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

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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-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'.

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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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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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.

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

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04 Nov 2017, 09:04

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