Arithmetic Sequences and Common Difference
d.For example, in the sequence 3, 7, 11, 15, 19, …, each term is 4 more than the last, so d = 4. In the sequence 20, 17, 14, 11, …, each term decreases by 3, so d = −3.
d = a₂ − a₁ = a₃ − a₂ = aₙ₊₁ − aₙ. If the differences between consecutive terms are not constant, the sequence is not arithmetic.Why arithmetic sequences matter
Arithmetic sequences model situations with steady, constant change — for example, saving the same amount of money each month, or a car accelerating by the same increment every second.
The terms grow linearly, which means if you plot the terms against their position number, the points lie on a straight line with slope d. This linear behavior makes arithmetic sequences the simplest and most predictable type of sequence in algebra.
Finding the nth Term (Arithmetic)
aₙ = a₁ + (n − 1)d, where a₁ is the first term, d is the common difference, and n is the position number. This formula lets you jump directly to any term without listing all the terms before it.Find the 50th term of 3, 7, 11, 15, ….
Here a₁ = 3 and d = 4.
So a₅₀ = 3 + (50 − 1)(4) = 3 + 196 = 199.
a₅₀ = 199The 5th term of an arithmetic sequence is 22 and the 12th term is 57. Find a₁ and d.
Using the formula: a₅ = a₁ + 4d = 22 and a₁₂ = a₁ + 11d = 57.
Subtract the first equation from the second: 7d = 35, so d = 5.
Then a₁ = 22 − 4(5) = 2.
The sequence is 2, 7, 12, 17, 22, …Is 301 a term in the sequence 5, 8, 11, …?
Set aₙ = 301: 301 = 5 + (n − 1)(3) = 5 + 3n − 3 = 3n + 2.
So n = 299/3 ≈ 99.67. Since n is not a whole number, 301 is not a term in this sequence.
301 is not a termThis technique of setting aₙ equal to a target and solving for n is useful for determining whether a specific value appears in a sequence.
Arithmetic Series Formula
n terms is Sₙ = n/2 · (a₁ + aₙ), or equivalently, Sₙ = n/2 · (2a₁ + (n − 1)d). Both formulas give the same result — use whichever is more convenient based on the information you have.The intuition behind the formula is elegant: if you pair up the first and last terms, the second and second-to-last terms, and so on, each pair has the same sum a₁ + aₙ, and there are n/2 such pairs.
Find the sum 2 + 5 + 8 + … + 302.
First, find how many terms: 302 = 2 + (n − 1)(3), so n = 101.
Then S₁₀₁ = 101/2 · (2 + 302) = 101/2 · 304 = 15352.
S₁₀₁ = 15352Find the sum of the first 200 positive integers.
Here a₁ = 1, a₂₀₀ = 200: S₂₀₀ = 200/2 · (1 + 200) = 100 · 201 = 20100.
This is the same trick famously attributed to young Gauss.
S₂₀₀ = 20100An arithmetic series has a₁ = 10, d = −2, and n = 15. Find the sum.
S₁₅ = 15/2 · (2(10) + 14(−2)) = 15/2 · (20 − 28) = 15/2 · (−8) = −60.
S₁₅ = −60A negative sum is perfectly valid — it means the later negative terms outweigh the earlier positive ones.
Geometric Sequences and Common Ratio
r.For example, in 2, 6, 18, 54, …, each term is multiplied by 3, so r = 3. In 100, 50, 25, 12.5, …, each term is multiplied by 1/2, so r = 1/2. In 4, −12, 36, −108, …, the ratio is r = −3, and the signs alternate.
r = a₂/a₁ = a₃/a₂ = aₙ₊₁/aₙ. If the ratios are not constant, the sequence is not geometric.Why geometric sequences matter
Geometric sequences model multiplicative growth and decay — compound interest, population growth, radioactive decay, and the depreciation of assets all follow geometric patterns. Unlike arithmetic sequences (which grow linearly), geometric sequences grow exponentially.
| Case | Condition | Behavior | Why |
|---|---|---|---|
| 1 | |r| > 1 | Grows rapidly toward infinity | Each term multiplies the previous by a factor larger than 1. |
| 2 | |r| < 1 | Shrinks toward zero | Each term is a fraction of the previous one. |
| 3 | r is negative | Alternates in sign | The terms bounce between positive and negative values. |
Finding the nth Term (Geometric)
aₙ = a₁ · r^(n−1), where a₁ is the first term, r is the common ratio, and n is the position number.Find the 8th term of 3, 6, 12, 24, ….
Here a₁ = 3 and r = 2.
So a₈ = 3 · 2⁷ = 3 · 128 = 384.
a₈ = 384Find the 6th term of 1000, 500, 250, ….
Here a₁ = 1000 and r = 1/2.
So a₆ = 1000 · (1/2)⁵ = 1000 · (1/32) = 31.25.
a₆ = 31.25The 3rd term of a geometric sequence is 36 and the 6th term is 972. Find a₁ and r.
Using the formula: a₃ = a₁ · r² = 36 and a₆ = a₁ · r⁵ = 972.
Divide the second by the first: r³ = 972/36 = 27, so r = 3.
Then a₁ = 36/r² = 36/9 = 4.
The sequence is 4, 12, 36, 108, 324, 972, …A useful comparison: in arithmetic sequences, knowing any two terms lets you find a₁ and d by subtraction. In geometric sequences, knowing any two terms lets you find a₁ and r by division. The operations mirror the nature of each sequence type — additive versus multiplicative.
Geometric Series and Sum Formula
n terms is Sₙ = a₁(1 − rⁿ) / (1 − r) when r ≠ 1. If r = 1, every term equals a₁, so Sₙ = n · a₁.Find the sum 3 + 6 + 12 + 24 + … + 3072.
Here a₁ = 3, r = 2. First find n: 3072 = 3 · 2^(n−1), so 2^(n−1) = 1024 = 2¹⁰ and n = 11.
Then S₁₁ = 3(1 − 2¹¹)/(1 − 2) = 3(1 − 2048)/(−1) = 3 · 2047 = 6141.
S₁₁ = 6141Find the sum of the first 5 terms of 100, 50, 25, ….
S₅ = 100(1 − (1/2)⁵)/(1 − 1/2) = 100(1 − 1/32)/(1/2) = 100(31/32)(2) = 193.75.
S₅ = 193.75Infinite geometric series
|r| < 1, the terms shrink toward zero, and the series converges. The sum of an infinite geometric series is S = a₁/(1 − r).For example, 10 + 5 + 2.5 + 1.25 + … = 10/(1 − 1/2) = 20.
S = 20This formula is widely used in finance (present value of perpetuities), physics (bouncing ball distance), and converting repeating decimals to fractions. For instance, 0.333… = 3/10 + 3/100 + 3/1000 + … = (3/10)/(1 − 1/10) = (3/10)/(9/10) = 1/3.
If you want to verify any sequence or series calculation, scan the problem with Solver AI for a complete step-by-step solution.