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§5Class 10, Chapter 5

Arithmetic Progressions: The Rhythm of Fixed Steps

Salaries that rise by a flat amount, ladder rungs that shrink by a fixed length, savings that grow at a steady pace — arithmetic progressions are the mathematics of constant, predictable steps. This companion walks through spotting them, jumping straight to any term, and summing them all at once, with tools, flashcards and a CBSE-style quiz to test yourself.

aa+da+2daₙSₙ

🪜5.1 Patterns hiding in everyday life

Nature is full of repeating patterns — sunflower petals, honeycomb cells, pineapple spirals. Everyday life has number patterns too. A new employee's salary might rise by a fixed amount each year: 8000, 8500, 9000, ... A ladder's rungs might shrink by a fixed amount each step: 45, 43, 41, 39, ... cm.

💰
Not every pattern is the same kind of pattern. A salary rising by a flat ₹500 every year is very different from a savings scheme that multiplies by 5/4 every 3 years (8000 → 10000 → 12500 → 15625...). One grows by constant addition, the other by constant multiplication. This chapter is entirely about the first kind.
45 cm43 cm41 cm39 cm37 cm35 cm33 cm31 cm

Some patterns aren't either kind — like the famous rabbit-breeding sequence 1, 1, 2, 3, 5, 8, ... (each term is the sum of the two before it, not a fixed add or multiply). Spotting which rule a sequence follows is the first skill this chapter builds.

📏5.2 What makes a list an Arithmetic Progression

Definition. An Arithmetic Progression (AP) is a list of numbers in which each term (after the first) is obtained by adding a fixed number to the term before it. That fixed number is called the common difference, usually written d. Crucially, d can be positive, negative, or even zero.
🎚️
Like a thermostat set to a fixed step. Imagine a dial that always clicks by the exact same amount each time you turn it — always +5, or always −3, never +5 one time and +7 the next. That rigid, unchanging step size is exactly what "common difference" means. The moment the step size varies, it stops being an AP.
General form of an AP: a, a+d, a+2d, a+3d, ...
(a = first term, d = common difference)

🔍How to test whether a list is an AP

Compute the difference between every pair of consecutive terms: a₂−a₁, a₃−a₂, a₄−a₃, and so on. If — and only if — every one of these differences comes out identical, the list is an AP.

6, 9, 12, 15, ...
a₂−a₁ = 9−6 = 3
a₃−a₂ = 12−9 = 3
a₄−a₃ = 15−12 = 3 → constant difference, so this IS an AP with a=6, d=3
⚠️Subtract in the right order. To find d, always subtract a term from the one that comes right after it (aₖ₊₁ − aₖ) — never the reverse, even if that makes the subtraction look "backwards" for a decreasing AP. For 6, 3, 0, −3, ..., you compute 3−6 = −3, not 6−3.

You only need to check one such difference once you already know the list is an AP — but to first confirm it's an AP at all, check at least two or three pairs, since a single matching difference could be a coincidence.

Finite vs infinite APs

Some APs stop at a fixed last term — like the heights of 11 students in a queue (147, 148, ..., 157). These are finite APs. Others, like 1, 2, 3, 4, ..., continue forever with no last term — these are infinite APs.

🔢5.3 Finding any term without listing them all

Reena's starting salary is ₹8000 with a ₹500 annual raise. What's her salary in year 25? You could add 500 twenty-four times — but there's a shortcut hiding in the pattern.

Salary, year 3 = 8000 + (3−1)×500 = 9000
Salary, year 4 = 8000 + (4−1)×500 = 9500
Salary, year 5 = 8000 + (5−1)×500 = 10000
... Salary, year 25 = 8000 + (25−1)×500 = 20000
nth term formula. For an AP with first term a and common difference d, the nth term is: aₙ = a + (n − 1)d. This is also called the general term.
🚶
Like counting steps from a starting line. If you start at position a and every step forward moves you exactly d units, then after (n−1) steps — to reach the nth position, including your starting spot — you've moved a total of (n−1)×d. That's the entire formula, in one sentence.

Worked example: 10th term of 2, 7, 12, ...

a=2, d=5, n=10
a₁₀ = 2 + (10−1)×5 = 2 + 45 = 47

Worked example: which term of 21, 18, 15, ... equals −81? Is any term 0?

SET UPa=21, d=−3. Solve −81 = 21+(n−1)(−3).
SOLVE−81 = 24 − 3n → −105 = −3n → n = 35. So the 35th term is −81.
CHECK FOR ZEROSolve 21+(n−1)(−3)=0 → 3(n−1)=21 → n=8. The 8th term equals 0.

