p^(2/a)=q^(2/b)=(p^b q^a) ^c then prove abc=1.
Answers
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:
Prove that {a \over P}\left( {q - r} \right) + {b \over q}\left( {r ... Sum of first p terms, when A is the first term and D is the ... 2). and rc(p−q)=(A+2(r−1)D)(p−q) ……..(3)