AS Mechanics Question

A particle of mass, M, which is subject to a constant resisting force, is observed to cover distances of s1, and sn respectively in the first and nth seconds of its motion. Find an expression in terms of s1 and sn for the magnitude of the resistance force. How do I make this equation?

So far I've got:
Part one:
s = s1, u = ?, v = ?, a = a, t= 1
(s + 1/2at^2) / t = v
v = s1 + a/2
Part two:
s = sn- s1, u = s1 + a/2, v = ?, a = a, t = n-1
(s - 1/2at^2)/t = u
((sn-s1) - a/2(n-1)^2) / (n-1) = u
As the final velocity of part one is the initial velocity of part two:
((sn-s1) - a/2(n-1)^2) / (n-1) = s1 + a/2
However, I can't seem to get the right answer from there?

"As the final velocity of part one is the initial velocity of part two:"

This is untrue. The final velocity for the first part will be the velocity at time t=1. This is not the same as the velocity at the start of the second interval. So, you can't use "v" to tie the two together.

Do you know what the answer is supposed to be?

The distance travelled in the first second is simply s1= u + a/2

What about the n'th second, the interval between t=n-1 and t=n (big hint)

Sn = u + 1/2a(2n-1)
u = S1 - a/2
Sn = (sn - a/2) + 1/2a(2n-1)
Sn - S1 = -1/2a(1-(2n-1))
(-2Sn+2S1) / (-2n+2) = a
a = (Sn-S1)/(n-1)?

That's what I got.

Thanks a lot man

Sn = u + 1/2a(2n-1)
u = S1 - a/2
Sn = (sn - a/2) + 1/2a(2n-1)
Sn - S1 = -1/2a(1-(2n-1))
(-2Sn+2S1) / (-2n+2) = a
a = (Sn-S1)/(n-1)?

Hi, I’m struggling with the same question, but I don’t understand where the 2n-1 in the first line came from. Can someone pls explain this to me?

Thanks