If pth , qth and rth term of an AP are a,b,c respectively , then show that

(a-b)r +(b-c)p + (c-a)q = 0

Please help me with this question asap .

Let A be the first term of the A.P. and D be the common difference of the A.P.

Given that,

a = pth term

Therefore,

b = qth term

Therefore,

c = rth term

Therefore,

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

= ar - br + bp - cp + cq - aq

= -(aq - ar) - (br - bp) - (cp - cq)

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

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