Arithmetic Progression Calculator
Calculate nth term, sum, and properties of arithmetic sequences and series. Enter values for instant results with step-by-step formulas.
Formula
a_n = a + (n-1)d | S_n = n/2 * (2a + (n-1)d)
Where a_n = nth term, a = first term, d = common difference, n = number of terms, and S_n = sum of first n terms. The nth term formula adds (n-1) increments of d to the first term. The sum formula uses the pairing method, multiplying the number of pairs (n/2) by the sum of each pair (first term + last term).
Worked Examples
Example 1: Salary Growth Over Time
Problem: An employee starts with a salary of $40,000 and receives a $2,500 annual raise. Find the salary in year 15 and the total earnings over 15 years.
Solution: First term (a) = 40,000, Common difference (d) = 2,500, n = 15\n15th term: a_15 = 40,000 + (15-1) x 2,500 = 40,000 + 35,000 = $75,000\nSum of 15 years: S_15 = 15/2 x (40,000 + 75,000)\n= 7.5 x 115,000 = $862,500\nAlternatively: S_15 = 15/2 x (2(40,000) + 14(2,500))\n= 7.5 x (80,000 + 35,000) = 7.5 x 115,000 = $862,500
Result: Year 15 salary: $75,000 | Total 15-year earnings: $862,500
Example 2: Stadium Seating Arrangement
Problem: A stadium section has 20 rows. The first row has 30 seats, and each subsequent row has 2 more seats. Find the total seating capacity and seats in the last row.
Solution: First term (a) = 30, Common difference (d) = 2, n = 20\nLast row (20th term): a_20 = 30 + (20-1) x 2 = 30 + 38 = 68 seats\nTotal seats: S_20 = 20/2 x (30 + 68)\n= 10 x 98 = 980 seats\nMiddle row (10th): a_10 = 30 + 9 x 2 = 48 seats\nArithmetic mean: (30 + 68)/2 = 49 = S/n = 980/20
Result: Last row: 68 seats | Total capacity: 980 seats | Average per row: 49
Frequently Asked Questions
What is an arithmetic progression and what defines it?
An arithmetic progression (AP), also called an arithmetic sequence, is an ordered list of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference, denoted by d. If the first term is a and the common difference is d, then the sequence is a, a+d, a+2d, a+3d, and so on. For example, the sequence 3, 7, 11, 15, 19 is an arithmetic progression with first term 3 and common difference 4. Arithmetic progressions are among the simplest and most important sequences in mathematics, appearing in everyday situations from counting by intervals to equally spaced time schedules.
How do you find the nth term of an arithmetic progression?
The nth term of an arithmetic progression is given by the formula a_n = a + (n - 1) times d, where a is the first term, d is the common difference, and n is the position number. This formula works because to reach the nth term, you start at a and add the common difference exactly (n-1) times. For example, in the AP 5, 8, 11, 14, ..., the 20th term is 5 + (20-1) times 3 = 5 + 57 = 62. This formula can also be rearranged to find other unknowns: if you know any three of a_n, a, n, and d, you can solve for the fourth. For instance, if the 15th term is 50 and the first term is 8, then d = (50 - 8) / (15 - 1) = 42/14 = 3.
What is the formula for the sum of an arithmetic progression?
The sum of the first n terms of an arithmetic progression can be calculated using two equivalent formulas: S_n = n/2 times (2a + (n-1)d), or equivalently S_n = n/2 times (a + l), where l is the last term. This formula was famously discovered by the young Carl Friedrich Gauss when asked to sum the numbers from 1 to 100 by his teacher. Gauss realized that pairing the first and last terms (1 + 100 = 101), second and second-to-last (2 + 99 = 101), and so on, creates 50 pairs each summing to 101, giving 50 times 101 = 5050. The general formula captures this pairing insight: n/2 pairs each summing to (a + l), which equals (a + a + (n-1)d) = (2a + (n-1)d).
How do you determine if a sequence is an arithmetic progression?
To verify that a sequence is an arithmetic progression, check whether the difference between consecutive terms is constant throughout the sequence. Calculate d = a2 - a1, then verify that a3 - a2, a4 - a3, and all subsequent differences equal the same value d. For example, the sequence 4, 9, 14, 19, 24 has differences 5, 5, 5, 5, confirming it is an AP with d = 5. Another test is that three consecutive terms a, b, c form an AP if and only if 2b = a + c, meaning the middle term equals the arithmetic mean of the outer terms. If any consecutive difference varies, the sequence is not arithmetic. Note that a constant sequence like 7, 7, 7, 7 is technically an AP with d = 0.
What is the arithmetic mean and how does it relate to arithmetic progressions?
The arithmetic mean of two numbers p and q is (p + q) / 2, and it is the value that, when placed between p and q, forms a three-term arithmetic progression. More generally, inserting k arithmetic means between two numbers a and b creates an arithmetic progression of k + 2 terms, where the common difference d = (b - a) / (k + 1). For example, inserting 3 arithmetic means between 2 and 18 gives d = (18 - 2) / 4 = 4, producing the sequence 2, 6, 10, 14, 18. In an arithmetic progression, every term (except the first and last) is the arithmetic mean of its two neighbors. This property is equivalent to the definition of constant common difference and provides a useful alternative characterization of APs.
What are some real-world applications of arithmetic progressions?
Arithmetic progressions appear in numerous practical applications across many fields. In finance, straight-line depreciation reduces an asset value by a fixed amount each period, forming an AP. Fixed monthly savings contributions create an AP of accumulated payments. In physics, an object under constant acceleration has velocities at equal time intervals forming an AP, and uniformly spaced distance markers along a road form an AP. In construction, staircase measurements use APs for equally spaced risers. Calendar calculations use APs for dates separated by fixed intervals. Seating arrangements in theaters with rows of increasing seats often follow APs. Production planning with constant output increases and salary schedules with fixed annual raises are additional everyday examples.