Interlacing an arithmetic and a geometric progression

Construction. Let $c$ be an even integer with $c>2(n-1)^2$ (for example $c=2n^2$) and set, for $k=1,\dots,n$,
$$b_k=c^{n-k}(c+1)^{k-1},\qquad a_k=c^{n-1}+\left(k-\tfrac12\right)c^{n-2}.$$
All $b_k$ are integers, in geometric progression with ratio $\frac{c+1}c$. The $a_k$ are in arithmetic progression with difference $c^{n-2}$, and they are integers because $c$ is even and $n\ge3$, so $\frac12c^{n-2}$ is an integer.

Normalising. Dividing by $c^{n-1}$: the claim $b_k<a_k<b_{k+1}$ becomes, with $x=\frac1c$,
$$(1+x)^{k-1}<1+\left(k-\tfrac12\right)x<(1+x)^k.$$

Right inequality. By Bernoulli, $(1+x)^k\ge1+kx>1+\left(k-\frac12\right)x$.

Left inequality. Let $m=k-1\le n-1$, so $mx\le\frac{n-1}c\le1$. Using $1+t\le e^t$ and $e^t\le1+t+t^2$ for $t\le1$:
$$(1+x)^m\le e^{mx}\le1+mx+m^2x^2.$$
We need $m^2x^2<\frac x2$, that is $c>2m^2$. Since $c>2(n-1)^2\ge2m^2$, the left inequality holds.

Example. The values of $c$ below are smaller than the one used in the proof, but they work, as one checks directly. For $n=3$ and $c=4$: $b=(16,20,25)$ and $a=(18,22,26)$, giving
$$16<18<20<22<25<26.$$
For $n=5$ and $c=14$: $b=(38416,\ 41160,\ 44100,\ 47250,\ 50625)$ and $a=(39788,\ 42532,\ 45276,\ 48020,\ 50764)$, and one checks
$$38416<39788<41160<42532<44100<45276<47250<48020<50625<50764.$$

Remark. Both sequences have a similar size, but the geometric one eventually overtakes any arithmetic one. The construction makes them dance around each other for exactly $n$ steps by choosing a ratio very close to $1$.