The third, fifth and eighth terms of an arithmetic progression are consecutive terms of a geometric progression. Given that the first term of the AP is 8, determine the common difference and the common ratio.

Let the first term of the AP be $a$ and the common difference be $d$.
Given $a = 8$.
The third term is $T_3 = a + 2d = 8 + 2d$.
The fifth term is $T_5 = a + 4d = 8 + 4d$.
The eighth term is $T_8 = a + 7d = 8 + 7d$.

Since $T_3, T_5, T_8$ are consecutive terms of a GP, the ratio of consecutive terms is constant.

Cross-multiply:

Subtract 64 from both sides and rearrange terms:

Factor out $2d$:

This gives two possible values for $d$:
$2d = 0 \implies d = 0$
or
$d - 4 = 0 \implies d = 4$.

Case 1: If $d = 0$
The terms of the AP are $8, 8, 8$. The GP terms are $8, 8, 8$.
The common ratio $r = \frac{8}{8} = 1$.
Common difference: $\boxed{d = 0}$, Common ratio: $\boxed{r = 1}$.

Case 2: If $d = 4$
The terms of the AP are:
$T_3 = 8 + 2(4) = 16$
$T_5 = 8 + 4(4) = 24$
$T_8 = 8 + 7(4) = 36$
The GP terms are $16, 24, 36$.
The common ratio $r = \frac{24}{16} = \frac{3}{2}$.
Common difference: $\boxed{d = 4}$, Common ratio: $\boxed{r = \frac{3}{2}}$.

Distance $L = 900 \text{ km}$.
Radius of Earth $R = 6370 \text{ km}$.
Latitude of A = $5^\circ N$.
The distance along a meridian is given by the formula:

where $\theta$ is the difference in latitude in degrees.

Substitute the given values into the formula:

Rearrange to solve for $\theta$:

Since A is due North of B, B must be South of A.
Latitude of B = Latitude of A - $\theta$
Latitude of B = $5^\circ N - 8.095^\circ = -3.095^\circ$.
This means B is $3.095^\circ$ South of the equator.

The latitude of B is $\boxed{3.1^\circ S}$.

The formula for compound appreciation is:

Where:
$A$ = Final amount = Ksh. 1049404
$P$ = Principal amount (current cost) = Ksh. 600000
$r$ = Annual rate of appreciation = 15% = 0.15
$n$ = Number of years (to be found)

Take the logarithm of both sides:

Using a calculator:

The number of years is $\boxed{4}$.

The relationship can be written as:

where $k$ and $c$ are constants.

Condition 1: When $B = 4, A = 22$.

Condition 2: When $B = 9, A = 42$.

From Equation 1, divide by 2:

Substitute this expression for $c$ into Equation 2:

Subtract 33 from both sides:

Now substitute $k = 3$ back into the expression for $c$:

Substitute the values of $k$ and $c$ into the original equation:

Substitute $B = 25$ into the complete variation equation:

When B = 25, A is $\boxed{100}$.

Nominal length $L = 4.8 \text{ m}$.
Nominal width $W = 3.2 \text{ m}$.
Nominal area $A = L \times W = 4.8 \text{ m} \times 3.2 \text{ m} = 15.36 \text{ m}^2$.

Maximum length $L_{max} = 4.8 + 0.05 = 4.85 \text{ m}$.
Minimum length $L_{min} = 4.8 - 0.05 = 4.75 \text{ m}$.
Maximum width $W_{max} = 3.2 + 0.05 = 3.25 \text{ m}$.
Minimum width $W_{min} = 3.2 - 0.05 = 3.15 \text{ m}$.

Maximum possible area $A_{max} = L_{max} \times W_{max} = 4.85 \text{ m} \times 3.25 \text{ m} = 15.7625 \text{ m}^2$.
Minimum possible area $A_{min} = L_{min} \times W_{min} = 4.75 \text{ m} \times 3.15 \text{ m} = 14.9625 \text{ m}^2$.

The maximum absolute error is the larger of $(A_{max} - A)$ and $(A - A_{min})$.
$A_{max} - A = 15.7625 - 15.36 = 0.4025 \text{ m}^2$.
$A - A_{min} = 15.36 - 14.9625 = 0.3975 \text{ m}^2$.
The maximum absolute error $\Delta A = 0.4025 \text{ m}^2$.

Percentage error = $\frac{\text{Maximum absolute error}}{\text{Nominal Area}} \times 100\%$
Percentage error = $\frac{0.4025 \text{ m}^2}{15.36 \text{ m}^2} \times 100\% \approx 2.61914\%$

Rounding to two decimal places:
The percentage error in the area is approximately $\boxed{2.62\%}$.

Let $x$ be the quantity (in kg) of Grade A tea.
Let $y$ be the quantity (in kg) of Grade B tea.

Cost of $x$ kg of Grade A tea = $80x$.
Cost of $y$ kg of Grade B tea = $60y$.
Total cost of the mixture = $80x + 60y$.
Total quantity of the mixture = $x + y$.
The desired price of the blend is Sh. 75 per kg.
So, the average cost per kg is 75:

Multiply both sides by $(x + y)$:

Rearrange the terms to group $x$ terms and $y$ terms:

Divide both sides by $5y$ to find the ratio $\frac{x}{y}$:

The ratio of Grade A to Grade B is $\boxed{3:1}$.