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

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Manager
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The second, the first and the third term of an AP whose comm  [#permalink]

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Updated on: 12 Jul 2013, 13:01
1
18
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Difficulty:

75% (hard)

Question Stats:

58% (02:40) correct 42% (02:17) wrong based on 132 sessions

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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|

Originally posted by Anasthaesium on 08 Dec 2011, 08:15.
Last edited by Bunuel on 12 Jul 2013, 13:01, edited 1 time in total.
Renamed the topic and edited the question.
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06 Jan 2012, 04:15
4
12
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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Karishma
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Manager
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08 Dec 2011, 09:44
1
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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08 Dec 2011, 10:00
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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08 Dec 2011, 10:13
1
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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03 Jan 2012, 04:43
I am getting the ratio as -1/2. Can someone please explain how it is -2?
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03 Jan 2012, 15:02
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:

Thanks,
Anu
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04 Jan 2012, 21:49
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:

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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05 Jan 2012, 21:23
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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06 Jan 2012, 04:17
1
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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06 Jan 2012, 04:56
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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07 Jan 2012, 02:32
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]

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18 Nov 2014, 10:43
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]

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18 Nov 2014, 23:37
2
2
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]

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27 Apr 2016, 09:33
this does not look like a GMAT question. Please let me know if im wrong.
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Re: The second, the first and the third term of an AP whose comm  [#permalink]

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27 Apr 2016, 19:46
abypatra wrote:
this does not look like a GMAT question. Please let me know if im wrong.

The concept could easily be tested this way in GMAT though the wording of the question would be much more explicit.
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Re: The second, the first and the third term of an AP whose comm  [#permalink]

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07 Feb 2019, 19:51
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Re: The second, the first and the third term of an AP whose comm   [#permalink] 07 Feb 2019, 19:51
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