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The first three terms of a series are 5^1/2, 5^1/3 and

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neelesh
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The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   12 Oct 2005, 20:24
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A
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The first three terms of a series are 5^1/2, 5^1/3 and 5^1/6...

The fourth term is....



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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   12 Oct 2005, 20:32
5^ 1/18 or 5 ^1/36

Just thinking aloud. 3rd term power is 1/6 which is a multiple of first and second term (1/2 * 1/3). Hence 4th term power is multiple of previous 2 (or) multiple of previous all).

Whatz OA?
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   12 Oct 2005, 20:47
Choices are:

A. 1
B. 5^1/7
C. 5^1/12
D. 5^1/8
E. 5^1/9
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 06:40
I am guessing here. I choose A. I think this is wrong but I can't seem to figure out where to go on this.

OK, 1/2-1/3=1/6. So I think we are deducting 1/6 from every power. Therefore 1/6-1/6=0, 5^0=1. Choice A

OA?
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 06:54
Agree with Chris. The sequence is subtracting 1/6 each time, so the next number will be 5^0 or 1.

By the way, does 0^0=1?
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 06:56
GMATT73 wrote:
Agree with Chris. The sequence is subtracting 1/6 each time, so the next number will be 5^0 or 1.

By the way, does 0^0=1?


0^0= 0^(1-1)= 0^1 : 0^1 ------> it seems unreasonable! :?
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 08:02
https://home.att.net/~numericana/answer/ ... htm#zeroth

At least according to this Math PhD: 0^0 =1.
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 17:55
Interestingly ... multiplication and subtraction yield the same answer for the given series.

1/2 * 1/3 and 1/2 - 1/3
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 19:57
sudhagar wrote:
Interestingly ... multiplication and subtraction yield the same answer for the given series.

1/2 * 1/3 and 1/2 - 1/3


Careful... just the first two terms... 1/3-1/6 is not the same as 1/3*1/6

Good problem.
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 20:49
5^1/2, 5^1/3 and 5^1/6 can also be written as

5^3/6, 5^2/6 and 5^1/6...and hence next term is 5^0/6 = 1
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 23:19
neelesh wrote:
The first three terms of a series are 5^1/2, 5^1/3 and 5^1/6...

The fourth term is....

I would go with 5 ^ 1/9.

The power of third term in the series is 1/3 the first term. Going by this rational, the power of the fourth term could be 1/3 of the second term. Hence it could be 5 ^ 1/9
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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink] New post   13 Oct 2005, 23:22
krisrini wrote:
neelesh wrote:
The first three terms of a series are 5^1/2, 5^1/3 and 5^1/6...

The fourth term is....

I would go with 5 ^ 1/9.

The power of third term in the series is 1/3 the first term. Going by this rational, the power of the fourth term could be 1/3 of the second term. Hence it could be 5 ^ 1/9


As I always do, I reviewed my answers with the rest, and I agree the answer should be 1.

Moral of the Story: Read and look closer into the problems.



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Re: The first three terms of a series are 5^1/2, 5^1/3 and [#permalink]
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