Arithmetic Progression • Topic 1 of 3

nth Term

An AP is fixed by just two numbers: the first term a and the common difference d. Every other term follows from a_n = a + (n − 1)d, which is simply "start, then add d a total of (n − 1) times". The smart CAT habit is to read d straight off the sequence (subtract any term from the next) and never count terms one by one. To find how many terms a finite AP has, use n = (last − first)/d + 1 — this is the formula that powers "how many multiples of 7 between 100 and 500" type questions. A second time-saver is counting from the back: the mth term from the end is l − (m − 1)d. If you are ever given two terms, subtract their equations to get d in one step, because the difference of the two equations kills a and leaves a clean multiple of d.

✅ Solved examples

1. Find the 15th term of the AP 7, 11, 15, 19, …
a = 7, d = 4. a_15 = 7 + (15 − 1)×4 = 7 + 56 = 63.
2. How many terms are there in the AP 12, 18, 24, …, 120?
a = 12, d = 6, last = 120. n = (120 − 12)/6 + 1 = 108/6 + 1 = 18 + 1 = 19.
3. The 4th term of an AP is 20 and the 9th term is 45. Find the common difference and the first term.
a_9 − a_4 = 5d = 45 − 20 = 25 ⇒ d = 5. Then a + 3d = 20 ⇒ a = 20 − 15 = 5.
4. How many multiples of 7 lie between 100 and 500?
First multiple ≥ 100 is 105, last ≤ 500 is 497. n = (497 − 105)/7 + 1 = 392/7 + 1 = 56 + 1 = 57.

✏️ Practice — try these, take hints as needed

1. Find the 20th term of the AP 3, 8, 13, 18, …
a = 3, d = 5.
a_n = a + (n − 1)d.
3 + 19×5.
98
2. How many terms are in the AP 5, 9, 13, …, 101?
d = 4.
n = (last − first)/d + 1.
(101 − 5)/4 + 1.
25
3. The 7th term of an AP is 34 and the 13th term is 64. Find the first term.
Subtract: 6d = 64 − 34.
d = 5.
a + 6d = 34.
4
4. Which term of the AP 4, 9, 14, … equals 129?
a = 4, d = 5.
4 + (n − 1)×5 = 129.
(n − 1)×5 = 125.
26th term
5. How many multiples of 6 lie between 50 and 250?
First multiple ≥ 50 is 54, last ≤ 250 is 246.
d = 6.
(246 − 54)/6 + 1.
33

📝 Topic test — 8 questions

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