What is the ratio of the lengths of two mathematical pendulums if the first of them makes 20

What is the ratio of the lengths of two mathematical pendulums if the first of them makes 20 oscillations, and the second during the same time 50 oscillations?

The oscillation period is such a period of time during which the body returns to the same point from which it began its movement.
T = t / n, where n is the number of oscillations, t is the time during which the body made these oscillations.
T = 2π * √ (l / g), where l is the length of the pendulum, g is the free fall acceleration of a body lifted above the Earth by 9.8 m / s².
Let’s square both sides of the formula to get rid of the root:
T² = 4π² * (l / g)
Let us express from this the length of the pendulum l:
l = (T² * g) / (4π²) = ((t / n) ² * g) / (4π²)
Length of the first pendulum:
l1 = ((t / n1) ² * g) / (4π²)
Second pendulum length:
l2 = ((t / n2) ² * g) / (4π²)
Divide l1 / l2:
l1 / l2 = ((t / n1) ² * g) / (4π²) / ((t / n2) ² * g) / (4π²) = t² * g * n2² * 4π² / (n1² * 4π² * t² * g ) = n2² / n1²
l1 / l2 = 50² / 20² = 6.25
Answer: 6.25 times.