Worked example: is 301 a term of 5, 11, 17, 23, ...?

a=5, d=6. Solve 301 = 5+(n−1)×6 → n = 302/6 = 151/3 ≈ 50.3
🚫n must be a positive whole number. Since n = 151/3 isn't a whole number, 301 is simply not a term of this AP — no amount of algebra can force a fractional "term number" into existence. This check (is n a positive integer?) is essential every time you solve for n.

Worked example: how many two-digit numbers are divisible by 3?

List: 12, 15, 18, ..., 99 — an AP with a=12, d=3, last term 99
99 = 12+(n−1)×3 → n = 30
So there are 30 such numbers.

🔄Counting from the other end

Find the 11th term from the last term of 10, 7, 4, ..., −62. First find how many terms exist in total (25 terms, here), then note that "11th from the last" is the (25−11+1) = 15th term from the start — NOT the 14th, a common slip.

a₁₅ = 10 + (15−1)(−3) = 10 − 42 = −32
🔁
An easier way: flip the AP around. Alternatively, treat the last term as a new "first term" (a = −62) and the common difference as its negative (d = +3, since going backward reverses the direction), then simply find the 11th term of that reversed sequence — same answer, less bookkeeping.

➕5.4 Adding up a whole AP at once

Shakila adds money to her daughter's box every birthday: 100, 150, 200, 250, ... How much has accumulated by the 21st birthday? Listing and adding 21 numbers by hand is exactly the kind of tedious task mathematicians love to shortcut.

🧮
The 10-year-old who outsmarted his teacher. As legend has it, a teacher asked young Carl Friedrich Gauss to add up 1 through 100 as busywork — expecting it to take him a long while. He answered almost instantly: 5050.
1
2
3
...
98
99
100
101
101
101
101
101
101
100
99
98
...
3
2
1

His trick: write the sum forwards, then write it again backwards, and add the two versions term by term. Every single pair (1+100, 2+99, 3+98, ...) adds up to exactly 101 — and there are 100 such pairs.

2S = 101 × 100 = 10100
S = 10100 / 2 = 5050
🪞
Why this trick works for ANY AP, not just 1-to-100. Pairing the 1st term with the last, the 2nd term with the second-last, and so on always produces the same sum every time — (a + l) — no matter what a and d actually are. That's not a coincidence specific to Gauss's numbers; it's a structural fact about how AP terms are built.
Sum of the first n terms. Sₙ = n/2 [2a + (n−1)d] Equivalently, if l is the last (nth) term: Sₙ = n/2 (a + l) Use the second form when a and l are both known directly — it avoids needing d at all.

Worked example: sum of first 22 terms of 8, 3, −2, ...

a=8, d=−5, n=22
S₂₂ = 22/2 [2×8 + 21×(−5)] = 11[16 − 105] = 11×(−89) = −979

Worked example: how many terms of 24, 21, 18, ... sum to 78?

SET UPa=24, d=−3. Solve 78 = n/2[48+(n−1)(−3)].
SIMPLIFY156 = n(51−3n) → 3n²−51n+156=0 → n²−17n+52=0 → (n−4)(n−13)=0.
CONCLUDEn = 4 or n = 13 — BOTH are valid answers.
➕➖Why two different answers both work here. Because a is positive and d is negative, the terms start positive and eventually cross into negative territory. It turns out the terms from the 5th through the 13th cancel out to zero — so the sum after 4 terms and the sum after 13 terms are identical (both 78). Whenever an AP's terms change sign partway through, watch for this kind of double answer.

Sum of the first n positive integers

1 + 2 + 3 + ... + n = n(n+1)/2 (the a=1, d=1, l=n special case)
🔗A useful shortcut: aₙ = Sₙ − Sₙ₋₁. The nth term is exactly the difference between the running total through n terms and the running total through n−1 terms — logical, since adding the nth term is precisely what changes one sum into the other.

📋5.5 Summary

  1. 🪜 An Arithmetic Progression (AP) is a list where each term (except the first) is obtained by adding a fixed common difference d to the term before it. d can be positive, negative, or zero.
  2. 🔍 A list a₁, a₂, a₃, ... is an AP exactly when a₂−a₁ = a₃−a₂ = a₄−a₃ = ... (the same value every time).
  3. 🔢 The nth term: aₙ = a + (n − 1)d.
  4. ➕ The sum of the first n terms: Sₙ = n/2 [2a + (n−1)d], or equivalently Sₙ = n/2 (a + l) when the last term l is known.
  5. ⚖️ If a, b, c are in AP, then b = (a+c)/2 — b is called the arithmetic mean of a and c.