Question

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

Answer 1

Let a1 be the first term of AP and d be the common difference We know that nth term of AP is given by

an = a1 + (n-1) d

Given pth , qth and rth terms are a , b and c respectively

ap = a1 + (p – 1) d = a1 + (pd – d) = a

→ (1) aq = a1 + (q – 1) d = a1 + (qd – d) = b

→ (2) ar = a1 + (r – 1) d = a1 + (rd – d) = c

→ (3) Subtract (3) from (2), we get (b - c) = ( q -r) d Hence (q – r) = (b – c) /d

→ (4) Subtract (1) from (3), we get (c – a) = (r – p) d Hence (r – p) = (c – a)/d

→ (5) Subtract (2) from (1), we get (a - b) = (p – q) d Hence (p – q) = (a – b)/d

→ (6) Consider the LHS : a(q – r) + b(r – p) + c(p – q) a (q – r) + b (r – p) + c(p – q) Substitute the above values from (4), (5) and )

6) we get a (q – r) + b (r – p) + c(p – q) = 0 = RHS

Answer 2

Answer:

Step-by-step explanation:

Let a = first term of the AP.

and 

Let d = common difference of the AP

Now

a = A+(p-1).d.......(1)

b = A+(q-1).d.......(2)

c = A+(r-1).d........(3)

Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we get

a-b = (p-q).d......(4)

b-c = (q-r).d........(5)

c-a = (r-p).d.......(6)

multiply 4,5,6 by c,a,b respectively we have

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

a.(b-c) = a.(q-r).d........(5)

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

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

Thanks!