[Algebra]
Problem: Sequence A has $n$th term $n^{2}-10 n+70$.

(The first three terms of sequence $\mathrm{A}$ are $61,54,49$. )

Sequence B is an arithmetic sequence with first term 5 and common difference 10. (The first three terms of sequence $\mathrm{B}$ are $5,15,25$.)

Determine all $n$ for which the $n$th term of sequence $\mathrm{A}$ is equal to the $n$th term of sequence B. Explain how you got your answer.

Solution: The $n$th term of sequence $\mathrm{A}$ is $n^{2}-10 n+70$.

Since sequence B is arithmetic with first term 5 and common difference 10 , then the $n$th term of sequence $\mathrm{B}$ is equal to $5+10(n-1)=10 n-5$. (Note that this formula agrees with the first few terms.)

For the $n$th term of sequence $\mathrm{A}$ to be equal to the $n$th term of sequence $\mathrm{B}$, we must have

$$
\begin{aligned}
n^{2}-10 n+70 & =10 n-5 \\
n^{2}-20 n+75 & =0 \\
(n-5)(n-15) & =0
\end{aligned}
$$

Therefore, $n=5$ or $n=15$. That is, 5 th and 15 th terms of sequence $\mathrm{A}$ and sequence $\mathrm{B}$ are equal to each other.

Final answer: 5,15