Extend numeric patterns, find rules/terms of arithmetic sequences, and resolve cyclic/repeating patterns via remainders (e.g., the nth digit of a repeating decimal, or the term at position n in a repeating cycle).
Free, timed SHSAT practice sets on patterns & sequences— every question includes a full worked explanation, and there's no account required.
Loading questions…
First find the rule.
The most common trap is being off by one. There are only \(n - 1\) steps from the 1st term to the \(n\)th term. To find which term equals a value, solve \(a_1 + (n - 1)d =\) value and remember the \(+1\).
Solve \(-3 + (n-1)(1) = 10\): \(n - 1 = (10 - (-3)) \div 1 = 13\), so \(n = 14\). (Forgetting to add the \(1\) back gives \(13\). Dividing \(10\) by \(1\) and ignoring the first term gives \(10\). Adding the first term instead of subtracting it, \((10 + (-3)) \div 1 + 1\), gives \(8\).)
Common ratio \(r = 3\). The \(n\)th term is \(a_1 \cdot r^{n-1} = 2 \cdot 3^{3} = 54\). (Using \(r^{4}\) instead of \(r^{3}\) gives \(162\). Multiplying \(2 \cdot 3 \cdot 3\) instead of using the power gives \(18\). Treating it as an arithmetic sequence with a common difference of \(4\) gives \(14\).)
The pattern repeats every 4 items. \(40 \div 4 = 10\) with remainder \(0\), so the 40th item is the last item, "yellow". (Counting one place too far gives "red". Counting one place too few gives "green". Using the quotient \(10\) instead of the remainder gives "blue".)