Lesson 24 of 46 · Arithmetic

Sequences of numbers

Find any term of an arithmetic or geometric sequence, count and add terms without off-by-one errors, and handle recursive and repeating patterns.

About 20 minutes

In this lesson

  • Write the nth term of an arithmetic or a geometric sequence and use it to find any term.
  • Count the terms in a list and add an arithmetic series with the first-plus-last method.
  • Work out terms of a sequence defined recursively.
  • Use the cycle length of a repeating sequence to find far-off terms and sums.

The idea

Notation

A sequence is an ordered list of numbers. Its terms are written a1,a2,a3,…a_1, a_2, a_3, \ldots, where the small number is the position and ana_n is the term in position nn. A question may define a sequence by a formula for ana_n, such as an=2n2−1a_n = 2n^2 - 1 (so a3=17a_3 = 17), or by a rule that builds each term from the one before.

Arithmetic sequences

In an arithmetic sequence, each term is the previous one plus a fixed common difference dd: 7, 11, 15, 19, … has d=4d = 4. To reach ana_n from a1a_1 you take n−1n - 1 steps, not nn:

an=a1+(n−1)da_n = a_1 + (n - 1)d

The same idea links any two terms: am−ak=(m−k)da_m - a_k = (m - k)d. If you know two terms anywhere in the sequence, divide their difference by the number of steps between them to get dd, then step to whatever term you need.

Counting terms

How many numbers are in the list 14, 17, 20, …, 98? The gap from first to last is 98−14=8498 - 14 = 84, which is 843=28\frac{84}{3} = 28 steps, and 28 steps join 29 numbers:

number of terms=last−firstd+1\text{number of terms} = \frac{\text{last} - \text{first}}{d} + 1

The "+1" is the fence-post rule: a fence with 28 gaps has 29 posts. For a range of whole numbers, "from 30 to 75 inclusive" contains 75−30+1=4675 - 30 + 1 = 46 integers. For multiples, first find the first and last multiple inside the range, then count steps.

Adding an arithmetic sequence

Pair the first term with the last, the second with the second-to-last, and so on: every pair has the same sum. So

sum=number of terms×first+last2\text{sum} = \text{number of terms} \times \frac{\text{first} + \text{last}}{2}

that is, the count times the average of the first and last terms. The sum of 1+2+⋯+n1 + 2 + \cdots + n is n(n+1)2\frac{n(n+1)}{2}; the sum of the first nn odd numbers is n2n^2.

Geometric sequences

In a geometric sequence, each term is the previous one times a fixed common ratio rr: 3, 6, 12, 24, … has r=2r = 2, and 80, 20, 5, 1.25, … has r=14r = \frac{1}{4}. Again there are n−1n - 1 steps to the nnth term:

an=a1rn−1a_n = a_1 r^{n-1}

To tell the two kinds apart, check differences and quotients of neighbouring terms: constant differences mean arithmetic, constant quotients mean geometric. A negative ratio makes the signs alternate: with a1=3a_1 = 3 and r=−2r = -2 you get 3, −6-6, 12, −24-24, …, so terms in odd positions are positive and terms in even positions negative.

Recursive sequences

A recursive definition gives the first term (or first few) and a rule for the next term: a1=2a_1 = 2 and an=3an−1−1a_n = 3a_{n-1} - 1 for n≥2n \ge 2. For the first few terms no shortcut is needed; write the terms out carefully, keeping track of which position each one fills. Most errors here are index slips, such as starting the rule at a1a_1 or labelling the third term computed as a3a_3 when it is a4a_4.

Repeating sequences

Some sequences cycle: 3, −1-1, 4, 2, 3, −1-1, 4, 2, … repeats every 4 terms. To find a far-off term, divide its position by the cycle length and use the remainder, as with remainders and repeating decimals: remainder 1 means the first term of the cycle, and remainder 0 means the last. For a sum of the first nn terms, count the complete cycles, multiply by the sum of one cycle, and add the leftover terms.

Strategy

Write out the first few terms whenever a sequence is defined abstractly. It costs seconds and shows the pattern, the sign changes and the index convention at once. Before you answer, check one value you can verify directly, such as the term you were given.

Worked examples

Example 1: From two terms to any term

In an arithmetic sequence, a3=14a_3 = 14 and a10=42a_{10} = 42. Find a1a_1 and a25a_{25}.

