The problem gives the formula $s = \frac{a}{1-r}$. We are asked to (i) find the value of $s$ when $a = 81$ and $r = \frac{1}{3}$, and (ii) make $r$ the subject of the formula.

1. Problem Description

The problem gives the formula $s = \frac{a}{1-r}$. We are asked to (i) find the value of $s$ when $a = 81$ and $r = \frac{1}{3}$, and (ii) make $r$ the subject of the formula.

2. Solution Steps

(i) We are given $s = \frac{a}{1-r}$, $a = 81$, and $r = \frac{1}{3}$. Substituting these values into the formula for $s$, we get:

$s = \frac{81}{1 - \frac{1}{3}}$

$s = \frac{81}{\frac{2}{3}}$

$s = 81 \times \frac{3}{2}$

$s = \frac{243}{2}$

$s = 121.5$

(ii) We are given $s = \frac{a}{1-r}$. We want to make $r$ the subject of the formula. First, multiply both sides by $(1-r)$:

$s(1-r) = a$

$s - sr = a$

Subtract $s$ from both sides:

$-sr = a - s$

Multiply both sides by $-1$:

$sr = s - a$

Divide both sides by $s$:

$r = \frac{s-a}{s}$

$r = 1 - \frac{a}{s}$

3. Final Answer

(i) $s = 121.5$

(ii) $r = 1 - \frac{a}{s}$ or $r = \frac{s-a}{s}$