Question
A particle is moving in a straight line with constant acceleration. It travels a distance of 
in first n seconds, 
in next n seconds and y in next n seconds.
Prove that,
α + γ = 2β.
Solution
Consider,
'u' be the initial velocity and 'a' be the acceleration of particle.
Particle travels
distance in first n seconds, β in next n seconds and 
in next n seconds.
Thus, particle travels α distance in first n seconds, α + β in first 2n seconds and α+β+γ in first 3n seconds.
Therefore, equation of kinematics can be written as
Subtracting (1) from (2),
For equation (3),
That is,
'u' be the initial velocity and 'a' be the acceleration of particle.
Particle travels
distance in first n seconds, β in next n seconds and in next n seconds. Thus, particle travels α distance in first n seconds, α + β in first 2n seconds and α+β+γ in first 3n seconds.
Therefore, equation of kinematics can be written as
Subtracting (1) from (2),
For equation (3),
That is,
