Arithmetic Sequence Problems

Arithmetic sequence problems ask you to find unknown terms, the common difference, the number of terms or sums, usually by writing one or two equations from a_n = a₁ + (n − 1)d and S_n = n(a₁ + a_n)/2. In brief: if a₄ = 14 and a₉ = 34, the difference is (34 − 14)/5 = 4, and a₁ = 2. The sum of the first 20 terms is 20(2 + 78)/2 = 800. Word problems are solved by identifying the first term, the difference and what is asked.

The formulas are simple, but problems come in many disguises: seats in a theatre, savings, stacks of logs, distances covered by a runner and numbers in sets. This guide gives a problem-solving routine and works through a wide range of typical problems, followed by a table that connects problem types with the formula to use.

How do you approach an arithmetic sequence problem?

1. Identify what is given: terms, sums, number of terms, difference.
2. Choose the unknowns, typically a₁ and d.
3. Write equations: a_n = a₁ + (n − 1)d, S_n = n[2a₁ + (n − 1)d]/2.
4. Solve the system.
5. Check with the original data.

Problems about terms and differences

Problem 1

In an arithmetic sequence a₂ + a₅ = 22 and a₃ · a₄ = 117. Find the sequence. From the first condition, (a₁ + d) + (a₁ + 4d) = 22, so a₁ = (22 − 5d)/2. Then a₃ = a₁ + 2d = (22 − d)/2 and a₄ = a₁ + 3d = (22 + d)/2. Their product is (484 − d²)/4 = 117, so d² = 16 and d = ±4. For d = 4: a₁ = 1 and the sequence is 1, 5, 9, 13, 17, … (check: 5 + 17 = 22 and 9 · 13 = 117). For d = −4: a₁ = 21 and the sequence is 21, 17, 13, 9, 5, … (check: 17 + 5 = 22 and 13 · 9 = 117). Both sequences satisfy the conditions.

Problem 2

The 7th term is 30 and the 12th term is 55. Find the first term, the difference and the 30th term. 5d = 25, so d = 5. a₁ = 30 − 6 · 5 = 0. a₃₀ = 0 + 29 · 5 = 145.

Problem 3

How many multiples of 7 lie between 100 and 500? The first multiple above 100 is 105 and the last below 500 is 497. Then 105 + 7(n − 1) = 497 gives n − 1 = 56, so n = 57.

Problems about sums

Problem 4

The sum of the first n terms is S_n = 3n² + 2n. Find a_n. a₁ = S₁ = 5. For n ≥ 2: a_n = S_n − S_n−1 = 3n² + 2n − 3(n − 1)² − 2(n − 1) = 6n − 1. For n = 1 the formula gives 5 ✓. So a_n = 6n − 1, an arithmetic sequence with d = 6.

Problem 5

The sum of the first 10 terms is 210 and the sum of the first 20 terms is 820. Find a₁ and d. S₁₀: 5(2a₁ + 9d) = 210, so 2a₁ + 9d = 42. S₂₀: 10(2a₁ + 19d) = 820, so 2a₁ + 19d = 82. Subtract: 10d = 40, d = 4. Then 2a₁ = 42 − 36 = 6, a₁ = 3.

Problem 6

Find the sum of all two-digit numbers divisible by 4. They are 12, 16, …, 96: n = (96 − 12)/4 + 1 = 22 terms. Sum = 22 · (12 + 96)/2 = 1,188.

Problem 7: the largest sum

For a_n = 50 − 4n, find the n for which S_n is largest. The terms are positive while 50 − 4n > 0, that is, for n ≤ 12 (a₁₂ = 2, a₁₃ = −2). Adding a positive term increases the sum, and adding a negative term decreases it. So S₁₂ = 12(46 + 2)/2 = 288 is the largest value. Check: S₁₁ = 11(46 + 6)/2 = 286 and S₁₃ = 288 − 2 = 286. The formula S_n = 46n − 2n² has its vertex at n = 11.5, between 11 and 12, and the integer n = 12 gives the larger value.

Word problems

Problem 8: stacked logs

A stack has 30 logs in the bottom row, and each higher row has one log fewer, the top row has 12 logs. The number of rows is 30 − 12 + 1 = 19, and the total is 19 · (30 + 12)/2 = 399 logs.

Problem 9: a runner

A runner trains for 20 days, running 3 km on day 1 and increasing by 0.4 km each day. On day 20 the distance is 3 + 19 · 0.4 = 10.6 km. The total is 20 · (3 + 10.6)/2 = 136 km.

Problem 10: a falling stone

A stone falls 4.9 m in the first second, 14.7 m in the second, 24.5 m in the third, and so on: the distance in each second increases by 9.8 m. In the sixth second it falls 4.9 + 5 · 9.8 = 53.9 m. The total distance after 6 seconds is 6 · (4.9 + 53.9)/2 = 176.4 m, which agrees with the formula 4.9t² = 4.9 · 36 = 176.4 ✓.

Problem 11: a loan

A loan of 6,000 dollars is repaid in 12 monthly installments, each 50 dollars lower than the previous one. Find the first installment. With the first installment x, the last one is x − 550 and the total is 12(2x − 550)/2 = 6000, so 2x − 550 = 1000 and x = 775. The installments go from 775 down to 225 dollars. Check: 12 · (775 + 225)/2 = 6,000 ✓.

Problems about divisibility and number sets

Problem 12

Find the sum of all numbers between 1 and 200 that leave the remainder 3 when divided by 5. They are 3, 8, 13, …, 198: n = (198 − 3)/5 + 1 = 40, sum = 40 · 201/2 = 4,020.

Problem 13

Show that the sum of any 5 consecutive integers is divisible by 5. They are m − 2, …, m + 2 with sum 5m.

Choosing the formula

Where do such problems come from?

Problems about arithmetic progressions appear in the Rhind papyrus (about 1650 BCE), in Babylonian tablets, in Chinese texts such as the Nine Chapters and in the Indian works of Aryabhata and Brahmagupta. They often involve distributing goods or computing wages over days.

Where are they used today?

Arithmetic sequences describe steady increases in salary or rent, linear schedules, evenly spaced data and the cost of tiered services. Spreadsheets generate them with the fill handle. Algorithms use the sum formula for counting steps in nested loops: a loop that performs 1 + 2 + … + n operations has n(n + 1)/2 steps.

Which mistakes are common?

Forgetting to check the answer

Substitute the answer back into the original conditions, as done in Problem 1, where both values of d have to be tested.

Counting the number of terms incorrectly

The number of terms between the first and last term is (last − first)/d + 1, one more than the number of steps.

Using a_n = S_n − S_n−1 for n = 1

For n = 1, use a₁ = S₁.

Ignoring integer constraints

In word problems, n must be a positive integer. Reject fractional solutions.