Arithmetic series

a1 = x

a2 = x + r

a3 = x + 2r

a4 = x + 3r

a5 = x + 4r

(x + r)/(x + 3r) = a2/a3

I don't think it's possible to figured out (x + 4r)/(x + 2r).

=^._.^=

a₂ / a ₄  =   3

So

a₄/ a₂  =   1/3

a₄  =  a₂ /  3

a₄  =   a₂  +  2d  =   a₂  / 3

So

a₂  +  2d  =  a₂  / 3

2d   =   -(2/3)a₂

d  =   -(1/3)a₂

a₃   =    a₂  +  d   =   a₂  -  (1/3)a₂ )   =    (2/3)a₂

a₅   =  a₂   +   3d   =   a₂  +  3( -(1/3)a₂)   =    a₂  -  a₂  =  0

So

a₅ / a₃     =     0

Let a1, a2, a3, ... be an arithmetic sequence.
If a2/a4 = 3, what is a5/a3?

\(\begin{array}{|rcll|} \hline \dfrac{a_2}{a_4} &=& 3 \\ a_2 &=& 3a_4 \\ a_1+d &=& 3(a_1+3d) \\ a_1 + d &=& 3a_1+9d \\ -2a_1 &=& 8d \\ \mathbf{a_1} &=& \mathbf{-4d} \\ \hline \dfrac{a_5}{a_3} &=& x \\ a_5 &=& x*a_3 \\ a_1+4d &=& x*a_3 \quad | \quad \mathbf{a_1=-4d} \\ -4d+4d &=& x*a_3 \\ 0 &=& x*a_3 \\ x &=& \dfrac{0}{a_3} \\ \mathbf{x} &=& \mathbf{0} \\ \hline \end{array}\)

\(\dfrac{a_5}{a_3} = 0\)

I always feel good  when my answer  matches up with heureka's   !!!!!!