Asked by Kaur
If 6th, 12th and 18th terms of a GP are P, Q and R respectively, then prove that P, Q and R are in G.P.
Answers
Answered by
Juice WRLD 999 forever RIP ILY Never forget u
Let's say a,b then
GM=
ab
=24 or ab=576
a=
b
576
HM=
a+b
2ab
=
5
72
a+b
2×576
=
5
72
We get
a+b=80
On substituting the value of a,
b
576
+b−80=0
b
2
−80b+576=0
b=
2×1
+80±
80
2
−4×1×576
b=40±32
b=8, 72
a=72,8
GM=
ab
=24 or ab=576
a=
b
576
HM=
a+b
2ab
=
5
72
a+b
2×576
=
5
72
We get
a+b=80
On substituting the value of a,
b
576
+b−80=0
b
2
−80b+576=0
b=
2×1
+80±
80
2
−4×1×576
b=40±32
b=8, 72
a=72,8
Answered by
oobleck
If the common ratio of the GR is r, then
Q/P = r^6
R/Q = r^6
since there is a common ratio (R/Q = Q/P), P,Q,R form a GP
Q/P = r^6
R/Q = r^6
since there is a common ratio (R/Q = Q/P), P,Q,R form a GP
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