A patient is prescribed a medication where the dosage follows an arithmetic progression. Ifthe dose on Day 3 is 15 mg and on Day 7 is 27 mg, lind the first dose and the common difference. (ii) What will be the dose on Day 12? Calculate the total medication administered in the first 2 weeks (14 days). Find the first four terns in the expansion of (1 - 2x)" in ascending power of x

a)

(i) Find the first dose and the common difference.
Let the first dose be $a_1$ and the common difference be $d$.
For an arithmetic progression, the $n$-th term is given by $a_n = a_1 + (n-1)d$.

Given:
Dose on Day 3 ($a_3$) is 15 mg.

$$a_3 = a_1 + (3-1)d \implies a_1 + 2d = 15 \quad (1)$$

Dose on Day 7 ($a_7$) is 27 mg.

$$a_7 = a_1 + (7-1)d \implies a_1 + 6d = 27 \quad (2)$$

Subtract equation (1) from equation (2):

$$(a_1 + 6d) - (a_1 + 2d) = 27 - 15$$ $$4d = 12$$ $$d = \frac{12}{4}$$ $$d = 3 \text{ mg}$$

Substitute $d=3$ into equation (1):

$$a_1 + 2(3) = 15$$ $$a_1 + 6 = 15$$ $$a_1 = 15 - 6$$ $$a_1 = 9 \text{ mg}$$

The first dose is 9 mg and the common difference is 3 mg.

(ii) What will be the dose on Day 12?
Using the formula $a_n = a_1 + (n-1)d$ with $a_1 = 9$ mg, $d = 3$ mg, and $n=12$:

$$a_{12} = 9 + (12-1)3$$ $$a_{12} = 9 + (11)3$$ $$a_{12} = 9 + 33$$ $$a_{12} = 42 \text{ mg}$$

The dose on Day 12 will be 42 mg.

(iii) Calculate the total medication administered in the first 2 weeks (14 days).
The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}(2a_1 + (n-1)d)$.
Here, $n=14$, $a_1 = 9$ mg, and $d = 3$ mg.

$$S_{14} = \frac{14}{2}(2(9) + (14-1)3)$$ $$S_{14} = 7(18 + (13)3)$$ $$S_{14} = 7(18 + 39)$$ $$S_{14} = 7(57)$$ $$S_{14} = 399 \text{ mg}$$

The total medication administered in the first 2 weeks is 399 mg.

b) Find the first four terms in the expansion of $(1 - 2x)^n$ in ascending power of x.
Using the binomial expansion formula $(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \binom{n}{3}a^{n-3}b^3 + \dots$
Here, $a=1$ and $b=-2x$.

First term ($k=0$):

$$\binom{n}{0}(1)^n(-2x)^0 = 1 \cdot 1 \cdot 1 = 1$$

Second term ($k=1$):

$$\binom{n}{1}(1)^{n-1}(-2x)^1 = n \cdot 1 \cdot (-2x) = -2nx$$

Third term ($k=2$):

$$\binom{n}{2}(1)^{n-2}(-2x)^2 = \frac{n(n-1)}{2!} \cdot 1 \cdot (4x^2) = \frac{n(n-1)}{2} \cdot 4x^2 = 2n(n-1)x^2$$

Fourth term ($k=3$):

$$\binom{n}{3}(1)^{n-3}(-2x)^3 = \frac{n(n-1)(n-2)}{3!} \cdot 1 \cdot (-8x^3) = \frac{n(n-1)(n-2)}{6} \cdot (-8x^3) = -\frac{4n(n-1)(n-2)}{3}x^3$$

The first four terms are:

$$\boxed{1 - 2nx + 2n(n-1)x^2 - \frac{4n(n-1)(n-2)}{3}x^3}$$