Eva owes her mom 570 dollars. She plans to write her one check per week to pay off the balance. The first week she pays 10 dollars. Each successive check will be for 5 dollars more than the previous check. In how many weeks will Eva finish paying her mother back?
The answer is at the end of 14 weeks. I couldn't arrive to this and got a "weird" answer.
What I did:
$$570 = \frac{n}{2}\Big(2\cdot 10+5(n-1)\Big).$$
This produces:
$$1140 = 5n^2+15n.$$
And I do not get 14, what am I doing wrong?
On
Here's how to check that the polynomial is correct if in doubt.
Denote $S$ to be the sum of money owed and how it's paid off.
$$S=570=10+15+20+25+...=10+(10+5)+(10+10)+(10+10+5)+...$$
$$S=10(1)+10\left(1+\frac{1}{2}\right)+10(1+1)+10\left(1+1+\frac{1}{2}\right)+...$$
$$S=10\left(1+\frac{3}{2}+2+\frac{5}{2}+...\right)=10\left(\sum_{m=1}^{t/2}m+\frac{1}{2}\sum_{n=2}^{1+t/2}(2n-1)\right)$$
The upper limits on the sums may seem funky but we know she has to pay once a week until the debt is paid, so she must pay $t$ times. The length of time paying the first sum (how many terms it contains) is exactly $t/2$, so the second sum must also have $t/2$ terms, but it is first indexed at $2$ because the first fraction that appears has a numerator of $3$. Therefore, the upper limit of the second sum is forced to be $\frac{t}{2}+1$. The values of these partial sums are known
$$\sum_{m=1}^{t/2}m=\frac{t/2}{2}\left(1+\frac{t}{2}\right)=\frac{t}{8}(t+2)$$
and
$$\frac{1}{2}\sum_{n=2}^{1+t/2}(2n-1)=\frac{1}{2}\left(\frac{t/2}{2}\left(4+t\right)\right)=\frac{t}{8}(t+4)$$
$$\therefore S=\frac{10}{8}(t^2+2t+t^2+4t)=\frac{5}{2}(t^2+3t)$$
And we are left with your original polynomial
$$1140=5(t^2+3t)$$
The answer is $14$ because she pays once a week and there is still money owed at the end of week $13$ due to the decimals. Therefore, the remainder is paid off on week $14$. This would still be true if $t$ turned out to be say $13.1$.
Dividing your equation by $5$; we get $a=1, b=3, c=-228$. Then, by the quadratic formula, we get \begin{align*} n&=\frac{-3\pm\sqrt{9-4(1)(-228)}}{2}\\ &=\frac{-3\pm\sqrt{921}}{2} \end{align*} Obviously we discard the negative solution since we cannot have a negative number of weeks; so we are left with $$n=\frac{-3+\sqrt{921}}{2}\approx13.67,$$ which we round up to $14$.