A man travels three-fifths of a distance \( AB \) at a speed \( 3a \), and the remaining two-fifths at a speed \( 2b \). Let total distance \( AB = d \).
Time from A to B:
First part: \( \frac{3d}{5} \) at speed \( 3a \) ⇒ Time = \( \frac{3d}{5 \cdot 3a} = \frac{d}{5a} \)
Second part: \( \frac{2d}{5} \) at speed \( 2b \) ⇒ Time = \( \frac{2d}{5 \cdot 2b} = \frac{d}{5b} \)
Total time from A to B = \( \frac{d}{5a} + \frac{d}{5b} \)
Time from B to A and back to B:
Total distance = \( 2d \) at speed \( 5c \) ⇒ Time = \( \frac{2d}{5c} \)
Given: Both times are equal:
\[ \frac{d}{5a} + \frac{d}{5b} = \frac{2d}{5c} \] Cancel \( d \) and multiply all terms by 5: \[ \frac{1}{a} + \frac{1}{b} = \frac{2}{c} \div 2 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \]
Answer: \( \boxed{\frac{1}{a} + \frac{1}{b} = \frac{1}{c}} \)