Find the value of $\sum_{n=1}^k u_n$, arithmetic series

Question

An arithmetic series has first term $80$ and common difference $5$.

The sum of the first $25$ terms of the series is $3500$.

The $n$th term of the series is $u_n$. Given that

$$33\left(\sum_{n=1}^{25} u_n-\sum_{n=1}^k u_n\right)=67\sum_{n=1}^k u_n$$

Find the value of $$\sum_{n=1}^k u_n$$

Note: I know that I could of done this is a much quicker way but I wanted to test another method

\begin{align} 33\left(\sum_{n=1}^{25} u_n-\sum_{n=1}^k u_n\right) & = 67\sum_{n=1}^k u_n \\ 33\left(3500-(\frac{k}{2}(2\cdot80-(k-1)5))\right) & = 67(\frac{k}{2}(2\cdot80-(k-1)5)) \\ 33(3505-75k) & = 67(75k+5) \\ 115665-2475k & = 5025k+335 \\ 7500k & = 115330 \\ k & = \frac{115330}{7500} \\ \end{align}

Then I use $k$ to find the value of $$\sum_{n=1}^k u_n$$

$$(\frac{k}{2}(2\cdot80-(k-1)5))$$

\begin{align} (\frac{k}{2}(2\cdot80-(k-1)5)) & = (\frac{\frac{115330}{7500}}{2}(2\cdot80-(\frac{115330}{7500}-1)5)) \\ & = \frac{115330}{15000}(160-\frac{10783}{150}) \\ & = \frac{18452800}{15000}-\frac{1243603390}{2250000}) \\ & = 677.47... \end{align}

Where did I go wrong?

For anyone interested the answer is

$$\sum_{n=1}^k u_n=1155$$

Answer 1

Please observe that for large $k$ your $2\cdot80-(k-1)5$ is negative, which is impossible.

But, if $y=\sum_{n=1}^k u_n$, we have $$ 33\cdot3500-33y=67y $$ and $100y=33\cdot3500$, hence $y=33\cdot35$.