Question: The first five terms of a sequence are $-2, 1, 4, 7, 10$. Find the next two terms, the $n$th term, the 55th term, and determine whether $359$ appears in the sequence.

Step 1. The common difference is $+3$, so the next terms are $13$ and $16$.

Step 2. For the $n$th term, start at $-2$ and add $3$, $(n-1)$ times:

$$T_n = -2 + (n-1) \times 3 = 3n - 5$$

Check on a known term: $T_3 = 9 - 5 = 4$. ✓

Step 3. 55th term:

$$T_{55} = 3(55) - 5 = 160$$

Step 4. Is $359$ a term? Solve $T_n = 359$:

$$3n - 5 = 359 \;\Rightarrow\; 3n = 364 \;\Rightarrow\; n = 121\tfrac{1}{3}$$

$n$ is not a positive whole number, so $359$ is not in the sequence. (With full working shown — the yes/no alone scores nothing.)

The same three moves — difference, formula, membership equation — run every arithmetic-sequence question in the paper.

⚠ Watch out: $T_n = a + (n-1)d$, not $a + nd$ — test the formula on $n = 1$ before using it. In the membership check, move the $-5$ correctly ($3n = 364$, not $354$), and the conclusion must cite "n is not a whole number" as the reason.

Question: A sequence of diagrams gives this table — Diagram $n$: unshaded squares $U$, shaded squares $S$: $(1, 1, 8), (2, 4, 12), (3, 9, 16)$. Find $x$ if $U = 36$ in Diagram $x$; find $U$ (called $y$) in Diagram 9; find $S$ (called $z$) in Diagram $x$.

[diagram in original sheet: growing square grids with a shaded border]

Step 1. Find each column's rule. Unshaded: $1, 4, 9$ — the perfect squares:

$$U = n^2$$

Shaded: $8, 12, 16$ — constant difference $4$:

$$S = 8 + (n-1) \times 4 = 4n + 4$$

Step 2. $U = 36$:

$$n^2 = 36 \;\Rightarrow\; x = 6$$

Step 3. Diagram 9's unshaded count:

$$y = 9^2 = 81$$

Step 4. Shaded squares in Diagram $x = 6$:

$$z = 4(6) + 4 = 28$$

Check. Geometry agrees: an $n \times n$ inner grid ($n^2$ unshaded) with a one-square border needs $4n + 4$ shaded squares. ✓

⚠ Watch out: Treat each column as its own sequence — one may be quadratic while its neighbour is linear, as here. Read the table's question marks carefully: $x$ is a diagram number, $y$ and $z$ are counts; answering 36 for $x$ mixes them up.

Question: Find the $n$th term of (a) $9, 16, 25, 36, \ldots$ and (b) $4, 12, 36, 108, \ldots$

Step 1. (a) The differences $7, 9, 11$ are not constant but growing — the signature of a square pattern. Compare with the square numbers:

$$9 = 3^2, \quad 16 = 4^2, \quad 25 = 5^2, \quad 36 = 6^2$$

Step 2. The base runs $3, 4, 5, 6$ while $n$ runs $1, 2, 3, 4$ — a shift of $2$:

$$T_n = (n+2)^2$$

Check: $T_1 = 3^2 = 9$. ✓

Step 3. (b) Each term is the previous one $\times 3$ — a constant ratio, so the pattern is exponential:

$$4, \quad 4 \times 3, \quad 4 \times 3^2, \quad 4 \times 3^3, \ldots$$

Step 4. The power lags $n$ by one:

$$T_n = 4 \times 3^{\,n-1}$$

Check: $T_2 = 4 \times 3 = 12$ ✓, $T_4 = 4 \times 27 = 108$ ✓.

⚠ Watch out: Diagnose by differences first: constant difference → $3n + c$ form; growing differences → try $n^2$ (shifted); constant ratio → $a \cdot r^{n-1}$. The two classic slips are writing $n^2$ without the shift (fails at $n=1$) and $3^n$ instead of $3^{n-1}$ (off by one factor).