If pth,qth and rth terms of an A.P. are a, b, c respectively, then show that (i) a(q-r)+b(r-p)+c(p-q)=0

Answer

N/a

Explanation

Let first term be x and common difference be y of A.P.
`T_(p) = a`
`rArr` x + (p - 1)y = a …(1)
`T_(q) = b`
`rArr` x + (q - 1)y = b …(2)
`T_(r) = c`
`rArr` x + (r - 1)y = c …(3)
Subtracting equation (2) from equation(1)
(p - q)y = a - b
`rArr` `y = (a - b)/(p - q)` ...(5)
From equations (4) and (5),
`(a - b)/(p - q) = (b - c)/(q - r)`
`rArr` a(q - r) - b (q - r) = b (p - q) - c (p - q)
`rArr` a (q - r) - b(q -r) - b (p - q) + c (p - q) = 0
`rArr` a (q - r) + b(-q + r - p + q) + c (p - q) = 0
`rArr` a (q - r) + b (r - p) + c (p - q) = 0
Hence Proved.