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 $a_{28} < a_{291}$?
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
$a_{3m + n; n=1,2,3}=\begin{cases}b_{m+1}&n=1\\c_{m+1}&n=2;\\d_{m+1}&n=3\end{cases}$