Problem (Algebra):
The $n$th term of an arithmetic sequence is given by $t_{n}=555-7 n$.

If $S_{n}=t_{1}+t_{2}+\ldots+t_{n}$, determine the smallest value of $n$ for which $S_{n}<0$.

Solution:
This is an arithmetic sequence in which $a=548$ and $d=-7$.

Therefore, $S_{n}=\frac{n}{2}[2(548)+(n-1)(-7)]=\frac{n}{2}[-7 n+1103]$.

We now want $\frac{n}{2}(-7 n+1103)<0$.

Since $n>0,-7 n+1103<0$

$$
n>157 \frac{4}{7}
$$

Therefore the smallest value of $n$ is 158 .

For this series we want, $\sum_{k=1}^{n} t_{k}<0$, or $\sum_{k=1}^{n}(555-7 k)<0$.

Rewriting, $555 n-7 \frac{(n)(n+1)}{2}<0$

$$
\begin{aligned}
1110 n-7 n^{2}-7 n & <0 \\
7 n^{2}-1103 n & >0 \\
\text { or, } n & >\frac{1103}{7} .
\end{aligned}
$$

The smallest value of $n$ is 158 .

We generate the series as $548,541,534, \ldots, 2,-5, \ldots,-544,-551$.

If we pair the series from front to back the sum of each pair is -3 .

Including all the pairs $548-551,541-544$ and so on there would be 79 pairs which give a sum of -237 .

If the last term, -551 , were omitted we would have a positive sum.

Therefore we need all 79 pairs or 158 terms.

Final answer: 158