The problem has two parts. (a) We need to express the sum of two fractions, $\frac{3}{2}$ and $\frac{4x-1}{3}$, as a single fraction in its simplest form. (b) We are given the formula $S = \frac{1}{2}n(r+l)$. (i) We need to calculate the value of $S$ when $n = 13$, $r = 7$, and $l = 11$. (ii) We need to make $r$ the subject of the formula, i.e., express $r$ in terms of $S$, $n$, and $l$.

1. Problem Description

The problem has two parts.

(a) We need to express the sum of two fractions, $\frac{3}{2}$ and $\frac{4x-1}{3}$, as a single fraction in its simplest form.

(b) We are given the formula $S = \frac{1}{2}n(r+l)$.

(i) We need to calculate the value of $S$ when $n = 13$, $r = 7$, and $l = 11$.

(ii) We need to make $r$ the subject of the formula, i.e., express $r$ in terms of $S$, $n$, and $l$.

2. Solution Steps

(a) To add the fractions $\frac{3}{2}$ and $\frac{4x-1}{3}$, we need to find a common denominator, which is

6. $$\frac{3}{2} + \frac{4x-1}{3} = \frac{3 \times 3}{2 \times 3} + \frac{2 \times (4x-1)}{3 \times 2} = \frac{9}{6} + \frac{8x-2}{6}$$

Now, we can add the numerators:

$$\frac{9}{6} + \frac{8x-2}{6} = \frac{9 + (8x - 2)}{6} = \frac{8x + 7}{6}$$

(b) (i) We substitute $n = 13$, $r = 7$, and $l = 11$ into the formula $S = \frac{1}{2}n(r+l)$:

$$S = \frac{1}{2}(13)(7+11) = \frac{1}{2}(13)(18) = 13 \times 9 = 117$$

(ii) To make $r$ the subject of the formula $S = \frac{1}{2}n(r+l)$, we first multiply both sides by 2:

$$2S = n(r+l)$$

Then, we divide both sides by $n$:

$$\frac{2S}{n} = r+l$$

Finally, we subtract $l$ from both sides to isolate $r$:

$$r = \frac{2S}{n} - l$$

3. Final Answer

(a) $\frac{8x+7}{6}$

(b) (i) $S = 117$

(ii) $r = \frac{2S}{n} - l$