# prove A.P from G.P.

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#1
1. Prove that 3x, 3y and 3z are successive terms of a G.P., if x, y and z are successive terms of an A.P.

2. If the pth, qth and rth terms of a G.P are in geometric progression, show that p, q and r are successive terms of an A.P.
0
6 years ago
#2
(Original post by cucumberpj)
...
Show attempt?

Second question seems like tautology to me unless I am misunderstanding question.
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6 years ago
#3
(Original post by lazy_fish)
Second question seems like tautology to me unless I am misunderstanding question.
Not a tautology.

e.g. take the GP 1, 2, 4, 8, 16, 32, 64, 128, 256, ...

pick 3 terms that are in geometric progression. E.g. 2, 16, 128 (common ratio is 8).
Note that these are the 2nd, 5th and 8th terms in the series, and 2, 5 and 8 are in AP.

Now prove this is the case for any GP and any 3 terms in geometric progression.

It's not actually hard.
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#4
x, y, z
=a, a+d, a+2d

3^a, 3^a+d, 3^a+2d
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6 years ago
#5
(Original post by cucumberpj)
1. Prove that 3x, 3y and 3z are successive terms of a G.P., if x, y and z are successive terms of an A.P.

2. If the pth, qth and rth terms of a G.P are in geometric progression, show that p, q and r are successive terms of an A.P.
Interesting.

1.

z = y + (y - x) = 2y - x

We have x,y,z
Which is x,y,2y - x

So basically the G.P is
3x, 3y and 32y - x

I need to show its a G.P and successful terms. Therefore a common ratio should exist.

Second term ÷ first term is = 3y ÷ 3x= 3y-x

Third term ÷ second term = 32y - x ÷ 3y = 3y-x

So a common ratio exists.
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#6
(Original post by TheKingOfTSR)
Interesting.

1.

z = y + (y - x) = 2y - x

We have x,y,z
Which is x,y,2y - x

So basically the G.P is
3x, 3y and 32y - x

I need to show its a G.P and successful terms. Therefore a common ratio should exist.

Second term ÷ first term is = 3y ÷ 3x= 3y-x

Third term ÷ second term = 32y - x ÷ 3y = 3y-x

So a common ratio exists.
Thanks. I have solved it.
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