What is an arithmetic progression?

An arithmetic progression (AP) is a list of numbers where the difference between any two consecutive terms (the common difference, d) is always the same.

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

aₙ = a + (n − 1)d, where a is the first term, d is the common difference, and n is the term's position. Substituting n=1 gives exactly the first term, since (1−1)d = 0.

What is the formula for the sum of the first n terms of an AP?

Sₙ = (n/2)[2a + (n−1)d], which can also be written as Sₙ = (n/2)(a + l) when the first term a and last term l are both known.

Why can solving for n in a sum equation give two different valid answers?

Sₙ is a quadratic expression in n, so solving Sₙ = (a target value) is solving a quadratic equation, which can have two positive-integer roots. Both are valid when the AP's terms change sign partway through, causing some terms to cancel each other out.

Why must n be a positive integer when finding a term's position?

Term positions represent 1st, 2nd, 3rd, and so on — only positive whole numbers make sense. If solving the nth-term equation for n gives a fraction or a negative number, the target value is not actually a term of that AP at all.

Chapter 5 · CBSE Class 10 Maths

Arithmetic Progressions

Two ways to learn — a crisp formula reference, or a story that makes the nth term formula feel obvious.

Common differencenth term formulaSum of n termsTwo-answer sum equations

Also useful:Chapter 4 — Quadratic Equations Chapter 6 — Triangles NCERT Solutions — Class 10 Maths 1-on-1 CBSE Class 10 Tuition

CBSEClass 10MathematicsChapter 5⏱ 8 min crisp · 12 min story

Four scenes that build the ideas from scratch, step by step.

How this works

Four scenes, one thread: adding the same fixed number over and over creates a pattern regular enough to skip straight to any term, or any sum, without ever listing the terms out. Read the story, then tap "The Maths" for the formal result.

Reena's Raise

Reena starts a job at ₹8000/month, with a ₹500 annual increment. Her salary goes 8000, 8500, 9000, 9500, ... Working out her 5th-year salary by adding 500 four times is easy enough. But what about her 25th-year salary?

Student

Adding 500 twenty-four times by hand seems slow. Is there a shortcut for jumping straight to any year?

Look at the pattern differently: her 3rd-year salary is 8000 + 2×500 (not 3×500 — she only gets the raise TWICE by year 3, on her way there). Her 15th-year salary is 8000 + 14×500. Her 25th-year salary is 8000 + 24×500.

3rd year: 8000 + (3−1)×500 = 9000

15th year: 8000 + (15−1)×500 = 15000

25th year: 8000 + (25−1)×500 = 20000

The pattern generalises: the nth term of any AP is the first term, plus the common difference added (n−1) times — never n times, since the first term itself needed zero additions.

The formula also runs backward. For the AP 21, 18, 15, ... (a=21, d=−3), is −81 actually a term?

−81 = 21 + (n−1)(−3)

−105 = −3n → n = 35

n = 35 is a positive whole number, so yes — the 35th term is −81. But if solving backward like this ever gives a fraction or a negative n, that means the target value is simply NOT a term of this AP at all, no matter how close it looks.

Counting from the End

Take the AP 10, 7, 4, ..., −62. What's the 11th term counting backward from the end (not from the start)?

Teacher

First figure out how many terms this AP has at all — you can't count "from the end" without knowing where the end is.

l = −62 = 10 + (n−1)(−3) → n = 25 (25 terms total)

11th from the last = position (25 − 11 + 1) = 15th term

a₁₅ = 10 + 14(−3) = −32

Notice the position is 25−11+1, not 25−11. If the AP has exactly 25 terms, the very LAST term (1st from the end) is position 25, not 24 — the same off-by-one care needed as Scene 1.

A cleaner alternative: reverse the whole AP. The new first term is the old last term, and the new common difference is the negative of the old one. Then just find the 11th term of THIS new sequence — no subtraction-from-the-total needed at all.

Gauss Strikes Again

Shakila puts ₹100 in her daughter's money box at age 1, ₹150 at age 2, ₹200 at age 3, and so on. How much is collected by the daughter's 21st birthday? Adding 21 numbers by hand works, but there's a faster way — the same trick young Gauss used to add 1 to 100 instantly.

Gauss

Write the sum forwards, then write it again backwards, and add the two lists term by term. Every pair adds up to the exact same total.

S = a + (a+d) + ⋯ + [a+(n−1)d]

S = [a+(n−1)d] + ⋯ + (a+d) + a

2S = n × [2a + (n−1)d] (n identical pairs)

Dividing by 2 gives the sum of ANY arithmetic progression, not just consecutive counting numbers — Gauss's trick for 1 to 100 was just this formula's simplest special case.

Shakila's money box: a=100, d=50, n=21

S₂₁ = (21/2)[2(100) + 20(50)] = (21/2)(1200)

When Two Answers Are Both Right

How many terms of the AP 24, 21, 18, ... (a=24, d=−3) must be added to get a sum of exactly 78?

Student

I set up the sum formula and solved for n — I got two different positive whole numbers, 4 and 13. That must mean I made an error somewhere, right?

78 = (n/2)[48 + (n−1)(−3)]

3n² − 51n + 156 = 0 → n² − 17n + 52 = 0

(n−4)(n−13) = 0 → n = 4 or n = 13

Both are genuinely correct — because Sₙ is a QUADRATIC in n, an equation like this can legitimately have two positive-integer solutions. Here's why both work: the first 4 terms already sum to 78. But continuing on, terms 5 through 13 happen to add up to exactly ZERO (the AP crosses from positive to negative partway through), so the sum of the first 13 terms is still 78.

This only happens because a is positive and d is negative — the terms eventually flip sign and start cancelling earlier terms out. Don't assume a sum equation always has one clean answer; solve the full quadratic and check every positive-integer root.

The Big Picture

Every idea in this chapter branches from one fact: adding a fixed number repeatedly makes both the nth term AND the sum predictable by formula — no listing terms out required, in either direction.

Is this list an AP?

Check aₖ₊₁ − aₖ is constant

Find any specific term?

aₙ = a + (n−1)d

Term position, working backward?

Solve for n — must be a positive integer

Sum the first n terms?

Sₙ = (n/2)[2a+(n−1)d] = (n/2)(a+l)

Sum equation gives two n values?

Both can be valid — Sₙ is quadratic in n

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