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by shashank.ism » Sat Feb 06, 2010 1:44 pm
If a1, a2, a3 are the first three terms of an arithmetic progression containing 100 terms, what is the possible common ratio when a2, a1 and a3 are in geometric progression?


a) - 2
b) 2
c) 1
d) both (1) and (3)
e) impossible
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by harsh.champ » Sat Feb 06, 2010 2:04 pm
shashank.ism wrote:If a1, a2, a3 are the first three terms of an arithmetic progression containing 100 terms, what is the possible common ratio when a2, a1 and a3 are in geometric progression?


a) - 2
b) 2
c) 1
d) both (1) and (3)
e) impossible
First of all,I would like to point that 2 more alternatives of this question can appear.
This question can be rephrased as :-
If a1, a2, a3 are the first three terms of an arithmetic progression containing 100 terms, what is the possible common difference of the A.P. when a2, a1 and a3 are in geometric progression?

Or

If a1, a2, a3 are the first three terms of an geometric progression containing 100 terms, what is the possible common ratio when a2, a1 and a3 are in arithmetic progression?

Now,solving the initial question:-
So,let the the no.s be (a2 - n),a2,(a2 + n).
Since they are in G.P.,hence , (a2)^2 = (a2 - n)(a2 + n )
hence, n=0.
So,all the numbers are same.
Hence,the common ratio is 1.Hence,C

The (D) answer choice should be both (A) and (C) not both (1) and (3).
The numeric option choice can be confusing at times.
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by Ian Stewart » Sat Feb 06, 2010 3:21 pm
harsh.champ wrote:
shashank.ism wrote:If a1, a2, a3 are the first three terms of an arithmetic progression containing 100 terms, what is the possible common ratio when a2, a1 and a3 are in geometric progression?


a) - 2
b) 2
c) 1
d) both (1) and (3)
e) impossible
First of all,I would like to point that 2 more alternatives of this question can appear.
This question can be rephrased as :-
If a1, a2, a3 are the first three terms of an arithmetic progression containing 100 terms, what is the possible common difference of the A.P. when a2, a1 and a3 are in geometric progression?

Or

If a1, a2, a3 are the first three terms of an geometric progression containing 100 terms, what is the possible common ratio when a2, a1 and a3 are in arithmetic progression?

Now,solving the initial question:-
So,let the the no.s be (a2 - n),a2,(a2 + n).
Since they are in G.P.,hence , (a2)^2 = (a2 - n)(a2 + n )
hence, n=0.
So,all the numbers are same.
Hence,the common ratio is 1.Hence,C

The (D) answer choice should be both (A) and (C) not both (1) and (3).
The numeric option choice can be confusing at times.
First, none of the above questions could appear as written on the GMAT; you don't need to know what the phrase 'geometric progression' means for the test. If you see a geometric sequence on the test, they'll explain what it is using different language. The wording is also somewhat backwards in the question; it should state as a fact that a_2, a_1, a_3 form a geometric progression, and then ask what the common ratio could be if a_1, a_2, a_3 are in arithmetic progression. It also needs to make clear that the terms are nonzero; otherwise the ratio can be anything at all.

Second, there is a mistake when you write: (a2)^2 = (a2 - n)(a2 + n ). I'm not sure how you arrived at this; I can only guess that you've assumed that the first term in the progression will be equal to the product of the second and third terms. That's not normally true, as you can see by looking at any familiar geometric series.

Finally, it should be clear that the terms can be equal, so '1' is a possible answer here. Now, is it possible that -2 is the ratio? We could just take a numerical example; a_2 is the first term of our geometric sequence, so we could make it equal to 1, making a_1 = -2 and a_3 = 4. We can verify that a_1, a_2, a_3 now forms an arithmetic sequence: -2, 1, 4. So the ratio could be -2.

Or we can prove this more formally: we know a_2, a_1, a_3 form a geometric sequence, so in this case we would have

a_1 = (-2)a_2
a_3 = (-2)a_1 = 4a_2

Now, a_1, a_2, a_3 needs to be an arithmetic sequence; that is, (-2)a_2, a_2, 4a_2 must form an arithmetic sequence. It clearly does; the difference between consecutive terms is 3a_2.

So D is the correct answer (we don't need to bother checking '2' since there is no answer choice 'all of the above').
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by Eldorjon » Sat Feb 06, 2010 4:10 pm
I agree with harsh.champ and disagree with Ian

harsh.champ made correct calculations. It is right that the difference of arifmetic progression here must be 0.

Ian mistakenly took -2, 1, 4 sequence for geometric progression. The formula used by harsh.champ a(n)^2 = a(n-1)*a(n+1) is correct.

The answer is C
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by Ian Stewart » Sat Feb 06, 2010 4:19 pm
Eldorjon wrote:I agree with harsh.champ and disagree with Ian

harsh.champ made correct calculations. It is right that the difference of arifmetic progression here must be 0.

Ian mistakenly took -2, 1, 4 sequence for geometric progression. The formula used by harsh.champ a(n)^2 = a(n-1)*a(n+1) is correct.

The answer is C
I certainly did not 'mistakenly' take -2, 1, 4 for a geometric progression; I said it was an arithmetic progression. The sequence 1, -2, 4 is, however, a geometric progression, since we get the next term by multiplying by -2.

It is quite easy to misread the question here (something I did at first as well). The order of the terms in the geometric sequence given is not the same as for the arithmetic sequence. Notice that the question says that a2, a1, a3 is a geometric sequence (starting with a2), but that a1, a2, a3 is an arithmetic sequence; the terms are in a different order. If a1 = -2, a2 = 1, and a3 = 4, you have an arithmetic sequence if you look at a1, a2, a3, and you have a geometric sequence if you look at a2, a1, a3, as the question asks us to do.
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by Ian Stewart » Sat Feb 06, 2010 4:32 pm
Eldorjon wrote: The formula used by harsh.champ a(n)^2 = a(n-1)*a(n+1) is correct.
I should add that I now understand where harsh.champ's calculation comes from; he was assuming that both the arithmetic sequence and the geometric sequence were in the same order. If that were the case, then his calculation is indeed correct; if we have a geometric sequence

a, ar, ar^2

then if we multiply the first and third terms, we do get the square of the second term: (a)(ar^2) = (ar)^2. So if the question had said that a_1, a_2, a_3 formed both an arithmetic sequence and a geometric sequence, then his solution is perfect.
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by Eldorjon » Sat Feb 06, 2010 5:31 pm
I apologize, misread the question. Need to be more careful!
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by Ian Stewart » Sat Feb 06, 2010 6:02 pm
Eldorjon wrote:I apologize, misread the question. Need to be more careful!
I did too at first! It's an easy mistake to make, but an important one to learn from; you don't want that to happen on test day.
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