# The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that(q – r )a + (r – p )b + (p – q )c = 0

Let A = first term of the AP.
and
Let d = common difference of the AP

Let n be the number of terms of the A.P.

It is known that the nth term of an A.P. is given by -

an = a + (n –1) d

∴ pth term of A.P. is given by -
a = A + (p - 1).d.......(1)

qth term of A.P. is given by -
b = A + (q - 1).d.......(2)

rth term of A.P. is given by -
c = A + (r - 1).d........(3)
Subtracting (2) from (1) , (3) from (2) and (1) from (3), we get
a - b = (p - q).d......(4)
b - c = (q - r).d........(6)
c - a = (r - p).d.......(6)

Multiply (4),(6) and (6) by c, a and b respectively, we have

c.(a - b) = c.(p - q).d......(4)
a.(b - c) = a.(q - r).d........(6)
b.(c - a) = b.(r - p).d.......(6)

Adding (4),(6) and (6), we get -

c.(a - b) + a.(b - c) + b.(c - a) = c.(p - q).d + a.(q - r).d + b.(r - p).d
ac - bc + ab - ac + bc - ab = a.(q - r).d + b.(r - p).d + c.(p - q).d

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

Now since d is common difference, it should be non zero
Hence
a(q - r) + b(r - p) + c(p - q) = 0

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