Arithmetic Sequences: Practice

First find $d$ from the two known terms, which are $15 - 7 = 8$ steps apart.

$d = \frac{91 - 43}{15 - 7} = 6$

Then back up six steps from $a_7$ to reach $a_1 = a_7 - 6d$.

$a_1 = 43 - 6\times 6 = 7$

The common difference is the change from one term to the next, $a_{n+1} - a_n$.

$a_{n+1} - a_n = [4(n+1) + 1] - [4n + 1] = 4$

The coefficient of $n$ is the common difference.

Set the formula equal to $89$ and solve for $n$.

$4n + 1 = 89 \Rightarrow 4n = 88 \Rightarrow n = 22$

Write $a_n = 100 - 7(n-1) = 107 - 7n$ and require it to stay positive.

$107 - 7n > 0 \Rightarrow n < 15.28$

So the terms stay positive up to $n = 15$ (where $a_{15} = 2$), giving $15$ positive terms.

The value climbs by $20$ over $7 - 3 = 4$ equal steps.

$d = \frac{(x + 20) - x}{7 - 3} = \frac{20}{4} = 5$

First find $d$ over the $5 - 2 = 3$ steps between the known terms.

$d = \frac{17 - 8}{5 - 2} = 3$

Then step from $a_5$ up $6$ more places to $a_{11} = a_5 + 6d$.

$a_{11} = 17 + 6\times 3 = 35$

The seat counts are arithmetic with $a_1 = 18$ and $d = 2$. Set the count equal to $60$.

$18 + 2(n-1) = 60 \Rightarrow 2(n-1) = 42 \Rightarrow n = 22$

Back up three steps from the fourth term, $a_1 = a_4 - 3d$, and keep the sign of $d$.

$a_1 = 6 - 3\times(-3) = 6 + 9 = 15$

The third term is two steps past the first, so $a_3 = 3 + 2d$. Set this equal to $11$.

$3 + 2d = 11 \Rightarrow 2d = 8 \Rightarrow d = 4$

The terms follow $a_n = 4 + 5(n-1) = 5n - 1$, so a value $V$ is a term only when $\tfrac{V+1}{5}$ is a whole number.

$5n - 1 = 30 \Rightarrow n = 6.2$

That is not a whole number, so $30$ is not a term (the other three all are).

Write each $n$th term and set them equal: $5 + 2(n-1) = 3 + 2n$ and $50 - 3(n-1) = 53 - 3n$.

$3 + 2n = 53 - 3n \Rightarrow 5n = 50 \Rightarrow n = 10$

Both reach $23$ at the tenth term.

The hourly amounts are arithmetic with $a_1 = 500$ and $d = -30$. Require the amount below $200$.

$500 - 30(n-1) < 200 \Rightarrow n - 1 > 10$

So $n = 12$ is the first hour below $200$ (at hour $11$ it is exactly $200$).