  1. From position 3 to position 10 is 10−3=710 - 3 = 7 steps, and the terms differ by 42−14=2842 - 14 = 28, so d=287=4d = \frac{28}{7} = 4.

  2. Step back from a3a_3: a1=14−2⋅4=6a_1 = 14 - 2 \cdot 4 = 6.

  3. Step forward: a25=6+24⋅4=102a_{25} = 6 + 24 \cdot 4 = 102.

  4. Check against a given term: a10=6+9⋅4=42a_{10} = 6 + 9 \cdot 4 = 42. A second route to a25a_{25}: 42+15⋅4=10242 + 15 \cdot 4 = 102.

Answer

a1=6a_1 = 6 and a25=102a_{25} = 102. Using 25 steps instead of 24 gives 106.

Example 2: Sum of the multiples of 7 between 100 and 200

What is the sum of all the multiples of 7 between 100 and 200?

  1. The first multiple of 7 above 100 is 105=7⋅15105 = 7 \cdot 15; the last below 200 is 196=7⋅28196 = 7 \cdot 28.

  2. Count them: 196−1057+1=13+1=14\frac{196 - 105}{7} + 1 = 13 + 1 = 14 terms.

  3. Sum =14×105+1962=7×301=2107= 14 \times \frac{105 + 196}{2} = 7 \times 301 = 2107.

  4. Second route: the sum is 7(15+16+⋯+28)7(15 + 16 + \cdots + 28), and 15+⋯+28=14×15+282=30115 + \cdots + 28 = 14 \times \frac{15 + 28}{2} = 301, so the sum is 7⋅301=21077 \cdot 301 = 2107.

Answer

The sum is 2,107. Counting 13 terms instead of 14 gives 13×150.5=1956.513 \times 150.5 = 1956.5; a sum of whole numbers cannot end in .5, which shows the count was wrong.

Example 3: A recursive rule

A sequence has a1=2a_1 = 2 and an=3an−1−1a_n = 3a_{n-1} - 1 for n≥2n \ge 2. What is a5a_5?

  1. a2=3⋅2−1=5a_2 = 3 \cdot 2 - 1 = 5.

  2. a3=3⋅5−1=14a_3 = 3 \cdot 5 - 1 = 14.

  3. a4=3⋅14−1=41a_4 = 3 \cdot 14 - 1 = 41.

  4. a5=3⋅41−1=122a_5 = 3 \cdot 41 - 1 = 122.

  5. Check with a pattern you can test on every term so far: an=3n+12a_n = \frac{3^n + 1}{2} gives 2, 5, 14, 41 for n=1n = 1 to 4, and 243+12=122\frac{243 + 1}{2} = 122 for n=5n = 5.

Answer

a5=122a_5 = 122. Stopping one step early gives 41, the fourth term.

Example 4: Sum of a repeating sequence

The sequence 3, −1-1, 4, 2, 3, −1-1, 4, 2, … repeats the same four terms forever. What is the sum of its first 50 terms?

  1. One cycle has 4 terms with sum 3−1+4+2=83 - 1 + 4 + 2 = 8.

  2. 50=4×12+250 = 4 \times 12 + 2: twelve complete cycles and two leftover terms.

  3. The leftover terms are the first two of a cycle, 3 and −1-1.

  4. Sum =12×8+3+(−1)=96+2=98= 12 \times 8 + 3 + (-1) = 96 + 2 = 98.

Answer

The sum is 98. Dividing 50 by 4 and multiplying 12.5 by 8 gives 100, which wrongly assumes the half cycle adds half of 8.

Common traps

  • Taking n steps to the nth term

    From a1a_1 to ana_n there are n−1n - 1 steps, so an=a1+(n−1)da_n = a_1 + (n - 1)d and an=a1rn−1a_n = a_1 r^{n-1}. Check with n=1n = 1: the formula must give a1a_1 back.

  • The fence-post error when counting

    The integers from 30 to 75 inclusive number 75−30+1=4675 - 30 + 1 = 46, not 45. Dividing the gap by the step counts gaps, not terms; add 1.

  • Calling a geometric sequence arithmetic

    2, 6, 18, 54 has differences 4, 12, 36, which are not constant, but quotients of 3, which are. Test both before you choose a formula.

