Plz Solve this and Explain also

Dear Student,
 
Let A and D be the first term and common difference of A.P.
ap = a  ⇒A + (p– 1) D = a    … (1)
aq = b ⇒A + (q – 1) D = b    … (2)
ar = c   ⇒ A + (r – 1) D = c      … (3)
 
a)
. a (qr) + b  (rp) + c (pq)
= [A + (p– 1) D] (qr) + [A + (q – 1) D] (rp) + [A + (r – 1) D] (pq)]
= A (qr)  + (p – 1) (qr)D + A (rp) + (q – 1) (r – p) D + A (p – q)  + (r – 1) (pq) D
= A × (q – r + r – p + p – q) + D × (pq – pr – q + r + qr – pq – r + p + pr – rq – p + q)
= A × 0 + D × 0
= 0
 
b)
 a (qr) + b (rp) + c (pq) = 0  (Proved)
aqar + brbp + cpcq = 0
r (–a + b) + p (–b + c) + q (–c + a) = 0
⇒ – r (ab) – p (b – c) – q (c – a) = 0
r (ab) – p(b – c) + q (c – a) = 0
 
Regards!

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