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?



*Let the numbers be $b-d,b,b+d,b+2d$.  Then $b-d,b^2,(b+d)^3,(b+2d)^4$ are to form  a geometric series.
We need $(b-d)(b+d)^3=(b^2)^2$.  Let $e=b/d$, then it simplifies to $2e^3-2e-1=0$.
We need $(b^2)(b+2d)^4=(b+d)^6$.  This simplifies to $2e^5+9e^4+12e^3+e^2-6e-1=0$.
Do long division to check that the cubic and the quintic do not share a root.  

*I didn't do; perhaps the two polynomials do share a root?
