nth term questions and answers

Worked solution

1. The common difference is 4, so the rule starts $4n$.
2. When $n = 1$, $4n = 4$, and the first term is 5, so add 1.
3. nth term $= 4n + 1$

Worked solution

1. The difference is $-3$, so the rule has $-3n$.
2. When $n = 1$, $-3n = -3$; to get 20 add 23.
3. nth term $= 23 - 3n$

Worked solution

1. $3(20) + 7 = 60 + 7 = 67$

Worked solution

1. $6n - 4 = 92$
2. $6n = 96$, so $n = 16$

Worked solution

1. $4n - 3 = 100$
2. $4n = 103$, so $n = 25.75$
3. A position must be a whole number, so 100 is not in the sequence. (The 25th term is 97 and the 26th is 101.)

Worked solution

1. $n = 1$: $1 + 4 = 5$
2. $n = 2$: $4 + 4 = 8$
3. $n = 3$: $9 + 4 = 13$

Worked solution

1. First differences: 3, 5, 7, 9. Second differences: 2, so the rule starts $n^2$ (half of 2 is 1).
2. Subtract $n^2$ (1, 4, 9, 16, 25): 3, 3, 3, 3, 3.
3. nth term $= n^2 + 3$

Worked solution

1. First differences: 5, 7, 9, 11. Second differences: 2, so the rule starts $n^2$.
2. Subtract $n^2$: $2 - 1 = 1$, $7 - 4 = 3$, $14 - 9 = 5$, $23 - 16 = 7$, $34 - 25 = 9$.
3. 1, 3, 5, 7, 9 has nth term $2n - 1$.
4. nth term $= n^2 + 2n - 1$

Worked solution

1. First differences: 7, 11, 15. Second differences: 4, so the rule starts $2n^2$.
2. Subtract $2n^2$ (2, 8, 18, 32): 1, 2, 3, 4.
3. That's $n$, so nth term $= 2n^2 + n$

Worked solution

1. $2n^2 - 3n - 104 = 0$
2. $(2n + 13)(n - 8) = 0$
3. $n = 8$ (the other solution, $-6.5$, isn't a position).
4. Check: $2(64) - 24 = 104$ ✓