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January 6th, 2016, 03:48 AM   #1
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Finding the nth term in a fractional sequence

Could someone help me with the process of figuring out # 25?

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January 6th, 2016, 04:56 AM   #2
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It can be observed that the denominators are powers of 3 in ascending order and the numerator seem to also be increasing in ascending powers of 2 so we can assume it is a geometric progression. Divide the 2nd term by the first.
(2/9)/(1/3)=2/3
Divide the 3rd by the 2nd to verify the ratio
(4/27)/(2/9)=2/3
Hence we can assume it is a geometric sequence with a common ratio 2/3 and a first term 1/3. Although we cannot be entirely sure until we have enough terms of the sequence. However if we assume that it is geometric, then we can substitute into the formula
u(n)=ar^(n-1), where u(n) is the nth term, a is the first term and r is the common ratio. We just substitute n=24 and n=26 and divide the result of u(24) by u(26) to find the quotient
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January 6th, 2016, 05:22 AM   #3
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So u(n) is the nth term?

I'm a little confused, what is u?

Is this the problem that need to be solved for the 24th term?
u(24)=1/3(2/3)^(24-1)
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January 6th, 2016, 07:30 AM   #4
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Lanthanide said "where u(n) is the nth term" of the sequence. As for "what is u?" there is no "u". There are only terms of the sequence, u(0), u(1), etc..

By the way, to "find the quotient if the 24th term is divided by the 26th term", you do NOT need to actually find the 24th and 26th terms. The reason this is a "geometric sequence" is that any term divided by the next term is 2/3. The 24th term divided by the 25th term is 2/3 and the 25th term divided by the 26th term is 2/3. The "24th term divided by the 26th term" is $\displaystyle \frac{u(24)}{u(26)}= \frac{u(24)}{u(25)}\frac{u(25)}{u(26)}$.

Last edited by Country Boy; January 6th, 2016 at 07:36 AM.
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January 6th, 2016, 11:32 AM   #5
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nth term = a * [m^(n-1)]

a = 1/3, m = 2/3, n = 24, 26


24th / 26th =
1/3[(2/3)^(23)]
==========
1/3[(2/3)^(25)]

Simplify...
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