please help

Let the first term be   a      and the common ratio  = r

ar^(13)  =  3755    ⇒    a  =  375 / r^13    (1)

And

ar^4013  =  135       (2)      sub (1) into (2)

(375 / (r^13)  ( r^4013)  =  135

375 r^4000 =  135

r^4000 =  135/375  =9/25

r =  (9/25)^(1/4000)

The 2014th term is

(375/r^13)[ r]^(2013)   =  

375 [ r]^2000  =

375  ( [ 9/25]^(1/4000)) ^2000 = 

375  ( 9/25) ^(1/2)  =

375 (3/5)  =

225

If the 4014th term of a geometric sequence of non-negative numbers is 135,

and the 14th term is 375,

what is the 2014th term?

\(\begin{array}{|rcll|} \hline i < j < k \\ a_i &=& ar^{i-1} \\ a_j &=& ar^{j-1} \\ a_k &=& ar^{k-1} \\ \hline \dfrac{a_k}{a_i} &=& r^{(k-1)-(i-1)} \\ \mathbf{\dfrac{a_k}{a_i}} &=& \mathbf{r^{k-i}} \\\\ \dfrac{a_k}{a_j} &=& r^{(k-1)-(j-1)} \\ \mathbf{\dfrac{a_k}{a_j}} &=& \mathbf{r^{k-j}} \\ \hline r= \left(\dfrac{a_k}{a_i}\right)^{\frac{1}{k-i}} &=& \left(\dfrac{a_k}{a_j}\right)^{\frac{1}{k-j}} \\\\ a_k^{\frac{1}{k-i}} a_j^{\frac{1}{k-j}} &=& a_k^{\frac{1}{k-j}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{1}{k-j}-\frac{1}{k-i}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{k-i-k+j}{(k-j)(k-i)}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{j-i}{(k-j)(k-i)}} a_i^{\frac{1}{k-i}} \\\\ \mathbf{a_j} &=& \mathbf{a_k^{\frac{j-i}{k-i}} a_i^{\frac{k-j}{k-i}} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_j} &=& \mathbf{a_k^{\frac{j-i}{k-i}} a_i^{\frac{k-j}{k-i}} } \\\\ && \boxed{a_{14} =375=a_i\quad i = 14\\ a_{2014}=a_j \quad j=2014\\ a_{4014}=135=a_k\quad k=4014} \\\\ a_{2014} &=& 135^{\frac{2014-14}{4014-14}} 375^{\frac{4014-2014}{4014-14}} \\\\ a_{2014} &=& 135^{\frac{2000}{4000}} 375^{\frac{2000}{4000}} \\\\ a_{2014} &=& 135^{\frac{1}{2}} 375^{\frac{1}{2}} \\\\ a_{2014} &=& \sqrt{135*375} \\ a_{2014} &=& \sqrt{50625} \\ \mathbf{a_{2014}} &=& \mathbf{225} \\ \hline \end{array}\)

The 2014th term is 225