Arithmetic Sequences

An arithmetic sequence has a constant difference between every pair of consecutive terms.

$9-5=4$, $13-9=4$, $17-13=4$, $21-17=4$. The difference is always 4, so this is arithmetic with $d=4$.

For 2, 4, 8, 16: differences are 2, 4, 8 — not constant. For 1, 4, 9, 16: differences are 3, 5, 7 — not constant. For 3, 6, 12, 24: differences are 3, 6, 12 — not constant.

$4-2=2$ but $8-4=4$. Differences are not equal, so this is geometric, not arithmetic.

These are perfect squares. $4-1=3$, $9-4=5$. Differences increase, so not arithmetic.

$6-3=3$ but $12-6=6$. The differences double each time, so this is geometric, not arithmetic.

5, 9, 13, 17, 21 is the only sequence with a constant difference ($d=4$), making it arithmetic.

The common difference is $d=a_{n+1}-a_n$, found by subtracting any term from the one that follows it.

$d=5-8=-3$. Verify: $2-5=-3$ and $-1-2=-3$. The common difference is confirmed.

$d=-3$. The sequence is decreasing because the common difference is negative.

If $d=3$, each term would increase: 8, 11, 14, … But the sequence is clearly decreasing, so $d=3$ is wrong.

If $d=-4$, the second term would be $8+(-4)=4$, not 5. So $d=-4$ is incorrect.

If $d=8$, the sequence would increase rapidly: 8, 16, 24, … This contradicts the given sequence.

$8+(-3)=5$ ✓, $5+(-3)=2$ ✓, $2+(-3)=-1$ ✓. The answer is $d=-3$.