The positive numbers a, b, c are the first three terms of a geometric progression. The numbers a, b, -3c are the first three terms of an arithmetic progression. (a) Show that a² − 10ac + 9c² = 0.

Given that $a, b, c$ are the first three terms of a geometric progression (GP), the common ratio is constant:

$$\frac{b}{a} = \frac{c}{b} \implies b^2 = ac \quad (*)$$

Given that $a, b, -3c$ are the first three terms of an arithmetic progression (AP), the common difference is constant:

$$b - a = -3c - b \implies 2b = a - 3c \quad (**)$$

From $(**)$, express $b$ in terms of $a$ and $c$:

$$b = \frac{a - 3c}{2}$$

Substitute this expression for $b$ into equation $(*)$:

$$\left(\frac{a - 3c}{2}\right)^2 = ac$$ $$\frac{a^2 - 6ac + 9c^2}{4} = ac$$

Multiply both sides by 4:

$$a^2 - 6ac + 9c^2 = 4ac$$

Rearrange the terms to match the required equation:

$$a^2 - 10ac + 9c^2 = 0$$

This completes part (a).

Given $a=9$, substitute this into the equation from Step 2:

$$9^2 - 10(9)c + 9c^2 = 0$$ $$81 - 90c + 9c^2 = 0$$

Divide the entire equation by 9:

$$9 - 10c + c^2 = 0$$

Rearrange into a standard quadratic form:

$$c^2 - 10c + 9 = 0$$

Factor the quadratic equation:

$$(c - 1)(c - 9) = 0$$

This gives two possible values for $c$: $c=1$ or $c=9$.

The problem states that $c$ takes the smaller of its two possible values, so $c=1$.
We are given $a=9$.
The numbers $a, b, c$ are positive. Use $b^2 = ac$ from Step 1:

$$b^2 = (9)(1)$$ $$b^2 = 9$$

Since $b$ must be positive, $b=3$.
So, the terms are $a=9$, $b=3$, $c=1$.

The GP is $9, 3, 1, \dots$.
The first term is $A = 9$.
The common ratio is $r = \frac{3}{9} = \frac{1}{3}$.
Since $|r| < 1$, the sum to infinity exists and is given by $S_{\infty} = \frac{A}{1 - r}$.

$$S_{\infty} = \frac{9}{1 - \frac{1}{3}} = \frac{9}{\frac{2}{3}} = 9 \times \frac{3}{2} = \frac{27}{2}$$ $$S_{\infty} = \boxed{\frac{27}{2}}$$

The AP is $a, b, -3c, \dots$. Using $a=9, b=3, c=1$, the terms are $9, 3, -3, \dots$.
The first term is $A_{AP} = 9$.
The common difference is $d = 3 - 9 = -6$.
The sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2A_{AP} + (n-1)d]$.
For $n=20$:

$$S_{20} = \frac{20}{2}[2(9) + (20-1)(-6)]$$ $$S_{20} = 10[18 + (19)(-6)]$$ $$S_{20} = 10[18 - 114]$$ $$S_{20} = 10[-96]$$ $$S_{20} = \boxed{-960}$$