Establish the relationship between mass (M), length (L), and diameter (D).

The mass of a wire varies jointly with its length and with the square of its diameter. This can be written as:

$$M = kLD^2$$

where $k$ is the constant of proportionality.

$M_1 = 31.5 \text{ kg}$
$L_1 = 500 \text{ m}$
$D_1 = 3 \text{ mm}$

$$31.5 = k \times 500 \times (3)^2$$ $$31.5 = k \times 500 \times 9$$ $$31.5 = 4500k$$ $$k = \frac{31.5}{4500} = 0.007$$

$L_2 = 1000 \text{ m}$
$D_2 = 2 \text{ mm}$

$$M_2 = kL_2D_2^2$$ $$M_2 = 0.007 \times 1000 \times (2)^2$$ $$M_2 = 0.007 \times 1000 \times 4$$ $$M_2 = 7 \times 4$$ $$M_2 = 28 \text{ kg}$$

The final answer is $\boxed{28 \text{ kg}}$.

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Gatua's annual salary from Ksh to K£.
Given annual salary = Ksh 80,000.
Since 1 K£ = 20 Ksh:

$$\text{Annual salary in K£} = \frac{80,000}{20} = 4,000 \text{ K£}$$

• First 1500 K£ taxed at 10%:

$$1500 \times 0.10 = 150 \text{ K£}$$

• Next 1500 K£ (from 1501 to 3000 K£) taxed at 15%:

$$1500 \times 0.15 = 225 \text{ K£}$$

• Remaining income: $4000 - 1500 - 1500 = 1000 \text{ K£}$. This falls into the 3000 - 4500 K£ bracket, taxed at 25%:

$$1000 \times 0.25 = 250 \text{ K£}$$

Total annual tax = $150 + 225 + 250 = 625 \text{ K£}$.

Given monthly personal relief = Ksh 1056.

$$\text{Annual personal relief} = 1056 \times 12 = 12,672 \text{ Ksh}$$

Annual PAYE = Total annual tax - Annual personal relief

$$\text{Annual PAYE} = 12,500 - 12,672 = -172 \text{ Ksh}$$

Since the annual personal relief is greater than the total annual tax, Mr. Gatua pays no tax. The PAYE cannot be negative.

Since the annual PAYE is 0, the monthly PAYE is also 0.
The final answer is $\boxed{\text{Ksh } 0}$.

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$$T_n = ar^{n-1}$$

where $a$ is the first term and $r$ is the common ratio.

The third term ($T_3$) is $3 \frac{1}{3} = \frac{10}{3}$:

$$T_3 = ar^{3-1} = ar^2 = \frac{10}{3} \quad (1)$$

The sixth term ($T_6$) is $11 \frac{1}{4} = \frac{45}{4}$:

$$T_6 = ar^{6-1} = ar^5 = \frac{45}{4} \quad (2)$$

$$\frac{ar^5}{ar^2} = \frac{\frac{45}{4}}{\frac{10}{3}}$$ $$r^3 = \frac{45}{4} \times \frac{3}{10}$$ $$r^3 = \frac{135}{40}$$ $$r^3 = \frac{27}{8}$$

Take the cube root of both sides:

$$r = \sqrt[3]{\frac{27}{8}}$$ $$r = \frac{3}{2}$$

$$ar^2 = \frac{10}{3}$$ $$a\left(\frac{3}{2}\right)^2 = \frac{10}{3}$$ $$a\left(\frac{9}{4}\right) = \frac{10}{3}$$ $$a = \frac{10}{3} \times \frac{4}{9}$$ $$a = \frac{40}{27}$$

The common ratio is $\frac{3}{2}$ and the first term is $\frac{40}{27}$.
The final answer is $\boxed{\text{common ratio } r = \frac{3}{2}, \text{ first term } a = \frac{40}{27}}$.