ACT · Pre-Algebra / Number Properties
Find the common difference, the next term, a term that is several steps away, and the difference between the mean and the median.
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Common difference
An arithmetic sequence adds the same number each time. That number is the common difference \(d\).
\(4,\ 9,\ 14,\ 19,\ \ldots\)
\(d = 9 - 4 = 5\)
The nth term
\(a_{n} = a_{1} + (n-1)d\)
First term 3. Common difference 2. Find the 10th term.
\(a_{10} = 3 + (10-1)(2)\)
\(a_{10} = 3 + 18\)
\(a_{10} = 21\)
Skipping terms
The 3rd term is 9 and the 6th term is 21.
Three additions (jumps) are needed to get to the sixth spot.
\(9 + 3d = 21\)
\(d = 4\)
Use \(d = 4\) to fill in the sequence.
Mean and median
In an arithmetic sequence the terms are equally spaced, so the mean and the median are the same.
\(\text{mean} - \text{median} = 0\)
This is true for an odd number of terms and for an even number of terms.
Find the common difference, then add it to the last given term.
Find the next term. \(4,\ 9,\ 14,\ 19,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 9 - 4\)
\(d = 5\)
Add the common difference to the last given term.
\(a = 19 + 5\)
\(a = 24\)
Answer: \(24\)
Integer terms and an integer common difference.
Find the next term. \(2,\ 6,\ 10,\ 14,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 6 - 2\)
\(d = 4\)
Add the common difference to the last given term.
\(a = 14 + 4\)
\(a = 18\)
Answer: \(18\)
Find the next term. \(9,\ 6,\ 3,\ 0,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 6 - 9\)
\(d = -3\)
Add the common difference to the last given term.
\(a = 0 + -3\)
\(a = -3\)
Answer: \(-3\)
Find the next term. \(1,\ 8,\ 15,\ 22,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 8 - 1\)
\(d = 7\)
Add the common difference to the last given term.
\(a = 22 + 7\)
\(a = 29\)
Answer: \(29\)
Find the next term. \(-5,\ -1,\ 3,\ 7,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = -1 - -5\)
\(d = 4\)
Add the common difference to the last given term.
\(a = 7 + 4\)
\(a = 11\)
Answer: \(11\)
Find the next term. \(12,\ 7,\ 2,\ -3,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 7 - 12\)
\(d = -5\)
Add the common difference to the last given term.
\(a = -3 + -5\)
\(a = -8\)
Answer: \(-8\)
Fractional terms and a fractional common difference.
Find the next term. \(\frac{1}{2},\ 1,\ \frac{3}{2},\ 2,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 1 - \frac{1}{2}\)
\(d = \frac{1}{2}\)
Add the common difference to the last given term.
\(a = 2 + \frac{1}{2}\)
\(a = \frac{5}{2}\)
Answer: \(\frac{5}{2}\)
Find the next term. \(\frac{5}{3},\ \frac{4}{3},\ 1,\ \frac{2}{3},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = \frac{4}{3} - \frac{5}{3}\)
\(d = -\frac{1}{3}\)
Add the common difference to the last given term.
\(a = \frac{2}{3} + -\frac{1}{3}\)
\(a = \frac{1}{3}\)
Answer: \(\frac{1}{3}\)
Find the next term. \(\frac{1}{4},\ \frac{3}{4},\ \frac{5}{4},\ \frac{7}{4},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = \frac{3}{4} - \frac{1}{4}\)
\(d = \frac{1}{2}\)
Add the common difference to the last given term.
\(a = \frac{7}{4} + \frac{1}{2}\)
\(a = \frac{9}{4}\)
Answer: \(\frac{9}{4}\)
Find the next term. \(-\frac{1}{2},\ 0,\ \frac{1}{2},\ 1,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = 0 - -\frac{1}{2}\)
\(d = \frac{1}{2}\)
Add the common difference to the last given term.
\(a = 1 + \frac{1}{2}\)
\(a = \frac{3}{2}\)
Answer: \(\frac{3}{2}\)
Find the next term. \(\frac{7}{6},\ \frac{5}{6},\ \frac{1}{2},\ \frac{1}{6},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = \frac{5}{6} - \frac{7}{6}\)
\(d = -\frac{1}{3}\)
Add the common difference to the last given term.
