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[1/a^2+1/a^4+1/a^6+...]>[1/a+1/a^3+1/a^5]

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[1/a^2+1/a^4+1/a^6+...]>[1/a+1/a^3+1/a^5] [#permalink] New post 18 Oct 2010, 04:03
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 1 sessions
[1/a^2+1/a^4+1/a^6+...]>[1/a+1/a^3+1/a^5]

B. One of the roots of the equation 4x2–4x+1 = 0 is a
Soln. (1) — Both the series are infinitely diminishing series.
For the first series: First term = 1/a2 and r = 1/a2
For the second series: First term = 1/a and r = 1/a2
The sum of the first series = (1/a2) / (1 – 1/a2) = 1 / (a2 – 1)
The sum of the second series = (1/a) / (1 – 1/a2) = a / (a2 – 1)

Now, from the first statement, the relation can be anything (depending on whether
a is positive or negative).
But the second statement tells us, 4a2 – 4a + 1 = 0 or a = ½. For this value of a, the
sum of second series will always be greater than that of the first.

first, I need explanation on how the answer calculates sum of series? I mean the formula.
secondly, in DS section, when asked about a certain for example inequality like the one above, if one of choices refutes the inequality and the other one insuffices, we can conclude that the one that stands against the inequality holds the right answer???

source of this question is a file that I downloaded from here which includes tough tricky questions.

Thanks
[Reveal] Spoiler: OA
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Re: DS: Sum of series [#permalink] New post 18 Oct 2010, 09:00
honeyhani wrote:
[1/a^2+1/a^4+1/a^6+...]>[1/a+1/a^3+1/a^5]

B. One of the roots of the equation 4x2–4x+1 = 0 is a
Soln. (1) — Both the series are infinitely diminishing series.
For the first series: First term = 1/a2 and r = 1/a2
For the second series: First term = 1/a and r = 1/a2
The sum of the first series = (1/a2) / (1 – 1/a2) = 1 / (a2 – 1)
The sum of the second series = (1/a) / (1 – 1/a2) = a / (a2 – 1)

Now, from the first statement, the relation can be anything (depending on whether
a is positive or negative).
But the second statement tells us, 4a2 – 4a + 1 = 0 or a = ½. For this value of a, the
sum of second series will always be greater than that of the first.

first, I need explanation on how the answer calculates sum of series? I mean the formula.
secondly, in DS section, when asked about a certain for example inequality like the one above, if one of choices refutes the inequality and the other one insuffices, we can conclude that the one that stands against the inequality holds the right answer???

source of this question is a file that I downloaded from here which includes tough tricky questions.

Thanks


It is a geometric progression series where the ratio of two consecutive terms will be same ...like a+ ar + ar^2 + ar^3+.......

Here the sum will be calculated in this way,S= a(1-(r^n))/ (1-r) where a is the first term and r is the common ratio.
If it is an infinite series then sum ,S=a/(1-r)



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Re: DS: Sum of series [#permalink] New post 18 Oct 2010, 10:48
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honeyhani wrote:
[1/a^2+1/a^4+1/a^6+...]>[1/a+1/a^3+1/a^5]

B. One of the roots of the equation 4x2–4x+1 = 0 is a

first, I need explanation on how the answer calculates sum of series? I mean the formula.
secondly, in DS section, when asked about a certain for example inequality like the one above, if one of choices refutes the inequality and the other one insuffices, we can conclude that the one that stands against the inequality holds the right answer???


A few things here:

* you never see questions on the GMAT which ask about the sum of an infinite series. That's out of the scope of the test, and so is the question above;

* the question above asks a Yes/No question. If from one of the two Statements you can be 100% certain the answer to the question is 'No', then that Statement is certainly sufficient, since we can answer the question using that Statement; we don't care whether the answer is 'Yes' or 'No'. So if, as you said, 'one of the choices refutes the inequality', then that choice is sufficient.

* Looking at the question above, if we know that a is a solution of 4x^2 - 4x + 1 = 0, then we know a is a solution of (2x - 1)^2 = 0, so a = 1/2. Now, if we know the value of a, we have no unknowns left at all; of course we can plug in this value of a into our inequality on both sides and it will be, at least in theory, possible to decide if the left side is greater than or less than the right side. You don't need to actually do the calculation, since we don't care if the answer is 'yes' or 'no', only whether the question can be answered.

* Finally, you can look at this without using a formula, if you actually need to do calculation for some reason. There are a few ways to do this; I can suggest two, but there are others. We are asked whether:

(1/a^2) + (1/a^4) + (1/a^6) + ... > (1/a) + (1/a^3) + (1/a^5) + ....

We can rephrase the question by getting zero on one side; we want to know if:

0 > (1/a) - (1/a^2) + (1/a^3) - (1/a^4) + (1/a^5) - (1/a^6) + ....

If we can write the right side as a product, this will just become a positive/negative question. So we'd like to factor. Notice how similar the terms are; we can, for example, factor out 1/a^2 from the first two terms, 1/a^4 from the next two, and so on:

0 > (1/a^2)(a - 1) + (1/a^4)(a - 1) + (1/a^6)(a - 1) + ...
0 > (a - 1) [ (1/a^2) + (1/a^4) + (1/a^6) + ... ]

We now have a product on the right side. Since in the factor [ (1/a^2) + (1/a^4) + (1/a^6) + ... ] all of the terms must be positive (the powers are even), then our product will be negative only if a - 1 is negative, or if a < 1.

* Or you can notice that the inequality clearly must be true if a is negative, since then the left side is positive (the powers are all even) and the right side is negative. So we only need to be concerned with the case where a is positive. Then we can factor immediately:

(1/a^2) + (1/a^4) + (1/a^6) + ... > (1/a) + (1/a^3) + (1/a^5) + ....
(1/a)[ (1/a) + (1/a^3) + (1/a^5) + ...] > (1/a) + (1/a^3) + (1/a^5) + ....

Now, since a is positive, we can divide on both sides by [ (1/a) + (1/a^3) + (1/a^5) + ...], since we can be sure that's positive:

1/a > 1
1 > a

So the inequality will only be true if 0 < a < 1, or if a < 0.
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Re: DS: Sum of series [#permalink] New post 20 Oct 2010, 05:59
did not understand the prob... some one can explain me this
Re: DS: Sum of series   [#permalink] 20 Oct 2010, 05:59
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