Sum of an Arithmetic Progression (AP): Formula and Examples

The sum of an arithmetic progression answers a natural follow-up question: once you can find any single term, what is the total of the first n terms? The young Carl Friedrich Gauss famously answered it for 1 + 2 + … + 100 by pairing numbers from opposite ends: 1 + 100 = 101, 2 + 99 = 101, and so on, making 50 pairs of 101, for a total of 5050. Every AP sum works on exactly this pairing idea.

AP sums matter whenever steady, repeated amounts accumulate. Total savings from a monthly plan that rises by a fixed step, the total fare across distance slabs, or the number of bricks in a stepped wall are all AP sums. The formula replaces a long addition with one short multiplication.

Key Concepts

• S_n: the sum of the first n terms of the AP.
• Pairing trick: in any AP, the first term plus the last term equals the second term plus the second-last term, and so on. Every pair has the same total.
• Average view: S_n = n times (average of first and last terms), because the n terms are symmetric around their middle.
• Two equivalent formulas: S_n = n/2 times [2a + (n - 1)d] and S_n = n/2 times (a + l), where l is the last term.

Formulas

• Sum from a and d: S_n = n/2 times [2a + (n - 1)d]. Use this when the last term is not known.
• Sum from first and last: S_n = n/2 times (a + l). This is usually the faster form when l is known.
• Assumptions: n is a positive integer and the common difference is constant across the summed terms.

Worked Example

First, the classic: the sum of the first 100 natural numbers. With a = 1 and l = 100, S_100 = 100/2 times (1 + 100) = 50 times 101 = 5050, exactly Gauss's result. Second, a savings plan: deposit 500 rupees in month one and raise the deposit by 50 rupees every month for 24 months. Here a = 500, d = 50, n = 24, so S_24 = 24/2 times [2 times 500 + 23 times 50] = 12 times [1000 + 1150] = 12 times 2150 = 25,800 rupees in total.

Common Misconceptions

• S_n = n times (a + d): wrong on two counts. The sum uses the average of the extremes, n/2 times (first + last), not n times a single step.
• The formula works for any sequence: it does not. The pairing trick depends on a constant difference; for a non-AP list the formula gives a wrong total.
• You must know the last term: not necessarily. The 2a + (n - 1)d form needs only the first term, the difference, and n.

Real-World Applications

• Finance: totals of recurring deposits or instalments that rise by a fixed step each period.
• Construction: counting bricks, tiles, or seats arranged in rows that grow by a constant amount.
• Sports: league tables where points per round follow a steady pattern.
• Combinatorics: triangular numbers 1, 3, 6, 10, … are AP sums of 1, 2, 3, … and count handshakes and pairs.

The sum of an AP is always n/2 times the sum of the first and last terms, or equivalently n/2 times [2a + (n - 1)d]. Gauss's pairing shows why: the terms fold into identical pairs. Use the (a + l) form when the last term is handy, the (2a + (n - 1)d) form when it is not, and never apply the formula to a sequence whose difference is not constant.