Question

If the distance described by any particle during the $p_{th}$, $q_{th}$ and $r_{th}$ second of its motion are $a,b,c$ respectively, prove that-
$a(q-r)+b(r-p)+c(p-q)=0$

Answer 1

Let Abe the first term and p be the common difference of the given A. P.

Then,

a = pth term

a = A + (p - 1) D

b = qth term

b = A + (q - 1) D

c = rth term

c = A + (r - 1)D

we have,

= a(q - r) + b (r - p) + c (p - q)

= [A + (p- 1) D] (q - r) + [A + (q - 1) D] (r - p) + [ A + (r - 1) D] (p - q)

= A [(q - r) + (r - p) + (p - q)] + D [(p - 1) (q - r) + (q- 1) (r - p) + (r - 1) (p - q)}

= A $\times$ 0 + D [p(q - r)] + q(r - p) + r (p - q) - (q - r) - (r - p) - (p - q)]

= 0 + D $\times$ 0

= 0