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

**Solution:**

Let t and d be the first term and common difference of the A.P respectively.

The nᵗʰ term of an A.P is given by, aₙ = t + (n - 1) d

aₚ = t + ( p - 1) d = a ....(1)

aq = t + (q - 1) d = b ....(2)

aᵣ = t + (r - 1) d = c ....(3)

Subtracting (2) from (1), we obtain

(p - 1 - q + 1) d = a - b

(p - q) d = a - b

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

Subtracting (3) from (2), we obtain

(q - 1 - r + 1) d = b - c

(q - r) d = b - c

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

From (4) and (5), we obtain

(a - b)/(p - q) = (b - c)/(q - r)

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

aq - ar - bq + br = bp - bq - cp + cq

bp - cp + cq - aq + ar - br = 0

(-aq + ar) + (bp - br) + (- cp + cq) = 0

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

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

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

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 15

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

**Summary:**

Using the formula aₙ= t + (n - 1) d and given the pᵗʰ, qᵗʰ and rᵗʰ terms of A.P we showed that (q - r) a + (r - p)b + (p - q)c = 0