The formula for the sum to n terms of an arithmetic series is given on the formula sheet (first page of exam).

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , where a is the first term, d is the common difference and n is the number of terms.

Use the formula to find an expression in terms of a and d for the first 48 terms of an arithmetic series (n = 48).

$S_{48} = \frac{48}{2} [2 a + (48 - 1) d] = 24 (2 a + 47 d)$

[1]

Use the formula to find an expression in terms of a and d for the first 36 terms of the same arithmetic series (n = 36).

$S_{36} = \frac{36}{2} [2 a + (36 - 1) d] = 18 (2 a + 35 d)$

S₄₈= 4 × S₃₆ , use this to form an equation in terms of a and d.

$24 (2 a + 47 d) = 4 (18 (2 a + 35 d))$

[1]

Simplify by canceling 4 from both sides.

$\begin{array}{rcl} \frac{24 (2 a + 47 d)}{4} & = & \frac{4 (18 (2 a + 35 d))}{4} \\ 6 (2 a + 47 d) & = & 18 (2 a + 35 d) \end{array}$

Simplify by canceling 6 from both sides.

$\begin{array}{rcl} \frac{6 (2 a + 47 d)}{6} & = & \frac{18 (2 a + 35 d)}{6} \\ 2 a + 47 d & = & 3 (2 a + 35 d) \end{array}$

Expand the brackets on the right-hand side.

$\begin{array}{rcl} 2 a + 47 d & = & 6 a + 105 d \end{array}$

Rearrange to make a (or d) the subject.

$\begin{array}{rcl} 2 a - 6 a & = & 105 d - 47 d \\ - 4 a & = & 58 d \\ a & = & \frac{58 d}{- 4} = - 14.5 d \end{array}$

[1]

Use the formula to find an expression in terms of a and d for the first 30 terms of an arithmetic series (n = 30).

$S_{30} = \frac{30}{2} [2 a + (30 - 1) d] = 15 (2 a + 29 d)$

[1]

Substitute a = -14.5d into the expression found for S_30.

$S_{30} = 15 (2 (- 14.5 d) + 29 d) = 15 (- 29 d + 29 d) = 15 \times 0$

$S_{30} = 0$ [1]

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