The eleventh term of an arithmetic progression is 33. The first, third and ninth terms of the arithmetic progression form consecutive terms of a geometric progression. Find the first term and common difference of the arithmetic progression.

Here's the solution to question 1.

1(a) Find the coordinates of B.

Step 1: Write down the equations of the two lines. Line $l_1$: $2y - x - 10 = 0$ (Equation 1) Line $l_2$: $y - 3x = 0$ (Equation 2)

Step 2: Solve the system of equations. From Equation 2, we can express $y$ in terms of $x$: $y = 3x$ Substitute this expression for $y$ into Equation 1: $2(3x) - x - 10 = 0$ $6x - x - 10 = 0$ $5x - 10 = 0$ $5x = 10$ $x = \frac{10}{5}$ $x = 2$

Step 3: Substitute the value of $x$ back into $y = 3x$ to find $y$. $y = 3(2)$ $y = 6$

Step 4: State the coordinates of B. The coordinates of B are $(x, y) = (2, 6)$. $\boxed{\text{B}(2, 6)}$

1(b) Hence or otherwise show that AB is perpendicular to $2y - x - 10 = 0$.

Step 1: Identify the coordinates of points A and B. Point A is $(3, 4)$. Point B is $(2, 6)$ (from part (a)).

Step 2: Calculate the gradient of the line segment AB. The gradient $m_{AB}$ is given by $\frac{y_2 - y_1}{x_2 - x_1}$. $m_{AB} = \frac{6 - 4}{2 - 3} = \frac{2}{-1} = -2$

Step 3: Calculate the gradient of the line $2y - x - 10 = 0$. Rearrange the equation into the form $y = mx + c$ to find the gradient $m$. $2y - x - 10 = 0$ $2y = x + 10$ $y = \frac{1}{2}x + 5$ The gradient of line $l_1$ is $m_{l1} = \frac{1}{2}$.

Step 4: Check the condition for perpendicularity. Two lines are perpendicular if the product of their gradients is $-1$. $m_{AB} \times m_{l1} = -2 \times \frac{1}{2} = -1$ Since the product of the gradients is $-1$, line AB is perpendicular to the line $2y - x - 10 = 0$. $\boxed{\text{Shown}}$