If the pth,qth and rth terms …

Given that the pth, qth and rth terms of an AP be a, b, c respectively.
Suppose, x be the first term and d be the common difference of the given arithmetic progression. Therefore,
T_p =  x+(p-1)d, T_q =x+(q-1)d and T_r =x+(r-1)d

Now, T_p = a {tex}\Rightarrow{/tex} x+(p-1)d = a ....(i)
T_q = b {tex}\Rightarrow{/tex} x+(q-1)d = b....(ii)
T_r = c {tex}\Rightarrow{/tex} x+(r-1)d = c....(iii)
On multiplying (i) by (q-r), (ii) by (r-p) and (iii) by (p-q), we get,
(q-r)[x+(p-1)d]=a(q-r)....(iv)
(r-p)[x+(q-1)d]=b(r-p)....(v)
(p-q)[x+(r-1)d]=c(p-q)....(vi)
Now adding we get,
a(q - r) + b(r - p) + c(p - q) = x{(q - r) + (r - p) + (p - q)} + d{(p - 1)(q - r)+ (q - 1)(r - p)+ (r - 1)(p - q)}
= (x {tex}\times{/tex} 0) + (d {tex}\times{/tex} 0) = 0
Therefore, a(q - r) + b(r - p) + c(p - q) = 0.
Hence proved.