A sequence of numbers starts with 1, 2, and 3. The fourth number of the sequence is the sum of the previous three numbers in the sequence: 1+2+3=6. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?
The sequence is defined by a clear repeating rule: each term is the sum of the previous three. Tool #5 (Look for a Pattern) fits exactly — we follow the rule that generates the pattern, term by term. Tool #2 (Make a Systematic List) keeps the work organized: write the terms in a single row, in order, so the "previous three" needed at each step are always the last three numbers on the list. With only $a_4$ through $a_8$ to find, this beats setting up algebra (tool #13).
Execute — Answer: D
#2 Make a Systematic List 4.OA.C.5Step 1
Write the three starting terms in order.
This is the seed of the list.
$$a_1, a_2, a_3 = 1, 2, 3$$
💡 Putting the known values on the list first means the "previous three" we need next is just the last three numbers visible.
#5 Look for a Pattern 3.NBT.A.2Step 2
Apply the rule to get $a_4$ by summing the last three terms on the list.
$$a_4 = 1 + 2 + 3 = 6$$
💡 The problem hands us this one as the worked example, so it doubles as a check that we read the rule correctly.
#5 Look for a Pattern 3.NBT.A.2Step 3
Now the list is $1, 2, 3, 6$.
Sum the last three to get $a_5$.
$$a_5 = 2 + 3 + 6 = 11$$
💡 Slide the "window of three" one step to the right and add.
#5 Look for a Pattern 3.NBT.A.2Step 4
List: $1, 2, 3, 6, 11$.
Sum the last three for $a_6$.
$$a_6 = 3 + 6 + 11 = 20$$
💡 Same window-slide move — the rule never changes, only the three numbers we add.
#5 Look for a Pattern 3.NBT.A.2Step 5
List: $1, 2, 3, 6, 11, 20$.
Sum the last three for $a_7$.
$$a_7 = 6 + 11 + 20 = 37$$
💡 By now the terms are growing fast — sanity-check each sum because a single addition error spreads forward.
#5 Look for a Pattern 3.NBT.A.2Step 6
List: $1, 2, 3, 6, 11, 20, 37$.
Sum the last three for $a_8$, which is the answer.
💡 One final window-slide gives the target term. The completed list is $1, 2, 3, 6, 11, 20, 37, 68$.
[1]
#2 4.OA.C.5Write the three starting terms in order. This is the seed of the list.
[2]
#5 3.NBT.A.2Apply the rule to get $a_4$ by summing the last three terms on the list.
[3]
#5 3.NBT.A.2Now the list is $1, 2, 3, 6$. Sum the last three to get $a_5$.
[4]
#5 3.NBT.A.2List: $1, 2, 3, 6, 11$. Sum the last three for $a_6$.
[5]
#5 3.NBT.A.2List: $1, 2, 3, 6, 11, 20$. Sum the last three for $a_7$.
[6]
#5 3.NBT.A.2List: $1, 2, 3, 6, 11, 20, 37$. Sum the last three for $a_8$, which is the answe
Review
Reasonableness: The terms grow but not crazily: each new term is roughly double the previous one (because it adds two smaller terms to the largest). From $a_7 = 37$, doubling gives about $74$ — and $68$ is right under that, which fits. Among the choices, $11, 20, 37$ are earlier terms in the sequence (classic distractors) and $99$ is too large to be a sum of $11 + 20 + 37$. Only $68$ fits, confirming (D).
Alternative: Tool #3 (Eliminate Possibilities) on the choices works too. The eighth term must equal the sum of the three terms before it, so it must be at least $a_5 + a_6 + a_7 \ge 11 + 20 + 37 = 68$ and exactly equal that. Choices $11, 20, 37$ are themselves earlier terms — too small — and $99$ exceeds the only valid sum. Only (D) $68$ survives.
CCSS standards used (min grade 4)
4.OA.C.5 Generate a number or shape pattern that follows a given rule (Reading the recurrence rule $a_n = a_{n-1} + a_{n-2} + a_{n-3}$ and using it to extend the pattern term by term from $a_3$ to $a_8$.)
3.NBT.A.2 Fluently add and subtract within 1000 (Computing each new term as a sum of three earlier terms: $1+2+3$, $2+3+6$, $3+6+11$, $6+11+20$, and $11+20+37$.)
⭐ This AMC 8 problem only needs a Grade 4 skill — follow a given pattern rule — plus plain addition you've known since Grade 3!
⭐ This AMC 8 problem only needs a Grade 4 skill — follow a given pattern rule — plus plain addition you've known since Grade 3!
