if (b+c-a) = p (c+a-b) = q (a+b+c) =r Find 1/p + 1/q + 1/r in term of z
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Answer:
Given a,b and c are respectively the p
th
,q
th
and r
th
terms of a harmonic progression.
⇒a=
A+(p−1)D
1
,b=
A+(q−1)D
1
and c=
A+(r−1)D
1
where A,D are first term and common difference of corresponding Arithmetic progression.
∣
∣
∣
∣
∣
∣
∣
∣
bc
p
1
ca
q
1
ab
r
1
∣
∣
∣
∣
∣
∣
∣
∣
=abc
∣
∣
∣
∣
∣
∣
∣
∣
1/a
p
1
1/b
q
1
1/c
r
1
∣
∣
∣
∣
∣
∣
∣
∣
=abc
∣
∣
∣
∣
∣
∣
∣
∣
A+(p−1)D
p
1
A+(q−1)D
q
1
A+(r−1)D
r
1
∣
∣
∣
∣
∣
∣
∣
∣
applying R
1
→R
1
−AR
3
gives
=abcD
∣
∣
∣
∣
∣
∣
∣
∣
p−1
p
1
q−1
q
1
r−1
r
1
∣
∣
∣
∣
∣
∣
∣
∣
applying R
1
+R
3
gives
=abcD
∣
∣
∣
∣
∣
∣
∣
∣
p
p
1
q
q
1
r
r
1
∣
∣
∣
∣
∣
∣
∣
∣
=0
∴
∣
∣
∣
∣
∣
∣
∣
∣
bc
p
1
ca
q
1
ab
r
1
∣
∣
∣
∣
∣
∣
∣
∣
=0
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