Question

If pth, qth,rth term of an A.P is a,b,c respectively, prove that a(q-r)+b(r-p)+c(p-q)=0.

Answer 1

✬ Given :-

• a = pth term
• b = qth term
• c = rth term

★ To prove :-

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

★ Solution :-

We know that;

aₙ = a + (n - 1)d

Let 'A' be the first term and 'd' be the Common difference

According to the question,

For pth term: a = A + (p - 1)d

For qth term: b = A + (q - 1)d

For rth term: c = A + (r - 1)d

Now,

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

= {A + (p - 1)d}[q - r] + {A + (q - 1)d}[r - p] + {A + (r - 1)d}[p - q]

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

= A[q - r + r - p + p - q] + d[p(q - r) + q(r - p) + r(p - q)]

= A[0] + d[pq - pr + qr - pq + pr - qr]

= A(0) + d(0)

= 0

HENCE PROVED