### Instruction:
Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$, where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$.

### Response:

Let's first find a general formula for $a_n$. First notice that $a_1 = \sin{1}$. And $a_2 = \sin{1} + \sin{2}$. So $a_2 = \sin{1} + 2 \sin{1} \cos{1}$. And $a_3 = \sin{1} + 2 \sin{1} \cos{1} + \sin{3}$. So $a_3 = \sin{1} + 2 \sin{1} \cos{1} + 3 \sin{1} - 4 \sin^3{1}$. And $a_4 = \sin{1} + 2 \sin{1} \cos{1} + 3 \sin{1} - 4 \sin^3{1} + \sin{4}$. Oh yeah. That's much easier. The formula is $\frac{\sin \left(\frac{n+1}{2}b \right) \sin \left(\frac{n}{2}b + a \right)}{\sin \left(\frac{b}{2} \right)}$. Now, we just need to plug in the numbers. So the sum is $\frac{\sin \left(\frac{n+1}{2} \right) \sin \left(\frac{n}{2} \right)}{\sin \left(\frac{1}{2} \right)}$. And the sum is equal to $a_n$, so $a_n = \frac{\sin \left(\frac{n+1}{2} \right) \sin \left(\frac{n}{2} \right)}{\sin \left(\frac{1}{2} \right)}$. Now, we just need to solve for $n$ in the equation $a_n < 0$. Notice that the denominator is positive. So the sign of $a_n$ is determined by the numerator. And the product of two sines is negative if and only if one of the sines is negative. So the numerator is negative when $cos \left( \frac{n+1}{2} - \frac{n}{2} \right) < cos \left( \frac{n+1}{2} + \frac{n}{2} \right)$. That is equivalent to $cos \left( \frac{1}{2} \right) < cos \left( \frac{2n+1}{2} \right)$. And $cos \left( \frac{1}{2} \right)$ is positive. So $\frac{2n+1}{2}$ has to be in the first or fourth quadrant. But the first quadrant is not an option because the cosine function is decreasing in the first quadrant. That's a good idea. We could use the periodicity of the cosine function. Yes, and we know that the cosine function is decreasing in the first quadrant, so there is no other solution in the first quadrant. Now, let's move to the fourth quadrant. In the fourth quadrant the cosine function is increasing. So there is only one solution in the fourth quadrant. We know that the cosine function is positive in the first and fourth quadrants. And so there is only one solution in the fourth quadrant. Let's find that solution. That's equivalent to $2x+1 = 4\pi - 1$. So, $2x = 4\pi - 2$. That means $x = 2\pi - 1$. So, $x = 2\pi - 1$ is the only solution in the fourth quadrant. We can add to $2\pi - 1$ any integer multiple of $2\pi$ and still have a solution. We already found that the interval $[2\pi - 1, 2\pi]$ works. Now, let's look for the intervals of the form $(2\pi-1+2\pi k,2\pi+2\pi k)$. Now, let's move to the next interval, $(4\pi-1,4\pi)$. In that interval the only integer is $n = 12$. And let's move to the next interval, $(6\pi-1,6\pi)$. So, we found $3$ values of $n$ that make $a_n < 0$. But we are looking for the $100^{\text{th}}$ value of $n$. And the floor function of $2\pi$ is $6$. And the floor function of $4\pi$ is $12$. And the floor function of $6\pi$ is $18$. The right end-point of the 100th interval is $200\pi$. We can notice that the right end of the 100th interval will be $200\pi$. And the floor value of $200\pi$ is $628$. So $n = 628$ is the index of the 100th term for which $a_n < 0$.

