Problem (Algebra):
Let $T=T N Y W R$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$, is arithmetic with $a_{16}=13$ and $a_{30}=20$. Compute the value of $k$ for which $a_{k}=T$.

Solution:
If $d$ is the common difference of the sequence, then the $n^{\text {th }}$ term of the sequence is $a_{n}=$ $a_{16}+d(n-16)$. The values $a_{16}=13$ and $a_{30}=20$ yield $d=(20-13) /(30-16)=1 / 2$, hence $a_{n}=13+(1 / 2)(n-16)$. If $a_{n}=T$, then $n=2(T-13)+16=2 T-10$. With $T=27 / 2$, it follows that $n=\mathbf{1 7}$.

Final answer: 17