a college entrance examination problem! how to find a geometric progression A Chinese college entrance examination math problem:

Given an arithmetic progression sequence $a_{1},a_{2},a_{3},a_{4}$, that are all positive real numbers, the common difference of the arithmetic progression is set as $d\;(d\in \mathbb{R}\setminus\{0\})$. Prove that:

*

*$2^{a_{1}},2^{a_{2}},2^{a_{3}},2^{a_{4}}$ is a geometric progression sequence.


*Whether there exists $a_{1}$ and $d$ such that $a_{1},{a_{2}}^2,{a_{3}}^3  $and $a_{4}^4$ can be formed as a geometric progression? Explain why.


*Whether there exists $a_{1}$ ,$d$, and $n$,$k\;$($n$,$k$ are positive integers) such that $a_{1}^n,a_{2}^{n+k},a_{3}^{n+3k},a_{4}^{n+5k}$ can be formed as a geometric progression? Explain why.

I think the first part is easy, and I am feeling struggled with the part (2) and part (3). This problem is from this year's China's college entrance examination of math. Can someone help me solve it?

$a_k=dk+a_0$ by definition of arithmetic progression, with $a_0:=a_1-d$. 
1) A geometric progression $b_k$ by definition satisfies $b_{k+1}/b_k=c$ for some $c\in \mathbb{R}$. Thus if $b_{k}=2^{a_k}$, you can verify this fact directly.
2) Define $b_k=a_k^k$. Then suppose $b_{k+1}/b_k=a_{k+1}^{k+1}/a^k_k=c$. Rearranging, we get:
$$c(dk+a_0)^k=(d(k+1)+a_0)^{k+1}=(dk+a_0+d)^{k+1}=(dk+a_0)^{k+1}(1+d/(dk+a_0))^{k+1}.$$
This gives:
$$c=(dk+a_0)((1+d/(dk+a_0))^{k+1}\geq (dk+a_0),$$
but $c$ must be the same for all $k$, so if $d>0$, this is impossible. Hence $d\leq 0$. But if $d<0$, then $a_k$ will be negative at some point. So $d=0$ gives $c=1$. 
3) Can you take it from here?
