Question

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

Answer 1

hey friend

here is answer. .

let x be the first term and d be the common difference of the given AP.then

Tp=x+(p-1)d,
Tq=x+(q-1)d
and ,tr=x+(r-1)d------



•°•x+(p-1)d=a------------1)


x+(q-1)d=b---------------2)

x+(r-1)d=c------------------3)

on multiplying 1 by (q-r) ,2 by (r-p) and 3 )by (p-q) and 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×0)+(d×0)=0

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

hope it helps you

@Rajukumar111

Answer 2

Answer:

Step-by-step explanation:

See the pic attested..