Let Gopal's initial salary be $S$.
After the first increment, his salary becomes:
\[ S_1 = S \times \left(1 + \frac{x}{100}\right) \]
where $x$ is the percentage increase in the first increment.
After the second increment, his salary becomes:
\[ S_2 = S_1 \times \left(1 + \frac{2x}{100}\right) \]
We are given that his final salary is 187.5. Thus, $S_2 = S \times 1.875$.
Substituting the expression for $S_2$ into the given equation:
\[ S \times \left(1 + \frac{x}{100}\right) \times \left(1 + \frac{2x}{100}\right) = S \times 1.875 \]
Divide both sides by $S$ and simplify the equation:
\[ \left(1 + \frac{x}{100}\right) \times \left(1 + \frac{2x}{100}\right) = 1.875 \]
Expand the left side:
\[ 1 + \frac{x}{100} + \frac{2x}{100} + \frac{2x^2}{10000} = 1.875 \]
Combine terms:
\[ 1 + \frac{3x}{100} + \frac{2x^2}{10000} = 1.875 \]
Subtract 1 from both sides:
\[ \frac{3x}{100} + \frac{2x^2}{10000} = 0.875 \]
Multiply the entire equation by 10000 to clear the denominators:
\[ 300x + 2x^2 = 87500 \]
Rearrange into standard quadratic form:
\[ 2x^2 + 300x - 87500 = 0 \]
Solve this quadratic equation using the quadratic formula:
\[ x = \frac{-300 \pm \sqrt{300^2 - 4 \times 2 \times (-87500)}}{4} \]
\[ x = \frac{-300 \pm \sqrt{790000}}{4} \implies x = \frac{-300 \pm 890}{4} \]
Calculate the positive solution for $x$: $x = \frac{590}{4} = 25$.
Therefore, the percentage increase in the first increment is 25.