Arithmetic Sequences and Series: Free Response

Position $5$ and position $12$ are separated by $12 - 5 = 7$ steps of size $d$, and those steps carry the value from $54$ up to $110$, a rise of $56$.

$d = \frac{110 - 54}{7} = \frac{56}{7} = 8.$

The first term is four steps before the fifth term, so back up from $a_5$ using $a_1 = a_5 - 4d$.

$a_1 = 54 - 4(8) = 54 - 32 = 22.$

Substitute $a_1 = 22$ and $d = 8$ into the explicit formula and simplify.

$a_n = 22 + (n-1)8 = 8n + 14.$

Evaluating at $n = 21$,

$a_{21} = 8(21) + 14 = 168 + 14 = 182.$

As a check, the formula still reproduces the given data: $a_5 = 8(5)+14=54$ and $a_{12}=8(12)+14=110$.

The weekly distances are arithmetic with $a_1 = 8$ and $d = 3$. Reaching week $11$ from week $1$ takes $n - 1 = 10$ steps.

$a_{11} = 8 + 10(3) = 8 + 30 = 38 \text{ kilometers}.$

The last week's distance is $L = 50$, so count how many steps of $3$ separate it from the first week's $8$, then convert steps back into weeks.

$n = \frac{50 - 8}{3} + 1 = \frac{42}{3} + 1 = 14 + 1 = 15 \text{ weeks}.$

The counting formula $n = \dfrac{L-a_1}{d}+1$ works by treating $L$ as a specific term of the sequence and solving for its position; it needs the division $(L-a_1)\div d$ to come out to a whole number. 'A distance exceeding $50$' is not one specific value, so there is no single $L$ to substitute, and the formula cannot be applied to it directly.

Instead, solve the inequality $8+(n-1)(3) > 50$ for $n$.

$8 + 3(n-1) > 50 \quad \Longrightarrow \quad 3(n-1) > 42 \quad \Longrightarrow \quad n - 1 > 14 \quad \Longrightarrow \quad n > 15.$

The smallest whole number satisfying $n > 15$ is $n = 16$. This differs from week $15$ in part B, because week $15$'s distance of $50$ kilometers exactly equals the threshold rather than exceeding it, so under this new stopping rule the block runs one week longer.

The weekly deposits form an arithmetic sequence with $a_1 = 13$ and $d = 7$ dollars, since each week adds a fixed $7$ dollars over the week before. Only $a_1$, $d$, and $n = 20$ are known directly, so use the $d$-based sum formula.

$S_{20} = \frac{20}{2}\Big(2(13) + (20-1)(7)\Big) = 10\big(26 + 133\big) = 10 \times 159 = 1590 \text{ dollars}.$

This range is its own arithmetic series, with its own first term, last term, and count. There are $20 - 9 + 1 = 12$ weeks in the range, since both the starting and ending week are counted. The range's end values are

$a_9 = 13 + 8(7) = 69, \qquad a_{20} = 13 + 19(7) = 146.$

Treat the range as a $12$-term series running from $69$ to $146$:

$S = \frac{12(69+146)}{2} = 6 \times 215 = 1290 \text{ dollars}.$

Apply the same $d$-based sum formula with $n = 8$.

$S_8 = \frac{8}{2}\Big(2(13) + 7(7)\Big) = 4\big(26+49\big) = 4 \times 75 = 300 \text{ dollars}.$

$S_{20}$ is the running total through week $20$, and $S_8$ is the running total through week $8$, so subtracting removes exactly the first $8$ weeks and leaves weeks $9$ through $20$ behind.

$S_{20} - S_8 = 1590 - 300 = 1290 \text{ dollars},$

which agrees with part B.

A term is the jump the running total makes when that term arrives, $a_n = S_n - S_{n-1}$.

$S_{n-1} = 5(n-1)^2 + 3(n-1) = 5(n^2-2n+1) + 3n - 3 = 5n^2 - 10n + 5 + 3n - 3 = 5n^2 - 7n + 2.$

Subtracting,

$a_n = S_n - S_{n-1} = (5n^2+3n) - (5n^2-7n+2) = 10n - 2.$

This subtraction is valid even at $n=1$, since $S_0$ is the empty partial sum, $S_0 = 0$ by convention, and the algebraic expression for $S_{n-1}$ evaluated at $n=1$ gives $5(0)^2+3(0)=0$ to match, so no special case is needed at the low end.

The formula $a_n = 10n-2$ is linear in $n$, and a sequence whose $n$th term is linear in $n$ is exactly an arithmetic sequence, with $d$ equal to the coefficient of $n$. So the sequence is arithmetic with $d = 10$.

Evaluate the formula from part A directly.

$a_1 = 10(1) - 2 = 8, \qquad a_8 = 10(8) - 2 = 78.$

As a check, $a_1$ should equal $S_1$ itself, and indeed $S_1 = 5(1)^2+3(1) = 8$.

Computing individual terms one at a time, $a_1, a_2, a_3, \ldots$, only ever shows that a finite list of differences happens to match; it can never rule out a change further along. Simplifying $a_n = S_n - S_{n-1}$ algebraically is different, because $n$ is never fixed to a specific number anywhere in that simplification: the result is a single formula valid for every positive integer $n$ simultaneously.

$a_n = S_n - S_{n-1} = 10n - 2 \quad \text{for every } n \ge 1,$

valid at $n=1$ too since $S_0=0$ by the usual empty-sum convention, as noted in part A. That formula being linear in $n$ is exactly the condition for the sequence to be arithmetic, and the coefficient of $n$ in it is $d$ by definition, so both facts, that the sequence is arithmetic and what its $d$ is, fall out of the same algebraic step. No individual term ever needs to be computed to reach either conclusion.

With $13$ terms, an odd count, the middle position is $7$, since $\tfrac{13+1}{2}=7$, and a sum of an odd number of arithmetic terms always equals the count times that middle term.

$a_7 = \frac{S_{13}}{13} = \frac{351}{13} = 27.$

Position $7$ is $7-1=6$ steps from position $1$, so

$a_7 = a_1 + 6d \quad \Longrightarrow \quad 27 = 3 + 6d \quad \Longrightarrow \quad d = 4.$

Position $13$ is $13-1=12$ steps from position $1$.

$a_{13} = a_1 + 12d = 3 + 12(4) = 51.$

Substituting the first and last term into the sum formula,

$S_{13} = \frac{13}{2}(3+51) = \frac{13}{2}(54) = 13 \times 27 = 351,$

which matches the sum given in the stem.