28}{12} = \frac{15}{6} - \frac{14}{6} = \frac{1}{6}$$ Hence $$a_5 = 6$$

Final Answer

$\boxed{6}$