\(a = \frac{1}{6} + -\frac{1}{3}\)
\(a = -\frac{1}{6}\)
Answer: \(-\frac{1}{6}\)
Fractions that do not share a denominator of 2 or 3 only.
Find the next term. \(\frac{2}{5},\ \frac{7}{10},\ 1,\ \frac{13}{10},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = \frac{7}{10} - \frac{2}{5}\)
\(d = \frac{3}{10}\)
Add the common difference to the last given term.
\(a = \frac{13}{10} + \frac{3}{10}\)
\(a = \frac{8}{5}\)
Answer: \(\frac{8}{5}\)
Find the next term. \(-\frac{3}{4},\ -\frac{1}{6},\ \frac{5}{12},\ 1,\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = -\frac{1}{6} - -\frac{3}{4}\)
\(d = \frac{7}{12}\)
Add the common difference to the last given term.
\(a = 1 + \frac{7}{12}\)
\(a = \frac{19}{12}\)
Answer: \(\frac{19}{12}\)
Find the next term. \(\frac{5}{8},\ \frac{1}{2},\ \frac{3}{8},\ \frac{1}{4},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = \frac{1}{2} - \frac{5}{8}\)
\(d = -\frac{1}{8}\)
Add the common difference to the last given term.
\(a = \frac{1}{4} + -\frac{1}{8}\)
\(a = \frac{1}{8}\)
Answer: \(\frac{1}{8}\)
Find the next term. \(\frac{1}{3},\ \frac{5}{12},\ \frac{1}{2},\ \frac{7}{12},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = \frac{5}{12} - \frac{1}{3}\)
\(d = \frac{1}{12}\)
Add the common difference to the last given term.
\(a = \frac{7}{12} + \frac{1}{12}\)
\(a = \frac{2}{3}\)
Answer: \(\frac{2}{3}\)
Find the next term. \(-\frac{5}{6},\ -\frac{1}{2},\ -\frac{1}{6},\ \frac{1}{6},\ \ldots\)
\(d = a_{2} - a_{1}\)
\(d = -\frac{1}{2} - -\frac{5}{6}\)
\(d = \frac{1}{3}\)
Add the common difference to the last given term.
\(a = \frac{1}{6} + \frac{1}{3}\)
\(a = \frac{1}{2}\)
Answer: \(\frac{1}{2}\)
Two terms are given, with gaps between them. Find \(d\), then step to the term you want.
The 3rd term is 5 and the 8th term is 55.
5 additions (jumps) are needed to get to the 8th spot.
\(5 + 5d = 55\)
\(5d = 50\)
\(d = 10\)
Use \(d = 10\) to fill in the sequence.
The 9th term is \(65\).
Answer: \(65\)
Integer terms. The common difference is a whole number. Find the term right after the later given term.
The 2nd term is 7 and the 6th term is 23.
4 additions (jumps) are needed to get to the 6th spot.
\(7 + 4d = 23\)
\(4d = 16\)
\(d = 4\)
Use \(d = 4\) to fill in the sequence.
The 7th term is \(27\).
Answer: \(27\)
The 1st term is 4 and the 5th term is 16.
4 additions (jumps) are needed to get to the 5th spot.
\(4 + 4d = 16\)
\(4d = 12\)
\(d = 3\)
Use \(d = 3\) to fill in the sequence.
The 6th term is \(19\).
Answer: \(19\)
The 4th term is 10 and the 9th term is 30.
5 additions (jumps) are needed to get to the 9th spot.
\(10 + 5d = 30\)
\(5d = 20\)
\(d = 4\)
Use \(d = 4\) to fill in the sequence.
The 10th term is \(34\).
Answer: \(34\)
The 3rd term is −2 and the 7th term is 14.
4 additions (jumps) are needed to get to the 7th spot.
\(-2 + 4d = 14\)
\(4d = 16\)
\(d = 4\)
Use \(d = 4\) to fill in the sequence.
