P^(2/a)=q^(2/b)=(p^b q^a) ^c then prove abc=1.

1.	P^(2/a)=q^(2/b)=(p^b q^a) ^c then prove abc=1.
Answer» <P>Answer:Given sum of first p, q, r term of an A.P are a, B, C respectively. Sum of first p terms, when A is the first term and D is the common difference Given s p =a ⇒ 2 p [2A+(p−1)D]=a Similarly s q =b ⇒ 2 q [2A+(q−1)D]=b and s r =c ⇒ 2 r [2A+(r−1)D]=c Now p a = 2 1 [2A+(p−1)D]=A+ 2 (p−1) D Multply by q−r, we GET p a (q−r)=(A+ 2 (p−1) D)(q−r) ……..(1) Similarly q b (r−p)=(A+ 2 (q−1) D)(r−p) ……..(2) and r c (p−q)=(A+ 2 (r−1) D)(p−q) ……..(3) ADDING (1),(2) and (3) p a (q−r)+ q b (r−p)+ r c (p−q)=A(q−r+r−p+p−q)+ 2 D [pq−pr−q+r+rq−pq−r+p+rp−rq−p−q]

Answer»

<P>Answer:Given sum of first p, q, r term of an A.P

are a, B, C respectively.

Sum of first p terms, when A is the first term and D is the common difference

Given s

⇒

[2A+(p−1)D]=a

Similarly s

⇒

[2A+(q−1)D]=b

and s

⇒

[2A+(r−1)D]=c

Now

[2A+(p−1)D]=A+

(p−1)

Multply by q−r, we GET

(q−r)=(A+

(p−1)

D)(q−r) ……..(1)

Similarly

(r−p)=(A+

(q−1)

D)(r−p) ……..(2)

and

(p−q)=(A+

(r−1)

D)(p−q) ……..(3)

ADDING (1),(2) and (3)

(q−r)+

(r−p)+

(p−q)=A(q−r+r−p+p−q)+

[pq−pr−q+r+rq−pq−r+p+rp−rq−p−q]

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