Question

If the pth,qth and rth term of an AP are a,b,c respectively then show that: a(q-r) + b(r-p) + c(p-q)=0

Answer 1

let first term =A,
common difference=d,

Tp=a,

[A+(p-1)d]=a,

Tq=b,

[A+(q-1)d]=b,

Tr=c,

[A+(r-1)d]=c,


taking L.H.S , and put the value of a, b and c in the L.H.S so we have


[A+(p-1)d]×(q-r) + [A+(q-1)d]×(r-p) + [A+(r-1)d]×(p-q),


then

A[q-r + r-p + p-q] + d[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)],

now solve the bracket then here all the terms comes out to be cancel out and you get

A×0 + d×0,

0,

therefore

L.H.S=R.H.S