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Considering that the numerical sequence $a_n, n = 1, 2, 3, ...$ shown in table I is constructed by intercalating the terms of the three sequences presented in table II, judge the following items.

The terms $3n - 1, n = 1, 2, 3, ...$ of the sequence $a_n$ form a geometric progression of ratio 2.

Is it correct to say that $a_{28} < a_{291}$?

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Somos
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funfun
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  • Could you explain me what does this statement means? The terms 3n - 1, n = 1, 2, 3, ... of the sequence an form a geometric progression of ratio 2. – Crystal Wolftron Feb 18 '20 at 16:48
  • What have you done so far? – saulspatz Feb 18 '20 at 16:48
  • @CrystalWolftron I believe it refers to the sequence $a_{3n-1},\ n=1,2,\dots$ – saulspatz Feb 18 '20 at 16:50
  • Neither $28$ nor $291$ are of the for $3n-1$ so neither of them are in the exponential sequence which outstrip the other sequences very fast. $28$ is of the form $3n+1$ is it is geometric sequence which increases fasert than the arithmetic sequence, which is what terms of the form $3n$ are (which $291$ is) so $a_{28=39+1}=(9+1)^2=100$ does surpass $a_{291=397}= 97$. But $a_{25}= 81$ is less than $a_{288}=96$. However $3*7-1=20$ puts $a_{20}$ in the exponentail sequence $a_{20}=2^7=128$ outstrips the other two. – fleablood Feb 18 '20 at 17:01

1 Answers1

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Well $28 \equiv 1 \pmod 3$ so $a_{28}$ is in the first sequence and $28=3*9 + 1$ so it is the $9+1=10$ term so $a_{28}=10^2=100$.

And $291\equiv 3\pmod 3$ so $a_{291}$ is in the third sequence. And $291 = 3*96 + 3$ so it is the $96+1=97$th term. So $a_{291} = 97$.

So, no, $a_{28} > a_{291}$.

Note if $b_k = k^2$ and $c_k = 2^k$ and $d_k = k$ then

$a_{3m + n; n=1,2,3}=\begin{cases}b_{m+1}&n=1\\c_{m+1}&n=2;\\d_{m+1}&n=3\end{cases}$

fleablood
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