a. Find a recurrence relation for the sequence \( \left\{d_{n}\right\} \) that gives the amount of drug in the blood after the nth dose, where \( d_{1}=80 \). A. \( d_{n+1}=2 d_{n}+160 \) B. \( d_{n+1}=\frac{1}{2} d_{n}+80 \) C. \( d_{n+1}=2 d_{n}-80 \) D. \( d_{n+1} \left\lvert\, \frac{1}{2} d_{n}-160\right. \) b. Using a calculator, determine the limit of the sequence. In the long run, how much drug is in the person's blood? Confirm the result by finding the limit of the sequence directly. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The limit of the sequence is \( \square \) mg. B. The limit does not exist.

The amount of drug in the blood after the \( n \)th dose is denoted by \( d_n \) with \( d_1 = 80 \). We assume that between doses the amount of drug decays by half, and then a new dose of \( 80 \) mg is added. 1. After the first dose, the blood contains \( d_1 = 80 \) mg. 2. Before the second dose, the amount decays to \( \frac{1}{2}d_1 \). Then, the dose of \( 80 \) mg is added, so: \[ d_2 = \frac{1}{2}d_1+80. \] 3. In general, before the \((n+1)\)th dose, the blood holds \( \frac{1}{2}d_n \) mg, and once the dose of \( 80 \) mg is administered: \[ d_{n+1} = \frac{1}{2}d_n+80. \] Thus, the recurrence relation is: \[ d_{n+1} = \frac{1}{2}d_n + 80. \] This corresponds to option B. Next, we determine the limit of the sequence. Let \( \lim_{n\to\infty} d_n = L \). In the limit, the recurrence becomes: \[ L = \frac{1}{2}L + 80. \] Now, solve for \( L \): \[ L - \frac{1}{2}L = 80, \] \[ \frac{1}{2}L = 80, \] \[ L = 160. \] Thus, the limit of the sequence is \( 160 \) mg, meaning that in the long run, the person will have \( 160 \) mg of the drug in their blood. This corresponds to option A for part (b) with the answer \( 160 \) mg.