Question

If the pth, qth, rth terms of an AP be x,y,z respectively , show that x (q-r)+y (r-p)+z (p-q)=0​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Let the first term be a and common difference of given AP be d.

Now, given that pth, qth, rth term of AP be x,y,z respectively, this means that

x = a + (p - 1)d .... (i)

y = a + (q - 1)d .... (ii)

z = a + (r - 1)d .... (iii)

From equation (i)

x(q - r) = a(q - r) + (q - r)(p - 1)d .... (iv)

From equation (ii)

y(r - p) = a(r - p) + (r - p)(q - 1)d .... (v)

From equation (iii)

z(p - q) = a(p - q) + (p - q))(r - 1)d .... (vi)

Adding equation (iv), (v) and (vi) we get

x(q - r) + y(r - p) + z(p - q) = a(q- r + r - p + p - q) + d((q - r)(p - 1) + (r - p)(q - 1) + (p - q))(r - 1))

x(q - r) + y(r - p) + z(p - q) = a(0) + d(qp - q - rp + r + rq - r - pq + q + pr - p - qr + q)

x(q - r) + y(r - p) + z(p - q) = a(0) + d(0)

x(q - r) + y(r - p) + z(p - q) = 0, this completes the proof.