Sequence in Sequence Challenge Suppose that there are two arithmetic sequences $a_n$ and $b_n$. Given that ${a_b}_{20}$ + ${b_a}_{14} = 2014$. What is the value of ${a_b}_{14}$ + ${b_a}_{20}$
 A: What an interesting question! And a nice one to commemorate the year!
Here's a rather neat solution (even if I do say so myself), without having to substitute everything from scratch.
Let $p,q$ be the common difference for $a_n, b_n$ respectively. 
Hence
$$\large {\begin{align}a_{b_{14}}+b_{a_{20}}&=[\color{red}{a_{b_{20}}}-(b_{20}-b_{14})p]+[\color{red}{b_{a_{14}}}+(a_{20}-a_{14})q]\\
&=\color{red}{a_{b_{20}}+b_{a_{14}}}-(b_{20}-b_{14})p+(a_{20}-a_{14})q\\
&=\color{red}{2014}-(6q)p+(6p)q\\
&=2014\qquad \blacksquare\end{align}} $$
A: Let $a_1=a$, common difference of the sequence is $d_a$, we have $a_n=(n-1)d_a+a$
Let $b_1=b$, common difference of the sequence is $d_b$, we have $b_n=(n-1)d_b+b$
$$a_{b_{20}}+b_{a_{14}}=2014$$
$$a+(b_{20}-1)d_a+b+(a_{14}-1)d_b=2014$$
$$a+(b+19d_b-1)d_a+b+(a+13d_a-1)d_b=2014$$
$$a+b+bd_a+ad_b+32d_ad_b-d_a-d_b=2014$$
Therefore
$$a_{b_{14}}+b_{a_{20}}=\\a+(b_{14}-1)d_a+b+(a_{20}-1)d_b=\\a+(b+13d_b-1)d_a+b+(a+19d_a-1)d_b=\\a+b+bd_a+ad_b+32d_ad_b-d_a-d_b=\\2014$$
