A particle starts with a velocity u with a uniform acceleration f. In the p-th, q-th and r-th seconds, it moves through distances a,b and c respectively. Prove that `a(q-r)+b(r-p)+c(p-q) = 0`

Explanation

Distance travelled in the n-th second,
`s_(n) = u +(1)/(2)f(2n-1)`
`:. " " a = u+(1)/(2)f(2p-1)`
`= u +fp- (f)/(2) = (u-(f)/(2))+fp`
Similarly, b= `(u-(f)/(2))+fp`
and `" " c=(u-(f)/(2))+fr`
Multiplying (1), (2) and (3) by (q-r), (r-p) and (p-q) respectively and then adding we get
a(q-r)+b(r-p)+c(p-q)
= `(u-(f)/(2))`[ (q-r)+(r-p)+(p-q)]
+f[p(q-r)+q(r-p)+r(p-q)]
=`(u-(f)/(2)).0+f.0`= 0