Arithmetic Sequences and Series: Additional Practice (Set 1)

The per-second distances form an arithmetic sequence with $a_1 = 4$ and $d = 6$. Reaching the 10th second takes $9$ steps, not $10$.

$a_{10} = 4 + 9 \cdot 6 = 58 \text{ cm}$

Now add the ten distances.

$S_{10} = \frac{10}{2}(4 + 58) = 5 \cdot 62 = 310 \text{ cm}$

Taking ten steps instead of nine gives the wrong last term $64$ and the wrong total $340$; forgetting the half gives $620$.

The terms $2k - 1$ step by $2$, but the sum starts at $k = 5$, so find the two ends first.

$2(5) - 1 = 9, \qquad 2(30) - 1 = 59$

Count the terms: the index runs from $5$ to $30$, so there are $30 - 5 + 1 = 26$ of them. The $+1$ is the same fence-post correction as always.

$S = \frac{26}{2}(9 + 59) = 13 \cdot 68 = 884$

Missing the $+1$ and using $25$ terms would give $850$.

The integers from $-12$ to $40$ are arithmetic with $d = 1$, so count them before adding.

$n = \frac{40 - (-12)}{1} + 1 = 52 + 1 = 53$

Now apply the sum.

$S = \frac{53}{2}(-12 + 40) = 53 \cdot 14 = 742$

A second route confirms it: everything from $-12$ to $12$ cancels in pairs, leaving $13 + 14 + \cdots + 40 = \frac{28}{2}(13 + 40) = 742$.

The rows hold $25, 24, \ldots, 10$ logs, an arithmetic sequence with $d = -1$. Count the rows first.

$n = \frac{10 - 25}{-1} + 1 = 15 + 1 = 16$

Now add the sixteen rows.

$S_{16} = \frac{16}{2}(25 + 10) = 8 \cdot 35 = 280$

Using $15$ rows (forgetting the $+1$) would give $262.5$, not even a whole number of logs, which is a useful signal that the count was wrong.

No last term is given, so use the form of the sum built from $a_1$ and $d$.

$S_{15} = \frac{15}{2}\big(2(-14) + 14d\big) = 15(-14 + 7d) = 0$

A product is zero only when a factor is, and $15 \neq 0$, so the bracket must vanish.

$-14 + 7d = 0 \;\Rightarrow\; d = 2$

The sequence is then $-14, -12, \ldots, 12, 14$, symmetric about its middle term $a_8 = 0$, which is exactly why the total cancels.

The first $n$ odd numbers are $1, 3, 5, \ldots, 2n - 1$, an arithmetic sequence with $d = 2$. Apply the sum.

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

So the sum of the first $n$ odd numbers is always a perfect square. Set it equal to $400$.

$n^2 = 400 \;\Rightarrow\; n = 20$

A count cannot be negative, so $n = -20$ is discarded.

Recover the shared common difference from the sequence you know completely. From the 1st term to the 20th is $19$ steps.

$d = \frac{62 - 5}{19} = 3$

Now run the second sequence out to its 20th term with that same $d$.

$b_{20} = 12 + 19 \cdot 3 = 12 + 57 = 69$

There is a shortcut worth seeing: the second sequence starts $7$ above the first and climbs at the same rate, so it stays exactly $7$ above it forever, and $62 + 7 = 69$.

Name the angles $m - d$, $m$, and $m + d$, centring them on the middle one. Their sum collapses to $3m$, and a triangle's angles total $180^\circ$.

$3m = 180^\circ \;\Rightarrow\; m = 60^\circ$

The middle angle is $60^\circ$ whatever $d$ turns out to be. The smallest angle is $40^\circ$, so $d = 20^\circ$, and the largest is $60^\circ + 20^\circ$.

$40^\circ + 60^\circ + 80^\circ = 180^\circ$

The largest angle is $80^\circ$.

The terms run from $a_1 = 5$ to $a_n = 2n + 3$, arithmetic with $d = 2$, so the sum is

$S_n = \frac{n}{2}\big(5 + (2n + 3)\big) = \frac{n(2n + 8)}{2} = n^2 + 4n.$

Set that equal to $320$ and factor.

$n^2 + 4n - 320 = 0 \;\Rightarrow\; (n - 16)(n + 20) = 0$

The positive solution is $n = 16$, and it checks: $256 + 64 = 320$.

The number of terms is odd, so the middle term (the 6th of $11$) is the average of the first and last, and the sum is the number of terms times that average.

$S_{11} = \frac{11}{2}(a_1 + a_{11}) = 11 \cdot a_6$

Now divide.

$a_6 = \frac{231}{11} = 21$

The common difference is never needed, so the 6th term is fully determined by the total alone.

The rule $5k - 2$ is linear in $k$, so the terms are arithmetic with $d = 5$. The sum formula needs the two ends and the count.

$a_1 = 5(1) - 2 = 3, \qquad a_{12} = 5(12) - 2 = 58$

There are $12$ terms, so

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

Multiplying $5$ by $\frac{12 \cdot 13}{2}$ would add only the $5k$ parts and forget the twelve $-2$s; writing $5 + 58$ mistakes $5$ for the first term, which is really $3$; and $12(3 + 58)$ drops the half, giving twice the sum.

Write each $n$th term and simplify.

$a_n = 7 + (n-1)4 = 4n + 3, \qquad b_n = 105 + (n-1)(-3) = 108 - 3n$

Now set them equal and solve.

$4n + 3 = 108 - 3n \;\Rightarrow\; 7n = 105 \;\Rightarrow\; n = 15$

Both sequences hit $63$ at their 15th term: $4(15) + 3 = 63$ and $108 - 3(15) = 63$.