The 8th term is \(18\).
Answer: \(18\)
The 5th term is 20 and the 11th term is 2.
6 additions (jumps) are needed to get to the 11th spot.
\(20 + 6d = 2\)
\(6d = -18\)
\(d = -3\)
Use \(d = -3\) to fill in the sequence.
The 12th term is \(-1\).
Answer: \(-1\)
Fractional terms and a fractional common difference. Find the next term after the later given term.
The 2nd term is 12 and the 6th term is 52.
4 additions (jumps) are needed to get to the 6th spot.
\(\frac{1}{2} + 4d = \frac{5}{2}\)
\(4d = 2\)
\(d = \frac{1}{2}\)
Use \(d = \frac{1}{2}\) to fill in the sequence.
The 7th term is \(3\).
Answer: \(3\)
The 1st term is 34 and the 5th term is 114.
4 additions (jumps) are needed to get to the 5th spot.
\(\frac{3}{4} + 4d = \frac{11}{4}\)
\(4d = 2\)
\(d = \frac{1}{2}\)
Use \(d = \frac{1}{2}\) to fill in the sequence.
The 6th term is \(\frac{13}{4}\).
Answer: \(\frac{13}{4}\)
The 3rd term is 23 and the 6th term is 116.
3 additions (jumps) are needed to get to the 6th spot.
\(\frac{2}{3} + 3d = \frac{11}{6}\)
\(3d = \frac{7}{6}\)
\(d = \frac{7}{18}\)
Use \(d = \frac{7}{18}\) to fill in the sequence.
The 7th term is \(\frac{20}{9}\).
Answer: \(\frac{20}{9}\)
The 4th term is −12 and the 8th term is 52.
4 additions (jumps) are needed to get to the 8th spot.
\(-\frac{1}{2} + 4d = \frac{5}{2}\)
\(4d = 3\)
\(d = \frac{3}{4}\)
Use \(d = \frac{3}{4}\) to fill in the sequence.
The 9th term is \(\frac{13}{4}\).
Answer: \(\frac{13}{4}\)
The 2nd term is 56 and the 6th term is −16.
4 additions (jumps) are needed to get to the 6th spot.
\(\frac{5}{6} + 4d = -\frac{1}{6}\)
\(4d = -1\)
\(d = -\frac{1}{4}\)
Use \(d = -\frac{1}{4}\) to fill in the sequence.
The 7th term is \(-\frac{5}{12}\).
Answer: \(-\frac{5}{12}\)
Fractional terms and a fractional common difference. The asked term is not the very next one.
The 2nd term is 14 and the 6th term is 134.
4 additions (jumps) are needed to get to the 6th spot.
\(\frac{1}{4} + 4d = \frac{13}{4}\)
\(4d = 3\)
\(d = \frac{3}{4}\)
Use \(d = \frac{3}{4}\) to fill in the sequence.
The 9th term is \(\frac{11}{2}\).
Answer: \(\frac{11}{2}\)
The 3rd term is 56 and the 8th term is −56.
5 additions (jumps) are needed to get to the 8th spot.
\(\frac{5}{6} + 5d = -\frac{5}{6}\)
\(5d = -\frac{5}{3}\)
\(d = -\frac{1}{3}\)
Use \(d = -\frac{1}{3}\) to fill in the sequence.
The 10th term is \(-\frac{3}{2}\).
Answer: \(-\frac{3}{2}\)
The 1st term is 23 and the 7th term is −43.
6 additions (jumps) are needed to get to the 7th spot.
\(\frac{2}{3} + 6d = -\frac{4}{3}\)
\(6d = -2\)
\(d = -\frac{1}{3}\)
Use \(d = -\frac{1}{3}\) to fill in the sequence.
The 4th term is \(-\frac{1}{3}\).
Answer: \(-\frac{1}{3}\)
The 4th term is 72 and the 10th term is 12.
6 additions (jumps) are needed to get to the 10th spot.
\(\frac{7}{2} + 6d = \frac{1}{2}\)
\(6d = -3\)
\(d = -\frac{1}{2}\)
Use \(d = -\frac{1}{2}\) to fill in the sequence.
