Arithmetic Progressions in Real Life: Seating, Salary Increments and Loans

An arithmetic progression, or AP, is a list of numbers in which each term is the previous one plus a fixed amount. That sounds abstract, but it describes a surprising number of ordinary things: the rows of a stadium, a salary that rises by the same amount each year, a taxi fare, a monthly saving plan, even the interest on a loan that is repaid in equal parts. When students see the pattern in these settings, the formulas stop being something to memorise and become a way of answering questions they might genuinely ask.

This guide builds the two formulas from scratch, applies them to six everyday situations, and then covers the mistakes that cost marks. For the underlying theory, see these notes on the introduction to arithmetic progressions and the nth term of an AP, and keep the chapter notes open for reference.

The Two Formulas You Need

An AP is written as a, a + d, a + 2d, a + 3d, … where a is the first term and d is the common difference, the fixed amount added each time. If d is positive the terms grow, if d is negative they shrink, and if d = 0 they stay the same.

The nth term: aₙ = a + (n − 1)d

Why n − 1? Because to reach the nth term from the first, you add d exactly n − 1 times. The first term has had d added zero times.

The sum of the first n terms: Sₙ = n/2 × [2a + (n − 1)d], or equivalently Sₙ = n/2 × (a + l), where l is the last term.

The second form comes from a famous idea. Pair the first term with the last, the second with the second last, and so on. Every pair adds up to the same total, a + l, and there are n/2 pairs. A story goes that the young Gauss added the numbers from 1 to 100 this way: 100/2 × (1 + 100) = 5050.

Six Real-Life Situations

Seating in a hall

A hall has 25 rows. The first row has 20 seats, and each row has 2 more seats than the one in front. How many seats are in the last row, and how many in the hall?

Here a = 20, d = 2, n = 25. The last row has a₂₅ = 20 + 24 × 2 = 68 seats. The total is S₂₅ = 25/2 × (20 + 68) = 25/2 × 88 = 1,100 seats.

Salary increments

A person starts at ₹3,00,000 a year and gets a fixed rise of ₹20,000 every year. What is the salary in the 10th year, and the total earned in 10 years?

a = 3,00,000, d = 20,000, n = 10. The 10th-year salary is 3,00,000 + 9 × 20,000 = ₹4,80,000. The total is S₁₀ = 10/2 × [2(3,00,000) + 9(20,000)] = 5 × 7,80,000 = ₹39,00,000.

A monthly saving plan

A child saves ₹100 in the first month and increases the saving by ₹50 each month. How much is saved in 12 months?

a = 100, d = 50, n = 12. S₁₂ = 12/2 × [2(100) + 11(50)] = 6 × 750 = ₹4,500.

Taxi fare

A taxi charges ₹50 for the first kilometre and ₹12 for every additional kilometre. The fare for 1 km, 2 km, 3 km and so on is 50, 62, 74, …, an AP with a = 50 and d = 12. The fare for a 10 km ride is a₁₀ = 50 + 9 × 12 = ₹158.

Stacked pipes

Pipes are stacked with 12 in the bottom row, 11 in the next, and so on up to 1 at the top. This is an AP with a = 12, d = −1, and 12 terms. The total is S₁₂ = 12/2 × (12 + 1) = 78 pipes. A negative d simply means the terms go down.

A loan repaid in equal parts

Suppose ₹1,20,000 is repaid in 12 equal monthly installments of principal, ₹10,000 each, and interest at 1 percent a month is charged on the balance outstanding. The interest in month 1 is ₹1,200, in month 2 it is ₹1,100 (because the balance has fallen), in month 3 it is ₹1,000, and so on. This is an AP with a = 1,200 and d = −100. The total interest is S₁₂ = 12/2 × [2(1,200) + 11(−100)] = 6 × 1,300 = ₹7,800.

A word of care: a typical home or vehicle loan with a fixed EMI is different. The EMI stays the same each month while the split between interest and principal changes, so the EMI itself is not an AP. The AP above fits loans where the principal portion is the same every month.

How to Recognise an AP

Which Formula Do I Use?

How many terms? For 3, 7, 11, …, 143, we have a = 3, d = 4, l = 143. So n = (143 − 3)/4 + 1 = 35 + 1 = 36 terms.

When the Sum Gives a Quadratic

Sometimes a question gives the total and asks for n. Because Sₙ contains n², you get a quadratic equation, and two values of n appear.

Example. How many terms of the AP 9, 17, 25, … give a sum of 636?

a = 9, d = 8. Sₙ = n/2 × [18 + 8(n − 1)] = n/2 × (8n + 10) = n(4n + 5) = 4n² + 5n. Set 4n² + 5n = 636, so 4n² + 5n − 636 = 0. D = 25 + 4 × 4 × 636 = 25 + 10,176 = 10,201 = 101². So n = (−5 ± 101)/8, which gives n = 12 or n = −53/4. The number of terms must be a positive whole number, so n = 12.

This is the same “reject the root that does not make sense” habit as in why a quadratic equation has two roots.

Slips That Cost Marks

• Using n instead of n − 1 in the nth-term formula.
• Forgetting that d can be negative, and dropping the sign.
• Using the sum formula when only one term was asked for, or the reverse.
• Taking an increase by a percentage as an AP.
• Not rejecting a fractional or negative n.
• Finding aₙ as Sₙ − Sₙ₋₁ and getting Sₙ₋₁ wrong by substituting n − 1 carelessly.

Progressions in Your Syllabus

Arithmetic Progressions is a chapter of the Class 10 Maths book in both the CBSE and the Karnataka SSLC textbooks. Practice with these chapter MCQs and the full chapter explanation with formulas and questions. The topic grows into sequences and series in Class 11, including the next kind of progression, in these notes on geometric progressions, which matter for JEE Main and KCET. Confirm the current portions with your own board.

Frequently Asked Questions

Q1. What is an arithmetic progression?

It is a list of numbers in which each term after the first is obtained by adding a fixed number, the common difference, to the previous term.

Q2. What is the formula for the nth term of an AP?

aₙ = a + (n − 1)d, where a is the first term and d is the common difference.

Q3. What is the formula for the sum of n terms?

Sₙ = n/2 × [2a + (n − 1)d], or Sₙ = n/2 × (a + l) when the last term l is known.

Q4. Is an EMI an arithmetic progression?

No. An EMI is a fixed amount, so it is a constant sequence. The interest part of a loan that is repaid in equal principal installments does form an AP, because it falls by the same amount each month.

Q5. What if the sum formula gives two values of n?

Reject any value that is negative or not a whole number, because the number of terms must be a positive integer.

A Pattern You Can Count On

Whenever something grows or shrinks by the same amount each step, an AP is hiding in it. Spot the first term and the difference, choose between the nth-term formula and the sum formula, and the answer follows in a few lines.