NCERT Solutions
Class 11 Maths
Sequences and Series
Ex.9.2 Q.11
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that
{a (q - r) ÷ p} + {b (r - p) ÷ q} + {c (p - q) ÷ r} = 0
Let a1 and d be the first term and the common difference of the A.P. respectively.
According to the given information,
Sp = (
)[2a + (p - 1) d] = a
=> 2a1 + (p - 1) d] = 
………… (1)
Sq = (
)[2a + (q - 1) d] = a
=> 2a1 + (q - 1) d] = 
………… (2)
Sr = (
)[2a + (r - 1) d] = a
=> 2a1 + (r - 1) d] = 
………… (3)
Subtract equation (2) from equation (1), we get
(p - 1) d – (q - 1) d = – 
=> (p - q) d = (2aq – 2bp) ÷ pq
=> d = (2aq – 2bp) ÷ {pq (p - q)} ………. (4)
Subtract equation (3) from equation (2), we get
(p - 1) d – (r - 1) d = – 
=> (q - r) d = (2br – 2qc) ÷ qr
=> d = (2br – 2qc) ÷ {qr (q - r)} ………. (5)
From equation (4) and (5), we get
(2aq – 2bp) ÷ {pq (p - q)} = (2br – 2qc) ÷ {qr (q - r)}
=> (aq – bp) ÷ {pq (p - q)} = (br – qc) ÷ {qr (q - r)}
=> qr (q - r) (aq – bp) = pq (p - q) (br – qc)
=> r (q - r) (aq – bp) = p (p - q) (br – qc)
=> (q - r) (aqr – bpr) = (p - q) (bpr – pqc)
Divide both side by pqr, we get
(– 
)(q - r) = (p - q) ( – 
)
=> (
)(q - r) – ( )(q - r) = (p - q) ( ) – (p - q) ( )
=> ( )(q - r) – ( )(q – r + p - q) + (p - q) ( ) = 0
=> ( )(q - r) + ( )(r - p) + (p - q) ( ) = 0