Animated Diagram — Arithmetic Progressions

First term a = 2

Common diff d = 3

Terms n = 8

💡 Each bar is a term of the AP. The height difference between consecutive bars is always the common difference d. If d > 0 the AP grows; if d < 0 it shrinks; if d = 0 all bars are equal.

Testing Whether a Sequence is an AP

Example: Is 3, 7, 11, 15, 19, … an AP?

Step 1

Recall the rule: a list is an AP if the difference between every pair of consecutive terms is the same.

Step 2

a₂ − a₁ = 7 − 3 = 4

Step 3

a₃ − a₂ = 11 − 7 = 4

Step 4

a₄ − a₃ = 15 − 11 = 4

Step 5

All differences are equal to 4. ✓

Step 6

∴ It IS an AP with first term a = 3 and common difference d = 4.

Counter-example

For 1, 4, 9, 16, …: differences are 3, 5, 7 — not equal → NOT an AP.

💡 Formula: d = aₙ − aₙ₋₁ for any n. Sign of d tells direction — positive (rising), negative (falling), zero (constant).

aₙ = a + (n − 1)d
The nth term of an AP with first term a and common difference d

nth Term Calculator

a (1st term)

d (common diff)

n (which term)

Click Find aₙ to compute the nth term step by step.

Reverse: Which term equals a given value?

value

Enter a value to check whether (and where) it appears in the AP.

Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)

a = 1

d = 1

n = 6

💡 Gauss trick: Pair the first term with the last, the second with the second-last… each pair sums to (a + l). There are n/2 such pairs, giving Sₙ = n/2 (a + l).

Choosing unknown terms symmetrically makes the d-terms cancel in the sum

3️⃣

Three Terms

a−d, a, a+d

Sum = 3a. Common difference is d.

4️⃣

Four Terms

a−3d, a−d, a+d, a+3d

Sum = 4a. Common difference is 2d.

5️⃣

Five Terms

a−2d, a−d, a, a+d, a+2d

Sum = 5a. Common difference is d.

💡 Click any real-life situation to see how it becomes an AP problem.

💰Savings & Salary

Pattern: A fixed amount is added each period.
aₙ = a + (n−1)d Example: Saves ₹200 first month, ₹50 more each month. Month 12?
a=200, d=50 → a₁₂ = 200 + 11×50 = ₹750

🎭Seats in Rows

Pattern: Each row has a fixed number more seats than the previous.
Row n seats = a + (n−1)d Example: Row 1 has 20, each next row +2. Row 10?
a=20, d=2 → a₁₀ = 20 + 9×2 = 38 seats

🪵Stacking Logs / Bricks

Pattern: Each row up has one fewer object → use Sum formula.
Sₙ = n/2 [2a + (n−1)d] Example: 20 logs bottom row, 19 next… 200 logs total.
200 = n/2[41−n] → n²−41n+400=0 → 16 rows (reject 25)

🔢Multiples in a Range

Pattern: Multiples of k form an AP with d = k.
Count: l = a + (n−1)d → solve for n Example: Multiples of 4 between 10 and 250: 12,16,…,248.
248 = 12 + (n−1)4 → 60 numbers

🏆Cash Prizes / Instalments

Pattern: Amounts change by a fixed step; total known → Sum formula.
Sₙ = n/2 [2a + (n−1)d] Example: ₹700 in 7 prizes, each ₹20 less than previous.
700 = 7/2(2a−120) → a = ₹160 (top prize)

📉Depreciation Steps

Pattern: A fixed value is lost each year (linear decrease).
Value after n years = a − (n−1)×loss Example: Machine ₹80,000, loses ₹5,000/yr. After 6 years?
a=80000, d=−5000 → a₆ = 80000 − 5×5000 = ₹55,000