[Algebra]
Problem: Suppose that $n>5$ and that the numbers $t_{1}, t_{2}, t_{3}, \ldots, t_{n-2}, t_{n-1}, t_{n}$ form an arithmetic sequence with $n$ terms. If $t_{3}=5, t_{n-2}=95$, and the sum of all $n$ terms is 1000 , what is the value of $n$ ?

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.)

Solution: Since the sequence $t_{1}, t_{2}, t_{3}, \ldots, t_{n-2}, t_{n-1}, t_{n}$ is arithmetic, then

$$
t_{1}+t_{n}=t_{2}+t_{n-1}=t_{3}+t_{n-2}
$$

This is because, if $d$ is the common difference, we have $t_{2}=t_{1}+d$ and $t_{n-1}=t_{n}-d$, as well as having $t_{3}=t_{1}+2 d$ and $t_{n-2}=t_{n}-2 d$.

Since the sum of all $n$ terms is 1000, using one formula for the sum of an arithmetic sequence gives

$$
\begin{aligned}
\frac{n}{2}\left(t_{1}+t_{n}\right) & =1000 \\
n\left(t_{1}+t_{n}\right) & =2000 \\
n\left(t_{3}+t_{n-2}\right) & =2000 \\
n(5+95) & =2000
\end{aligned}
$$

and so $n=20$.
Suppose that the arithmetic sequence with $n$ terms has first term $a$ and common difference $d$.

Then $t_{3}=a+2 d=5$ and $t_{n-2}=a+(n-3) d=95$.

Since the sum of the $n$ terms equals 1000, then

$$
\frac{n}{2}(2 a+(n-1) d)=1000
$$

Adding the equations $a+2 d=5$ and $a+(n-3) d=95$, we obtain $2 a+(n-1) d=100$.

Substituting, we get $\frac{n}{2}(100)=1000$ from which we obtain $n=20$.

Final answer: 20