Arithmetic Sequences: Practice

Here $a_1 = 5$ and $d = 3$. The 20th term uses $n - 1 = 19$ steps.

$a_{20} = 5 + 19\times 3 = 62$

Between the 4th and 9th terms there are $9 - 4 = 5$ equal steps. Divide the change in value by the number of steps.

$d = \frac{42 - 17}{9 - 4} = \frac{25}{5} = 5$

The 12th term is $11$ steps from the first, so $46 = 2 + 11d$. Subtract $2$ and divide by $11$.

$11d = 44 \Rightarrow d = 4$

Here $a_1 = 5$ and $d = 3$, so set $5 + 3(n-1) = 50$ and solve for $n$.

$3(n-1) = 45 \Rightarrow n - 1 = 15 \Rightarrow n = 16$

There are $5 - 2 = 3$ steps between these terms. Divide the change in value by the number of steps.

$d = \frac{21 - 9}{5 - 2} = \frac{12}{3} = 4$

Here $a_1 = 3$ and $d = 4$, so set $3 + 4(n-1) = 85$ and solve for $n$.

$4(n-1) = 82 \Rightarrow n - 1 = 20.5$

Because $n$ is not a whole number, $85$ is not a term of this sequence.

The cost by gigabyte is arithmetic with $a_1 = 20$ and $d = 6$. For $8$ gigabytes use $7$ steps.

$a_8 = 20 + 7\times 6 = 62$

The first term is nine steps below the tenth, so $a_1 = a_{10} - 9d$.

$a_1 = 50 - 9\times 4 = 14$

Take $6$ steps of size $5$ from the first term.

$a_7 = -8 + 6\times 5 = 22$

Subtract a term from the one that follows it. The list decreases, so $d$ is negative.

$d = 93 - 100 = -7$

The weekly distances are arithmetic with $a_1 = 8$ and $d = 3$. Set $8 + 3(n-1) = 32$.

$3(n-1) = 24 \Rightarrow n - 1 = 8 \Rightarrow n = 9$

The first term is three steps below the fourth, so $a_1 = a_4 - 3d$.

$a_1 = 11 - 3\times 3 = 2$