Arithmetic Sequences: Additional Practice (Set 1)

First find $d$ over the $14 - 6 = 8$ steps between the known terms.

$d = \frac{52 - 20}{14 - 6} = 4$

Then back up five steps from $a_6$ to $a_1 = a_6 - 5d$.

$a_1 = 20 - 5\times 4 = 0$

The first term comes from $n = 1$, and the common difference is the coefficient of $n$.

$a_1 = 3(1) - 2 = 1, \quad d = 3$

Write $a_n = 45 - 4(n-1) = 49 - 4n$ and find where it drops below zero.

$a_{12} = 49 - 48 = 1, \quad a_{13} = 49 - 52 = -3$

The twelfth term is still positive, so the first negative term is $a_{13} = -3$.

First find $d$ over the $8 - 3 = 5$ steps between the known terms.

$d = \frac{29 - 14}{8 - 3} = 3$

Then step from $a_8$ up $17$ more places to $a_{25} = a_8 + 17d$.

$a_{25} = 29 + 17\times 3 = 80$

There are $10 - 4 = 6$ steps between the terms, and the value falls.

$d = \frac{4 - 22}{10 - 4} = \frac{-18}{6} = -3$

After month $1$ the total is $60 + 35 = 95$, and each later month adds $35$, so $a_1 = 95$ and $d = 35$.

$a_6 = 95 + 5\times 35 = 270$

First find $d$ over the $6 - 2 = 4$ steps between the known terms.

$d = \frac{17 - 5}{6 - 2} = 3$

Then step back one place from $a_2$ to $a_1 = a_2 - d = 5 - 3 = 2$.

Write $a_n = 2 + 4(n-1) = 4n - 2$ and require it above $80$.

$4n - 2 > 80 \Rightarrow n > 20.5$

So $n = 21$, and $a_{21} = 82$ is the first term past $80$ (with $a_{20} = 78$).

Back up six steps from the seventh term, $a_1 = a_7 - 6d$.

$a_1 = 22 - 6\times \tfrac{3}{2} = 22 - 9 = 13$

First find $d$ and $a_1$: over $9 - 4 = 5$ steps the value climbs $15$, so $d = 3$ and $a_1 = 19 - 3\times 3 = 10$.

$10 + 3(n-1) = 61 \Rightarrow 3(n-1) = 51 \Rightarrow n = 18$

The seat counts are arithmetic with $a_1 = 22$ and $d = 4$. Set the last row equal to $150$.

$22 + 4(n-1) = 150 \Rightarrow 4(n-1) = 128 \Rightarrow n = 33$

Between the 5th and 9th terms there are $4$ steps, and the value does not change.

$4d = 12 - 12 = 0 \Rightarrow d = 0$

The sequence is constant, which is arithmetic with a zero step.