Pattern Rules From Sequences - SSAT Middle Level: Quantitative

Multiply by $-1$ each term. Alternating signs are produced by negating each previous term, creating an oscillating pattern.

Add $4$ each term. The sequence forms an arithmetic progression with a common difference of $4$, increasing steadily by this constant each time.

Divide by $3$ each term. This geometric sequence decreases as each term is one-third of the previous, achieved by repeated division by $3$.

Multiply by $\frac{1}{3}$ each term. Each term is obtained by multiplying the prior one by $\frac{1}{3}$, forming a decreasing geometric sequence.

Add $5$ each term. An arithmetic sequence with a constant difference of $5$ added to each preceding term.

Add consecutive odd numbers: $+3, +5, +7, +9, \dots$. Consecutive odd integers beginning from $3$ are added to build each subsequent term.

Add $3, 4, 5, 6, \dots$ (increasing by $1$ each time). Differences start at $3$ and increase by $1$ each step, accumulating to form the sequence.

Subtract $3$ each term. An arithmetic sequence with a common difference of $-3$, resulting in a steady decrease.

Alternate $+2$ and $-2$. The sequence switches between $1$ and $3$ by alternately adding and subtracting $2$ from the previous term.

Alternate $0$ and $1$. The pattern oscillates between $0$ and $1$ in a repeating binary fashion for every pair of terms.

Multiply by $2$ each term. Doubling occurs between terms, establishing a geometric sequence with a constant ratio of $2$.

Multiply by $\frac{1}{2}$ each term. The sequence halves each time, creating a geometric progression with a common ratio of $\frac{1}{2}$.

Each term is the sum of the previous two terms. This resembles the Fibonacci sequence but starts at $2$ and $3$, with each subsequent term summing the prior two.

Multiply by $3$ each term. Tripling happens between terms, creating a geometric sequence with ratio $3$.

Each term is the sum of the previous two terms. This is the Fibonacci sequence starting with two $1$s, where terms sum the previous two.

$a_n = 5 \cdot 2^{n - 1}$. This expresses the $n$th term of a geometric sequence using the initial term and common ratio.

$a_n = 7 + 3(n - 1)$. The formula calculates the $n$th term of an arithmetic sequence using the first term and common difference.

Alternate $\times 2$ and $-2$. Operations cycle between multiplying by $2$ and subtracting $2$, generating the sequence's progression.

Alternate $2$ and $1$. The pattern repeatedly switches between $2$ and $1$ in an alternating manner for each term.

Add consecutive even numbers: $+4, +6, +8, +10, \dots$. Differences between consecutive terms are even numbers increasing by $2$ starting from $4$, building the sequence cumulatively.

Squares: $n^2$ for $n = 1, 2, 3, \dots$. Each term corresponds to the square of its position index, forming the sequence of perfect squares.

Cubes: $n^3$ for $n = 1, 2, 3, \dots$. Terms are cubes of consecutive positive integers, starting from $1^3$ and increasing accordingly.

Powers of $2$: $2^1, 2^2, 2^3, \dots$. Terms represent powers of $2$ where the exponent matches the term's position in the sequence.

Add $1, 2, 3, 4, 5, \dots$. The sequence is generated by adding consecutive positive integers starting from $1$ to the previous term.