Arithmetic Progression Word Problems: Method with Worked Examples

An AP word problem gives quantities and relationships in words; identify the first term a, fixed change d and term count or index n, noting which is unknown. Every example below starts from prose, fixes the given numbers with an explicit index (which payment, year, row or layer is term 1), and solves for what is missing: a step value uses the nth term, an accumulated total uses the sum, and an unknown count is found from the nth term first. The two are never interchangeable, and the check at the end of each example shows the difference. The class 10 grounding is the CBSE 2026-27 mathematics curriculum, which lists the application of the nth term and the sum of the first n terms of an AP in solving daily life problems.

Reading a situation as an AP

a, the first term

the first payment, the year-1 salary, row 1, or the bottom layer - as assigned index 1 in your model; write the index down before computing

d, the common difference

the fixed change per step; negative when each step is smaller than the last

n

the number of terms, or the position index of the term asked about; year 9 means n = 9 when year 1 is the first term, because the AP reaches term n after n - 1 equal changes. When the count itself is unknown, solve for n from the nth-term formula first, as in Example 4

step value

one term of the sequence: a_n = a + (n - 1)d - use it for "the salary in year 9" or "the seats in row 14"

accumulated total

the sum of the first n terms: S_n = n/2 x (2a + (n - 1)d), or S_n = n/2 x (a + l) when the last term l is known - use it for "total saved" or "total logs"

Worked examples

Example 1: a salary with a fixed yearly increment

A job pays Rs 28,000 per month in the first year, and the monthly salary rises by a fixed Rs 2,500 every year. What is the monthly salary in year 9?

1. Index: year 1 is the first term, so a = 28,000, d = 2,500, and year 9 means n = 9.
2. The question asks for one step value, not a total: use a_n = a + (n - 1)d.
3. a_9 = 28,000 + (9 - 1) x 2,500 = 28,000 + 20,000 = 48,000.
4. Check by stepping: 28,000; 30,500; 33,000; ... each year adds 2,500, and 8 steps land on 48,000.

The monthly salary in year 9 is Rs 48,000.

Example 2: savings that grow by a fixed amount

Meera saves Rs 500 in week 1 and increases her weekly saving by a fixed Rs 100 every week. How much has she saved in total after 12 weeks?

1. Index: week 1 is the first term, so a = 500, d = 100, and 12 weeks means n = 12.
2. The question asks for an accumulated total, not one week's deposit: use S_n = n/2 x (2a + (n - 1)d).
3. S_12 = 12/2 x (2 x 500 + 11 x 100) = 6 x 2,100 = 12,600.
4. Check: the week-12 deposit is a_12 = 500 + 11 x 100 = 1,600, and the other formula agrees: S_12 = 12/2 x (500 + 1,600) = 12,600.

Meera has saved Rs 12,600 in total after 12 weeks.

Example 3: seats in a theatre, step value and total

A theatre has 22 seats in row 1, and each row after that has 3 more seats than the row in front of it. How many seats are in row 14, and how many seats do the 14 rows hold altogether?

1. Index: row 1 is the first term, so a = 22, d = 3, and row 14 means n = 14.
2. Seats in row 14 are a step value: a_14 = 22 + 13 x 3 = 61.
3. The 14-row total is an accumulated total: S_14 = 14/2 x (22 + 61) = 7 x 83 = 581.
4. Check: the same total from the other formula, S_14 = 14/2 x (44 + 13 x 3) = 7 x 83 = 581. Same question, two different answers wanted - 61 for one row, 581 for all of them.

Row 14 has 61 seats, and the 14 rows hold 581 seats in all.

Example 4: stacked logs with a shrinking layer

A stack of logs has 30 logs in its bottom layer, and each layer above holds 2 fewer logs, up to a top layer of 10 logs. How many layers are there, and how many logs are in the stack?

1. Index: the bottom layer is the first term, so a = 30; the stack shrinks, so d = -2; the last term is l = 10.
2. Find the layer count from the nth term: 10 = 30 + (n - 1) x (-2), so n - 1 = 10 and n = 11 layers.
3. The total number of logs is an accumulated total: S_11 = 11/2 x (30 + 10) = 11 x 20 = 220.
4. Check by listing: 30, 28, 26, 24, 22, 20, 18, 16, 14, 12, 10 - eleven layers; the five pairs 30 + 10, 28 + 12, ... each make 40, and the middle term 20 stands alone: 5 x 40 + 20 = 220.

There are 11 layers and 220 logs in the stack.

Summary

Fix the index first

the first payment, year 1, row 1 or the bottom layer is term 1; read the known parameters off the prose, then solve for the unknown term, total or count

Step value vs total

one term asks for a_n = a + (n - 1)d; an accumulated total asks for S_n = n/2 x (2a + (n - 1)d) or n/2 x (a + l)

n - 1, not n

the first term is already there before any step, so term n is n - 1 changes away

Negative d

a shrinking pattern has d negative; the formulas are unchanged

Check

recompute with the other sum formula, or step the sequence by hand for a few terms

Frequently asked questions

How do I know whether to use the nth term or the sum?

Match the question. A step value, like the salary in one year or the seats in one row, is the nth term a + (n - 1)d. An accumulated total, like total savings or total logs, is the sum of the first n terms.

Why does the nth term use n - 1 instead of n?

The first term is already counted before any change happens. An AP reaches term n after exactly n - 1 equal steps, so a_n = a + (n - 1)d; using n instead shifts every answer one step too far.

What should I take as the first term?

Use the quantity assigned index 1 in your model: for example, the first deposit or row 1. State the direction of counting. The first quantity mentioned in the prose is not necessarily the first term.

Can the common difference be negative?

Yes. A stack that loses 2 logs per layer has d = -2. The formulas do not change; write the negative sign explicitly so the terms decrease instead of increase.

Where can I practise the formulas themselves?

The linked class 10 chapter 5 page drills the nth term, both sum formulas and membership checks on given sequences, and the class 9 chapter 8 page introduces sequences and rules, including a taxi-fare try-it item with its printed answer.

Sources and scope

Formulas checked against Wikipedia, Arithmetic progression, retrieved 15 November 2026. Class 10 grounding: the CBSE Class X 2026-27 mathematics curriculum lists the application of the nth term and the sum of the first n terms of an AP "in solving daily life problems", at cbseacademic.nic.in, retrieved 15 November 2026. This page teaches prose-first applications only: it does not derive the formulas, drill bare sequences, or cover geometric progressions, fares, percentages or interest.

Prerequisites on this site: arithmetic progressions (class 10, chapter 5) and sequences and progressions (class 9, chapter 8).