The 12th term is \(-\frac{1}{2}\).
Answer: \(-\frac{1}{2}\)
The 2nd term is −35 and the 6th term is 95.
4 additions (jumps) are needed to get to the 6th spot.
\(-\frac{3}{5} + 4d = \frac{9}{5}\)
\(4d = \frac{12}{5}\)
\(d = \frac{3}{5}\)
Use \(d = \frac{3}{5}\) to fill in the sequence.
The 9th term is \(\frac{18}{5}\).
Answer: \(\frac{18}{5}\)
\(a_{n} = a_{1} + (n-1)d\). Substitute, then simplify.
\(a_{1} = 5,\ d = 3\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{8} = 5 + (8-1)(3)\)
\(a_{8} = 5 + (7)(3)\)
\(a_{8} = 26\)
Answer: \(26\)
Integer first term and integer common difference.
\(a_{1} = 3,\ d = 4\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{8} = 3 + (8-1)(4)\)
\(a_{8} = 3 + (7)(4)\)
\(a_{8} = 31\)
Answer: \(31\)
\(a_{1} = 10,\ d = -2\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{9} = 10 + (9-1)(-2)\)
\(a_{9} = 10 + (8)(-2)\)
\(a_{9} = -6\)
Answer: \(-6\)
\(a_{1} = -1,\ d = 5\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{6} = -1 + (6-1)(5)\)
\(a_{6} = -1 + (5)(5)\)
\(a_{6} = 24\)
Answer: \(24\)
\(a_{1} = 7,\ d = 7\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{5} = 7 + (5-1)(7)\)
\(a_{5} = 7 + (4)(7)\)
\(a_{5} = 35\)
Answer: \(35\)
\(a_{1} = 0,\ d = 3\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{12} = 0 + (12-1)(3)\)
\(a_{12} = 0 + (11)(3)\)
\(a_{12} = 33\)
Answer: \(33\)
Fractional first term and fractional common difference.
\(a_{1} = \frac{1}{2},\ d = \frac{1}{2}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{7} = \frac{1}{2} + (7-1)(\frac{1}{2})\)
\(a_{7} = \frac{1}{2} + (6)(\frac{1}{2})\)
\(a_{7} = \frac{7}{2}\)
Answer: \(\frac{7}{2}\)
\(a_{1} = \frac{2}{3},\ d = -\frac{1}{3}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{8} = \frac{2}{3} + (8-1)(-\frac{1}{3})\)
\(a_{8} = \frac{2}{3} + (7)(-\frac{1}{3})\)
\(a_{8} = -\frac{5}{3}\)
Answer: \(-\frac{5}{3}\)
\(a_{1} = -\frac{1}{4},\ d = \frac{1}{4}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{9} = -\frac{1}{4} + (9-1)(\frac{1}{4})\)
\(a_{9} = -\frac{1}{4} + (8)(\frac{1}{4})\)
\(a_{9} = \frac{7}{4}\)
Answer: \(\frac{7}{4}\)
\(a_{1} = \frac{5}{6},\ d = \frac{1}{3}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{6} = \frac{5}{6} + (6-1)(\frac{1}{3})\)
\(a_{6} = \frac{5}{6} + (5)(\frac{1}{3})\)
\(a_{6} = \frac{5}{2}\)
Answer: \(\frac{5}{2}\)
\(a_{1} = \frac{3}{5},\ d = -\frac{1}{5}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{11} = \frac{3}{5} + (11-1)(-\frac{1}{5})\)
\(a_{11} = \frac{3}{5} + (10)(-\frac{1}{5})\)
\(a_{11} = -\frac{7}{5}\)
Answer: \(-\frac{7}{5}\)
Fractions, and some problems ask for the first term or for \(d\).
