# Sum of the first p, q and r terms of an A.P., are a, b and c respectively. Prove that: a/p (q - r) + b/q (r - p) + c/r (p - q) = 0

**Solution:**

Let a_{1} and d be the first term and the common difference of the given arithmetic progression respectively.

We know that S_{n} = n/2 [2a + (n - 1) d]

According to the given information,

S_{p} = p/2 [2a_{1} + (p - 1) d]

a = p/2 [2a_{1} + (p - 1) d]

2a_{1} + (p - 1) d = 2a/p ....(1)

S_{q} = q/2 [2a_{1} + (q - 1) d]

b = q/2 [2a_{1} + (q - 1) d]

2a_{1} + (q - 1) d = 2b/q ....(2)

S_{r} = r/2 [2a_{1} + (r - 1) d]

c = r/2 [2a_{1} + (r - 1) d]

2a_{1} + (r - 1) d = 2c/r ....(3)

Subtracting (2) from (1)

⇒ ( p - 1) d - (q - 1) d = 2a/p - 2b/q

d ( p - 1- q + 1) = (2aq - 2bp)/pq

d ( p - q) = (2aq - 2bp)/pq

⇒ d = 2(aq - bp)/pq (p - q) ....(4)

Subtracting (3) from (2)

(q - 1) d - (r - 1) d = 2b/q - 2c/r

⇒ d (q - 1- r + 1) = 2b/q - 2c/r

⇒ d (q - r) = (2br - 2cq)/qr

⇒ d = 2 (br - qc)/qr (q - r) ....(5)

Equating both the values of d obtained in (4) and (5) , we obtain

⇒ aq - bp/pq (p - q) = (br - qc)/qr (q - r)

⇒ qr (q - r)(aq - bp) = pq (p - q)(br - qc)

⇒ r (aq - bp)(q - r) = p (br - qc)(p - q)

⇒ (aqr - bpr)(q - r) = (bpr - pqc)(p - q)

Dividing both sides by pqr, we obtain

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

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

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

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

Hence, proved

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 11

## Sum of the first p, q and r terms of an A.P., are a, b and c respectively. Prove that: a/p (q - r) + b/q (r - p) + c/r (p - q) = 0

**Summary:**

We know that the sum of the first p, q, and r terms of the series are a, b, c. We have proved that a/p (q - r) + b/q (r - p) + c/r ( p - q) = 0

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