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Byju's Answer
Standard XII
Physics
Idea of Symmetry
If a,b,c ar...
Question
If
a
,
b
,
c
are
p
t
h
,
q
t
h
,
r
t
h
terms respectively of a G.P. then
∣
∣ ∣
∣
loga
p
1
logb
q
1
logc
r
1
∣
∣ ∣
∣
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Solution
a
=
p
t
h
term, let common ratio be k and first term be L
a
=
L
k
p
−
1
⇒
l
o
g
a
=
l
o
g
L
+
(
p
−
1
)
l
o
g
k
b
=
L
k
q
−
1
⇒
l
o
g
b
=
l
o
g
L
+
(
q
−
1
)
l
o
g
k
c
=
L
(
k
)
(
r
−
1
)
⇒
l
o
g
L
=
l
o
g
L
+
(
r
−
1
)
l
o
g
k
∣
∣ ∣
∣
l
o
g
a
p
1
l
o
g
b
q
q
l
o
g
c
r
r
∣
∣ ∣
∣
−
l
o
g
a
[
q
−
r
]
−
p
[
l
o
g
b
−
l
o
g
c
]
+
1
[
r
/
l
o
g
b
−
q
l
o
g
c
]
=
[
l
o
g
L
+
(
p
−
1
)
l
o
g
k
]
(
q
−
r
)
q
−
p
[
(
q
−
r
)
l
o
g
k
]
+
r
l
o
g
L
+
r
(
q
−
1
)
l
o
g
k
−
q
l
o
g
L
−
q
(
r
−
1
)
l
o
g
k
=
l
o
g
L
[
q
−
r
−
q
+
r
]
+
l
o
g
k
[
(
q
−
r
)
(
p
−
1
)
−
p
(
q
−
r
)
+
r
(
q
−
1
)
−
q
(
r
−
1
)
]
l
o
g
c
(
0
)
+
l
o
g
k
[
r
−
q
+
r
q
−
r
−
q
r
+
r
]
∣
∣ ∣
∣
l
o
g
a
p
1
l
o
g
b
q
1
l
o
g
c
r
1
∣
∣ ∣
∣
=
0
.
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Similar questions
Q.
If
a
,
b
,
c
are
p
t
h
,
q
t
h
and
r
t
h
terms of a GP, then
⎡
⎢
⎣
log
a
p
1
log
b
q
1
log
c
r
1
⎤
⎥
⎦
is equal to -
Q.
If a, b, c (all +ve) are the pth, qth and rth terms respectively of a geometric progression, then prove that
∣
∣ ∣
∣
l
o
g
a
p
1
l
o
g
b
q
1
l
o
g
c
r
1
∣
∣ ∣
∣
= 0
Q.
If
a
,
b
,
c
are positive and are the pth, qth and rth terms respectively of a G.P., then
Δ
=
∣
∣ ∣
∣
log
a
p
1
log
b
q
1
log
c
r
1
∣
∣ ∣
∣
is
Q.
If
a
,
b
,
c
are positive and are the
p
t
h
,
q
t
h
and
r
t
h
terms, respectively, of a G.P., then
Δ
=
∣
∣ ∣
∣
log
a
p
1
log
b
q
1
log
c
r
1
∣
∣ ∣
∣
is
Q.
If
a
>
0
,
b
>
0
,
c
>
0
are respectively the
p
t
h
,
q
t
h
,
r
t
h
terms of a G.P., then the value of the deteminant
∣
∣ ∣
∣
log
a
log
b
log
c
p
q
r
1
1
1
∣
∣ ∣
∣
is
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