Arithmetic Sequences: Free Response

The first term is the first number listed, $a_1 = 23$. Find $d$ by subtracting an earlier term from the one right after it:

$d = 19 - 23 = -4.$

Checking another gap confirms it, $15 - 19 = -4$ too, so the sequence is arithmetic with $d = -4$.

Reaching the tenth term takes $n - 1 = 9$ steps of size $d$ from the first term.

$a_{10} = 23 + 9 \times (-4) = 23 - 36 = -13.$

The recursive rule restates the definition of arithmetic directly: each term is the one before it plus the common difference.

$a_n = a_{n-1} - 4 \quad (n \ge 2), \qquad a_1 = 23.$

The rule $a_n = a_{n-1} - 4$ by itself only says how to step from one term to the next; it says nothing about where to start. The sequence $100, 96, 92, \ldots$ obeys this exact same recursive rule but is a completely different sequence from this one, so the first term must be stated alongside the rule to identify one specific sequence.

Between position $4$ and position $10$ there are $10 - 4 = 6$ steps of size $d$.

$a_{10} = a_4 + (10 - 4)d \quad \Longrightarrow \quad 47 = 17 + 6d \quad \Longrightarrow \quad d = 5.$

The first term is three steps before the fourth term, so back up using $a_1 = a_4 - 3d$:

$a_1 = 17 - 3 \times 5 = 17 - 15 = 2.$

Apply the between-terms formula straight from $a_{10}$ to position $20$, a jump of $20 - 10 = 10$ steps.

$a_{20} = a_{10} + (20 - 10)d = 47 + 10 \times 5 = 97.$

The formula $a_k = a_m + (k-m)d$ never requires $m = 1$; any known term can serve as the starting point, since the constant step $d$ carries you between any two positions, not specifically from the first term. Going through $a_1$ first, as in part B, is one valid route but never the only one.

The recursive rule steps by $-6$ each time from $a_1 = 34$, so this sequence is arithmetic with $a_1 = 34$ and $d = -6$. Substituting into the explicit formula:

$a_n = 34 + (n-1)(-6) = 34 - 6(n-1).$

Reaching position $16$ takes $n - 1 = 15$ steps.

$a_{16} = 34 - 6 \times 15 = 34 - 90 = -56.$

From $a_1 = 34$ to $a_2 = 28$ is a single application of the recursive rule, one step of $d = -6$:

$a_2 = a_1 - 6 = 34 - 6 = 28,$

so the second term needed only one step, not two. The first term $a_1$ itself needed zero steps, since it is where you start. Following the same counting, the $n$th term is reached after exactly $n-1$ steps, one fewer than the position number, which is exactly why the explicit formula multiplies $d$ by $(n-1)$ and not by $n$; multiplying by $n$ would add one step too many, overshooting every term after the first.

The tile counts are arithmetic with $a_1 = 14$ and $d = 3$. Reaching step $9$ takes $n - 1 = 8$ steps.

$a_9 = 14 + 8 \times 3 = 14 + 24 = 38.$

Each step's tile count is the step below it plus the fixed increase of $3$ tiles, so

$a_n = a_{n-1} + 3 \quad (n \ge 2), \qquad a_1 = 14.$

In this context $d = 3$ is not just a number, it is the physical fact that each step uses exactly $3$ more tiles than the step below it. The explicit formula relies on multiplying that one fixed value by the number of steps taken:

$a_n = a_1 + (n-1)d = 14 + 3(n-1).$

That single formula works only because the increase is assumed constant at every step. If the increase changed partway up the hillside, say growing from $3$ to $5$ tiles per step, a formula built on one constant $d$ would predict the wrong tile count for every step past where the increase changed, because it would still be multiplying by the old, no-longer-accurate difference.

Set the explicit formula equal to $110$ and solve for $n$.

$12 + 7(n-1) = 110 \quad \Longrightarrow \quad 7(n-1) = 98 \quad \Longrightarrow \quad n - 1 = 14.$

So $n = 15$, a positive whole number, meaning $110$ is the fifteenth term.

$12 + 7(n-1) = 150 \quad \Longrightarrow \quad 7(n-1) = 138 \quad \Longrightarrow \quad n - 1 = \tfrac{138}{7}.$

Since $\tfrac{138}{7}$ is not a whole number, there is no position $n$ landing on $150$, so $150$ is not a term of this sequence.

Solving $a_n = V$ always produces some real number for $n$, but that number only names an actual position in the sequence when it belongs to the counting numbers:

$n \in \{1, 2, 3, \ldots\}.$

A fractional or negative result means no step of the sequence ever lands exactly on $V$. This is exactly what happened with $150$ in part B: solving gave $n - 1 = \tfrac{138}{7}$, a fraction, so no whole step reaches it, and $150$ is not a term, even though it is bigger than the first term $a_1 = 12$ and well within the sequence's increasing range. Size alone never decides membership; only the positive-whole-number test does.