Arithmetic Sequences and Series: Practice

The four steps from term $3$ to term $7$ raise the value by $16$, so $d = \frac{16}{4} = 4$.

Now walk backwards from the 3rd term to the 1st, which is $2$ steps down.

$a_1 = a_3 - 2d = 11 - 2 \cdot 4 = 3$

Check it forwards: $a_3 = 3 + 2(4) = 11$ and $a_7 = 3 + 6(4) = 27$.

Step back from the 5th term to the 1st, a distance of $4$ steps. Each forward step adds $-3$, so each backward step adds $3$.

$a_1 = a_5 - 4d = 17 - 4(-3) = 29$

Now build the rule and simplify.

$a_n = 29 + (n-1)(-3) = 29 - 3n + 3 = 32 - 3n$

Check the given term: $a_5 = 32 - 15 = 17$.

The common difference is $d = 5$, and reaching the 12th term takes $11$ steps.

$a_{12} = 3 + 11 \cdot 5 = 58$

Now add the twelve terms.

$S_{12} = \frac{12}{2}(3 + 58) = 6 \cdot 61 = 366$

Using $12$ steps by mistake gives $a_{12} = 63$ and the wrong total $396$; forgetting the half gives $732$.

Write the general term first.

$a_n = 40 + (n-1)(-6) = 46 - 6n$

Now ask when it drops below zero: $46 - 6n < 0$ means $n > \frac{46}{6}$, which is about $7.67$, so the first whole number that works is $n = 8$.

Check the two candidates. $a_7 = 46 - 42 = 4$ is still positive, and $a_8 = 46 - 48 = -2$ is negative.

Use the form of the sum that needs only $a_1 = 2$ and $d = 3$.

$S_n = \frac{n}{2}\big(2(2) + (n-1)3\big) = \frac{n(3n + 1)}{2}$

Set it equal to $155$ and clear the fraction.

$3n^2 + n - 310 = 0 \;\Rightarrow\; (n - 10)(3n + 31) = 0$

The only positive whole-number solution is $n = 10$, and it checks out: $S_{10} = 5(4 + 27) = 155$.

The even numbers $2, 4, \ldots, 100$ are arithmetic with $d = 2$, so count them first.

$n = \frac{100 - 2}{2} + 1 = 49 + 1 = 50$

Now add the fifty terms.

$S_{50} = \frac{50}{2}(2 + 100) = 25 \cdot 102 = 2550$

Forgetting the half gives $5{,}100$, exactly twice the answer.

From the 1st term to the 20th is $19$ steps, not $20$, and the total climb is $102 - 7 = 95$.

$d = \frac{102 - 7}{20 - 1} = \frac{95}{19} = 5$

Dividing by $20$ gives $4.75$, which is the error of counting terms rather than the gaps between them.

The terms $2k - 1$ are the odd numbers, an arithmetic sequence with $d = 2$ running from $a_1 = 1$ to $a_{12} = 23$.

$S_{12} = \frac{12}{2}(1 + 23) = 6 \cdot 24 = 144$

That is $12^2$, and it is no coincidence. The first $n$ odd numbers always sum to $\frac{n}{2}\big(1 + (2n-1)\big) = n^2$.

The rows form an arithmetic sequence with $a_1 = 18$ and $d = 3$. Reaching row $12$ from row $1$ takes $11$ steps, not $12$.

$a_{12} = 18 + 11 \cdot 3 = 18 + 33 = 51$

The tempting $18 + 12 \cdot 3 = 54$ counts one row too many and describes row $13$.

The rows hold $18, 21, 24, \ldots$ seats. Row $12$ is $11$ steps past row $1$, so it holds $18 + 11(3) = 51$ seats.

Now add the twelve rows.

$S_{12} = \frac{12}{2}(18 + 51) = 6 \cdot 69 = 414$

Using $54$ for the last row gives $432$, and forgetting the half gives $828$.

Subtract $a_1$ from both sides of the $n$th-term formula.

$a_n - a_1 = (n-1)d$

Now divide by the number of steps between the two terms, which is $n - 1$.

$d = \frac{a_n - a_1}{n - 1}$

Dividing by $n$ would spread the climb over too many steps, and reversing the subtraction flips the sign of $d$.

The multiples of $7$ below $100$ run $7, 14, \ldots, 98$, an arithmetic sequence with $d = 7$. Count them first.

$n = \frac{98 - 7}{7} + 1 = 13 + 1 = 14$

Now add them.

$S_{14} = \frac{14}{2}(7 + 98) = 7 \cdot 105 = 735$

If you carelessly include $105$, which is past $100$, you get $840$ instead.