  • Splitting a cycle evenly

    When a sum stops part-way through a cycle, add the actual leftover terms. The first two terms of a cycle rarely add up to half of the cycle's sum.

Check yourself

Answer each question before you open its answer.

Question 1: The 20th term

Sequences

In an arithmetic sequence, the 4th term is 19 and the 9th term is 44. What is the 20th term?

  1. A.94
  2. B.99
  3. C.100
  4. D.104
  5. E.114
Show the answer

Answer: B. 99

Working

From the 4th to the 9th term is 5 steps covering 44−19=2544 - 19 = 25, so d=5d = 5. The first term is 19−3⋅5=419 - 3 \cdot 5 = 4, and the 20th is 4+19⋅5=994 + 19 \cdot 5 = 99. Check another way: from the 9th term, 44+11⋅5=9944 + 11 \cdot 5 = 99.

Why the other options are wrong

  • A. This takes 18 steps from the first term instead of 19.
  • C. This multiplies the position by dd, 20×520 \times 5, as if the sequence were 5n5n; that would need a first term of 5, but it is 4.
  • D. This takes 20 steps from the first term; the 20th term is only 19 steps away.
  • E. This starts from the 4th term, 19, and adds 19 steps of 5; from the 4th term to the 20th is 16 steps.

What carries over

Count the steps between positions: from aka_k to ama_m is m−km - k steps.

Question 2: An alternating geometric sequence

Quantitative Comparison

In a geometric sequence, the first term is 3 and the common ratio is −2-2.

Quantity A Quantity B
The 8th term The 7th term
  1. A.Quantity A is greater.
  2. B.Quantity B is greater.
  3. C.The two quantities are equal.
  4. D.The relationship cannot be determined from the information given.
Show the answer

Answer: B. Quantity B is greater.

Working

an=3(−2)n−1a_n = 3(-2)^{n-1}. The 7th term is 3(−2)6=3⋅64=1923(-2)^6 = 3 \cdot 64 = 192, and the 8th is 3(−2)7=3⋅(−128)=−3843(-2)^7 = 3 \cdot (-128) = -384. So 192>−384192 > -384. Check by listing: 3, −6-6, 12, −24-24, 48, −96-96, 192, −384-384.

Why the other options are wrong

  • A. This compares sizes only: the 8th term has the larger absolute value, 384, but it is negative.
  • C. This assumes the alternating signs leave neighbouring terms the same size; each term is twice the size of the one before.
  • D. The first term and the ratio fix every term, so both quantities are definite numbers.

What carries over

With a negative ratio, odd and even positions have opposite signs; check the sign before the size.

Question 3: Sum of a run of integers

Sequences

What is the sum of the integers from 41 to 80, inclusive?

  1. A.2,379
  2. B.2,400
  3. C.2,420
  4. D.3,240
  5. E.4,840
Show the answer

Answer: C. 2,420

Working

There are 80−41+1=4080 - 41 + 1 = 40 terms, with average 41+802=60.5\frac{41 + 80}{2} = 60.5, so the sum is 40×60.5=242040 \times 60.5 = 2420. Check: 1+⋯+80=80⋅812=32401 + \cdots + 80 = \frac{80 \cdot 81}{2} = 3240, 1+⋯+40=40⋅412=8201 + \cdots + 40 = \frac{40 \cdot 41}{2} = 820, and 3240−820=24203240 - 820 = 2420.

Why the other options are wrong

  • A. This subtracts the sum from 1 to 41, which also removes 41 itself; subtract the sum from 1 to 40.
  • B. This multiplies 40 terms by the average of 40 and 80; the first term is 41, so the average is 60.5.
  • D. This is the sum from 1 to 80; the integers from 1 to 40 were never subtracted.
  • E. This multiplies the count by first plus last, 40×12140 \times 121, without dividing by 2.

What carries over

The sum of an arithmetic run is the number of terms times the average of the first and last.

Key points

  • Arithmetic: an=a1+(n−1)da_n = a_1 + (n - 1)d. Geometric: an=a1rn−1a_n = a_1 r^{n-1}.

  • Number of terms =last−firstd+1= \frac{\text{last} - \text{first}}{d} + 1.

  • Sum of an arithmetic sequence = number of terms × average of first and last.

  • For a repeating sequence, divide the position by the cycle length and read the remainder.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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