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The second, the first and the third term of an AP whose comm

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The second, the first and the third term of an AP whose comm [#permalink] New post 08 Dec 2011, 07:15
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The second, the first and the third term of an AP whose common difference is non zero but lesser than 200, form a GP in that order. What is the common ration of that GP?

A. 1
B. -1
C. 2
D. -2
E. |1|
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2013, 12:01, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Progessions Ap/Gp/both?? [#permalink] New post 08 Dec 2011, 08:44
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its D

think A.P to be a-d,a,a+d.
G.P will be a,a-d,a+d

which means (a-d)^2 = a^2 + ad
that means d=3
and G.P is -1,2,-4 and A.P is 2,-1,-4. so G.P common ratio is -2 .....
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Re: Progessions Ap/Gp/both?? [#permalink] New post 08 Dec 2011, 09:00
avenkatesh007 wrote:
its D


which means (a-d)^2 = a^2 + ad
that means d=3


Can you please elaborate on how you solved the quad with two unknowns variables.
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Re: Progessions Ap/Gp/both?? [#permalink] New post 08 Dec 2011, 09:13
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(a-d)^2 = a^2 +ad
a^2-2ad+d^2=a^2+ad
d^2=3ad
d=3a
here I assumed 'a' to be 1 coz in G.P ratio will be a+d:a kind of.so it wont matter what 'a' is.......else u can substitute direct values interms of 'a'.
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Re: Progessions Ap/Gp/both?? [#permalink] New post 03 Jan 2012, 03:43
I am getting the ratio as -1/2. Can someone please explain how it is -2?
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Re: Progessions Ap/Gp/both?? [#permalink] New post 03 Jan 2012, 14:02
avenkatesh007 wrote:
its D

think A.P to be a-d,a,a+d.
G.P will be a,a-d,a+d

which means (a-d)^2 = a^2 + ad
that means d=3
and G.P is -1,2,-4 and A.P is 2,-1,-4. so G.P common ratio is -2 .....



Hi,

Can you please explain how did u get:

(a-d)^2 = a^2+ad

Thanks,
Anu
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Re: Progessions Ap/Gp/both?? [#permalink] New post 04 Jan 2012, 20:49
anuu wrote:
avenkatesh007 wrote:
its D

think A.P to be a-d,a,a+d.
G.P will be a,a-d,a+d

which means (a-d)^2 = a^2 + ad
that means d=3
and G.P is -1,2,-4 and A.P is 2,-1,-4. so G.P common ratio is -2 .....



Hi,

Can you please explain how did u get:

(a-d)^2 = a^2+ad

Thanks,
Anu


for GP we use b^2=ac

so using that (a-d)^2 = a(a+d)

by solving this we get d=3a

but the common difference is comming to be -1/2
can anyone please comment on this.
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Re: Progessions Ap/Gp/both?? [#permalink] New post 05 Jan 2012, 20:23
subhajeet, I have cited the same problem above. Even I am getting the ratio as -1/2. Wonder if we are missing something vital here.
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Re: Progessions Ap/Gp/both?? [#permalink] New post 06 Jan 2012, 03:15
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Anasthaesium wrote:
The second, the first and the third term of an AP whose common difference is non zero but lesser than 200, form a GP in that order. What is the common ration of that GP?

a)1
b)-1
c)2
d)-2
e)|1|


Detailed algebraic explanation:

Let the 3 terms of the AP be (a-d), a and (a+d)
Terms of the GP: a, (a-d), (a+d) in that order.
In a GP, terms next to each other have the same ratio.
So, \(\frac{(a-d)}{a} = \frac{(a+d)}{(a-d)}\)

\((a-d)^2 = a(a+d)\)

\(d^2 - 2ad = ad\)

\(d^2 - 3ad = 0\)

\(d(d - 3a) = 0\)

We know that d is not 0 from the question. So d = 3a

Common ratio \(= \frac{(a-d)}{a} = \frac{(a - 3a)}{a} = -2\)
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Re: Progessions Ap/Gp/both?? [#permalink] New post 06 Jan 2012, 03:17
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siddharthmuzumdar wrote:
subhajeet, I have cited the same problem above. Even I am getting the ratio as -1/2. Wonder if we are missing something vital here.


You probably got d = 3a but after that, substituted d in a/(a-d) as one would naturally since (a-d) is smaller than a. But, the terms in the GP are a, (a-d), (a+d) in that order. So the common ratio is (a-d)/a or (a+d)/(a-d)
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Re: Progessions Ap/Gp/both?? [#permalink] New post 06 Jan 2012, 03:56
VeritasPrepKarishma wrote:
siddharthmuzumdar wrote:
subhajeet, I have cited the same problem above. Even I am getting the ratio as -1/2. Wonder if we are missing something vital here.


You probably got d = 3a but after that, substituted d in a/(a-d) as one would naturally since (a-d) is smaller than a. But, the terms in the GP are a, (a-d), (a+d) in that order. So the common ratio is (a-d)/a or (a+d)/(a-d)


Karishma: U got me right. I was indeed making the same mistake as you have mentioned here. Thanks for the reply.
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Re: Progessions Ap/Gp/both?? [#permalink] New post 07 Jan 2012, 01:32
VeritasPrepKarishma wrote:
siddharthmuzumdar wrote:
subhajeet, I have cited the same problem above. Even I am getting the ratio as -1/2. Wonder if we are missing something vital here.


You probably got d = 3a but after that, substituted d in a/(a-d) as one would naturally since (a-d) is smaller than a. But, the terms in the GP are a, (a-d), (a+d) in that order. So the common ratio is (a-d)/a or (a+d)/(a-d)

Grrr....I am just cursing myself for such silly mistakes. :(
Thanks a ton for pointing it out. :)
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The second, the first and the third term of an AP whose comm [#permalink] New post 18 Nov 2014, 09:43
Hi Karishma,

I am bit confuse with this AP and GP, is it Arithmetic progression and geometric progression. And how do we decide this sequence of a,a-d, a+d.
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Re: The second, the first and the third term of an AP whose comm [#permalink] New post 18 Nov 2014, 22:37
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taleesh wrote:
Hi Karishma,

I am bit confuse with this AP and GP, is it Arithmetic progression and geometric progression. And how do we decide this sequence of a,a-d, a+d.


Yes, AP is Arithmetic Progression, GP is Geometric Progression.

The second first and third terms of an AP form a GP when put in that order.

How do we express the terms of AP? Three terms can be expressed as
a-d, a, a+d (with d as the common difference)

When you put them in this order: second, first and third
a, a-d, a+d - this is a GP

A GP has common ratio so (a-d)/a = (a+d)/(a-d) = Common Ratio

More on AP and GP:

http://www.veritasprep.com/blog/2012/03 ... gressions/
http://www.veritasprep.com/blog/2012/03 ... gressions/
http://www.veritasprep.com/blog/2012/04 ... gressions/
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Re: The second, the first and the third term of an AP whose comm   [#permalink] 18 Nov 2014, 22:37
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