What is the algebraic form of an arithmetic sequence?

Every arithmetic sequence can be written as xₙ = an + b, where a is the common difference and b is a fixed number (equal to the first term minus a). This single expression gives any term directly by substituting its position for n.

How do you find a and b in xₙ = an + b?

a is always just the common difference of the sequence. b is the first term minus a (since substituting n = 1 into an + b gives a + b, which must equal the first term).

What is the formula for the sum of an arithmetic sequence?

If the sequence is xₙ = an + b, the sum of the first n terms is Sₙ = (1/2)an(n+1) + bn. Equivalently, if you know the first and last (nth) term, Sₙ = (n/2) × (first term + last term) — the same shortcut Gauss used to add 1 to 100 in seconds.

Why is the sum of an arithmetic sequence always a quadratic in n?

Expanding Sₙ = (1/2)an(n+1) + bn always simplifies to the form Sₙ = pn² + qn, with no constant term. This is a fingerprint of arithmetic sequences — if a given sum formula has a leftover constant when n = 0, the sequence behind it isn't arithmetic.

How do you find the original sequence from its sum formula?

The first term equals S₁ (the sum of just one term). Every term after that is xₙ = Sₙ − Sₙ₋₁, since the difference between consecutive partial sums is exactly the new term being added.

Chapter 3 · Kerala SSLC Class 10 Maths

Arithmetic Sequences and Algebra

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

Algebraic form xₙ=an+bSum formulaGauss's trickSequence from a sum

Also useful:Chapter 1 — Arithmetic Sequences Practice Problems — Kerala SSLC Maths 1-on-1 Kerala SSLC Maths Tuition

Kerala SSLCClass 10MathematicsChapter 3⏱ 9 min crisp · 13 min story

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

How this works

Each scene turns a sequence problem into one clean piece of algebra. Read the story, then tap "The Maths" to see the formal result.

The 500th Delivery

A warehouse packs parcels onto a conveyor belt. The first parcel is numbered 12. Each parcel after that is numbered 11 higher than the one before — 12, 23, 34, 45, and so on.

Supervisor

I need the number printed on the 500th parcel. Don't write out all 500 — just tell me the number.

Writing 500 terms is out of the question. But there is a pattern: to reach the 500th parcel from the 1st, you add the common difference 499 times (not 500 — you're already standing on the 1st).

500th parcel = 1st term + (499 × common difference)

= 12 + (499 × 11)

= 12 + 5489 = 5501

That works for the 500th term specifically. But what if the supervisor asks for the 501st tomorrow, then the 502nd the day after? Writing "add (position − 1) times 11" every single time gets old fast. So instead, replace the position with a letter, n, once and for all.

nth parcel = 12 + (n − 1) × 11

= 12 + 11n − 11

= 11n + 1

Now any parcel number is one substitution away. Want the 500th? Put n = 500. Want the 5000th? Put n = 5000. No more repeating the same steps by hand.

The Shortcut

Once you know the trick, expanding (n − 1) × d by hand every time is unnecessary. There is a faster route straight to a and b.

Teacher

Find the algebraic form of the sequence starting at 1/2 and adding 1/3 each time. No expansion — just two quick steps.

Step one: a is always just the common difference — no work needed there. Step two: since the first term is a + b (substitute n = 1 into an + b), b must be the first term minus a.

a = common difference = 1/3

b = first term − a = 1/2 − 1/3 = 1/6

xₙ = (1/3)n + 1/6

Check it: put n = 1 into (1/3)n + 1/6 → 1/3 + 1/6 = 1/2. That matches the first term, so the shortcut checks out.

Ten-Year-Old Gauss

A famous classroom story: a teacher, wanting to keep her students busy, asks them to add every number from 1 to 100. Most children start at 1 and trudge forward. One ten-year-old boy, Gauss, finishes in seconds.

Gauss

1 and 100 make 101. So do 2 and 99. And 3 and 98. Every pair like that makes 101 — and there are 50 such pairs.

1 + 100 = 101

2 + 99 = 101

… (50 pairs total)

Total = 50 × 101 = 5050

The same pairing works for any run of consecutive natural numbers, not just up to 100 — pair the first with the last, the second with the second-last, and so on. Each pair adds to the same total.

Watch how it scales up. The warehouse parcels from Scene 1 were xₙ = 11n + 1. To sum the first n of them, add up 11 times each natural number, plus b added n times:

Sₙ = (11×1 + 1) + (11×2 + 1) + ⋯ + (11×n + 1)

= 11×(1 + 2 + ⋯ + n) + n×1

= 11 × n(n+1)/2 + n

Working Backwards

A textbook question gives you the sum formula directly and asks the reverse question: given that Sₙ = 3n² + n, what is the original sequence?

Student

There's no list of terms here — just an expression for the sum. How do I get the sequence out of that?

The trick: the sum of the first 1 term IS the first term. And the difference between the sum of the first 2 terms and the sum of the first 1 term must be exactly the 2nd term.

S₁ = 3(1)² + 1 = 4 → 1st term = 4

S₂ = 3(2)² + 2 = 14

2nd term = S₂ − S₁ = 14 − 4 = 10

Two terms is already enough to find the common difference: 10 − 4 = 6. So the sequence is 4, 10, 16, … and its algebraic form must be xₙ = 6n − 2.

Quick check: does Sₙ = pn² + qn ALWAYS have no constant term (no number left over when n = 0)? Yes — that's a fingerprint of every arithmetic sequence's sum. If a sum formula has a leftover constant, the sequence behind it isn't arithmetic.

The Big Picture

Every idea in this chapter branches from one fact: any arithmetic sequence can be written as xₙ = an + b. Once you have that, both the sequence and its sum become plain algebra.

Write any term directly?

xₙ = an + b

Find a and b fast?

a = common difference, b = first term − a

Sum the first n terms?

Sₙ = (1/2)an(n+1) + bn = (n/2)(x₁+xₙ)

Shape of the sum formula?

Sₙ = pn² + qn (always, no constant term)

Reverse — sequence from sum?

xₙ = Sₙ − Sₙ₋₁, x₁ = S₁

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