A

Step 1: Analyze the physics of a particle thrown vertically upwards. The particle will have an initial speed (u), will decelerate due to gravity (g), and eventually reach a maximum height where its velocity becomes zero.
Step 2: At maximum height, using the kinematic equation:
$$ v^2 = u^2 - 2gh $$ where v = 0 at maximum height, we find that:
$$ 0 = u^2 - 2gh \Rightarrow u^2 = 2gh $$
Hence, maximum height (h) is given by:
$$ h = \frac{u^2}{2g} $$
Step 3: Now, let's analyze the provided statements:

• Option A: Speed at time \( t \) is dependent on the time elapsed, and it decreases linearly until max height; therefore, it cannot be a constant value during upward motion.
• Option B: Speed decreasing to zero at max height is valid.
• Option C: At equal intervals of time during ascent, the height increases but at decreasing rates, hence heights at the same time on the way up and down are the same.
• Option D: Velocities at two symmetric points (same height) are equal in magnitude but opposite in direction.

Therefore, Option A is incorrect; speed does not remain constant.
Therefore, A.