Hello Chandrakanth,

We know that distance travelled in t seconds is given by the relation:

S = u * t + ½ * a * t^2

Where u = initial velocity in m/s

A = acceleration in m/s^2

T = time in s

Here, u = 0, a = a, t = n (given)

Therefore, distance travelled after n seconds is:

Sn = 0 * n + ½ * a * n^2 = ½ * a * n^2

also, distance travelled in n-2 seconds will be:

S(n-2) = 0 * (n – 2) + ½ * a * (n-2) = ½ * a * (n–2)^2

Therefore, distance travelled in last 2 seconds will be:

Sn – S(n-2) = ½ * a * n^2 - ½ * a * (n–2)^2

Solving the equation, we get,

Sn – S(n-2) = 2a * (n-1) (Answer) (Equation 1)

Also, we know that v = u + at

Substituting the values, we get, v = an

Substituting the value of a in equation 1, we get

Sn – S(n-2) = 2v/n * (n-1) (Answer)

Hope this is clear!