1. We are given the sequence: 1, 2, 4, 7, 11, \ldots, 821, and we need to find the term number (position) of 821 in this sequence. 2. First, observe the differences between consecutive terms: $$2 - 1 = 1,$$ $$4 - 2 = 2,$$ $$7 - 4 = 3,$$ $$11 - 7 = 4,$$ We see the differences increase by 1 each time. 3. This means the sequence is generated by starting at 1, then adding an increasing difference starting at 1: $$a_1 = 1,$$ $$a_2 = a_1 + 1 = 1 + 1 = 2,$$ $$a_3 = a_2 + 2 = 2 + 2 = 4,$$ $$a_4 = a_3 + 3 = 4 + 3 = 7,$$ $$a_5 = a_4 + 4 = 7 + 4 = 11,$$ and so on. 4. The $n$th term can be expressed as $$a_n = 1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}.$$ 5. We want to find $n$ such that $a_n = 821$. So: $$821 = 1 + \frac{(n-1)n}{2}$$ $$821 - 1 = \frac{(n-1)n}{2}$$ $$820 = \frac{n^2 - n}{2}$$ $$1640 = n^2 - n$$ 6. Rearranging as a quadratic equation: $$n^2 - n - 1640 = 0.$$ 7. Solve quadratic using the formula: $$n = \frac{1 \pm \sqrt{1 + 4 \times 1640}}{2} = \frac{1 \pm \sqrt{1 + 6560}}{2} = \frac{1 \pm \sqrt{6561}}{2}.$$ Since $\sqrt{6561} = 81$, we get: $$n = \frac{1 \pm 81}{2}.$$ 8. Taking the positive root (since term number must be positive): $$n = \frac{1 + 81}{2} = \frac{82}{2} = 41.$$ 9. Therefore, the term containing 821 is the 41st term in the sequence. 10. Final answer: $$\boxed{41}$$