\(a_{1} = -\frac{2}{5},\ d = \frac{3}{10}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{9} = -\frac{2}{5} + (9-1)(\frac{3}{10})\)
\(a_{9} = -\frac{2}{5} + (8)(\frac{3}{10})\)
\(a_{9} = 2\)
Answer: \(2\)
\(a_{1} = \frac{7}{12},\ d = -\frac{1}{6}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{10} = \frac{7}{12} + (10-1)(-\frac{1}{6})\)
\(a_{10} = \frac{7}{12} + (9)(-\frac{1}{6})\)
\(a_{10} = -\frac{11}{12}\)
Answer: \(-\frac{11}{12}\)
\(a_{1} = -\frac{5}{8},\ d = \frac{1}{4}\)
\(a_{n} = a_{1} + (n-1)d\)
\(a_{8} = -\frac{5}{8} + (8-1)(\frac{1}{4})\)
\(a_{8} = -\frac{5}{8} + (7)(\frac{1}{4})\)
\(a_{8} = \frac{9}{8}\)
Answer: \(\frac{9}{8}\)
\(a_{5} = \frac{11}{4},\ d = \frac{1}{2}\)
\(a_{n} = a_{1} + (n-1)d\)
\(\frac{11}{4} = a_{1} + (5-1)(\frac{1}{2})\)
\(\frac{11}{4} = a_{1} + 2\)
\(a_{1} = a_{n} - (n-1)d\)
\(a_{1} = \frac{11}{4} - 2\)
\(a_{1} = \frac{3}{4}\)
Answer: \(\frac{3}{4}\)
\(a_{1} = \frac{1}{3},\ a_{7} = \frac{5}{3}\)
\(d = \frac{a_{n} - a_{1}}{n-1}\)
\(d = \frac{\frac{5}{3} - \frac{1}{3}}{7-1}\)
\(d = \frac{\frac{4}{3}}{6}\)
\(d = \frac{2}{9}\)
Answer: \(\frac{2}{9}\)
In an arithmetic sequence the mean and the median are equal, so their difference is 0.
The sequence has 73 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
73 is odd, so the median is the middle term (term 37).
In an arithmetic sequence, the middle term is also \(\frac{\text{first} + \text{last}}{2}\).
\(\text{median} = \text{mean}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
An odd number of terms. You do not need the actual values.
The sequence has 15 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
15 is odd, so the median is the middle term (term 8).
In an arithmetic sequence, the middle term is also \(\frac{\text{first} + \text{last}}{2}\).
\(\text{median} = \text{mean}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
The sequence has 101 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
101 is odd, so the median is the middle term (term 51).
In an arithmetic sequence, the middle term is also \(\frac{\text{first} + \text{last}}{2}\).
\(\text{median} = \text{mean}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
An even number of terms, or extra numbers that you do not need.
The sequence has 80 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
80 is even, so the median is the average of terms 40 and 41.
Those two middle terms are equally far from the ends, so their average is still \(\frac{\text{first} + \text{last}}{2}\).
\(\text{median} = \text{mean}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
The sequence has 200 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
\(\text{mean} = \frac{3 + 399}{2}\)
200 is even, so the median is the average of terms 100 and 101.
Those two middle terms are equally far from the ends, so their average is still \(\frac{\text{first} + \text{last}}{2}\).
\(\text{median} = \text{mean}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
A listed sequence, or fractions in the given terms. The difference is still 0.
The sequence is \(\frac{1}{2},\ 1,\ \frac{3}{2},\ 2,\ \frac{5}{2},\ 3\).
The sequence has 6 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
\(\text{mean} = \frac{\frac{1}{2} + 3}{2}\)
\(\text{mean} = \frac{7}{4}\)
6 is even, so the median is the average of terms 3 and 4.
\(\text{median} = \frac{\frac{3}{2} + 2}{2}\)
\(\text{median} = \frac{7}{4}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
The sequence has 73 terms.
An arithmetic sequence is equally spaced, so it is symmetric.
\(\text{mean} = \frac{\text{first} + \text{last}}{2}\)
\(\text{mean} = \frac{-5 + 43}{2}\)
73 is odd, so the median is the middle term (term 37).
In an arithmetic sequence, the middle term is also \(\frac{\text{first} + \text{last}}{2}\).
\(\text{median} = \text{mean}\)
\(\text{mean} - \text{median} = 0\)
Answer: \(0\)
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