Arithmetic Progression
An arithmetic progression (AP) is a sequence in which each term differs from the one before it by a fixed amount called the common difference. That single idea — constant difference — is one of the most quietly useful tools in CAT Quant. Whenever a question hides a pattern that grows by equal steps (seats per row in an auditorium, instalments that rise by a fixed sum, days needed when output climbs steadily, even the count of multiples of a number in a range), an AP is sitting underneath it. CAT rarely asks you to "find the 10th term" in plain words; it dresses the AP up as a word problem and rewards students who can spot the linear pattern, pick the right formula, and avoid arithmetic slips. This chapter builds that fluency: the nth-term formula a_n = a + (n−1)d, the two faces of the sum formula S_n = n/2[2a + (n−1)d] = n/2(first + last), the average-of-neighbours property, and the symmetric-term trick (a−d, a, a+d) that collapses messy simultaneous equations into one line. Each topic comes with worked CAT-style examples, the fastest method, and the traps that cost careless aspirants easy marks.
Topics
⚡ CAT shortcuts & speed methods
The fastest ways to crack this chapter under time pressure — the techniques that separate a 95+ percentiler from the rest.
- Read d straight off the sequence (any term minus the previous one) — never count up term by term.
- For evenly spaced lists, count terms with n = (last − first)/d + 1; this nails "how many multiples" questions.
- Two terms given? Subtract their equations to kill a and get d instantly: a_p − a_q = (p − q)d.
- Sum fast with S_n = n/2 (first + last) = n × (average term) whenever both end terms are known.
- Given the sum (and product) of terms in AP, centre them: 3 terms (a − d, a, a + d), 4 terms (a − 3d, a − d, a + d, a + 3d).
- Keep 1 + 2 + … + n = n(n + 1)/2 and 1 + 3 + … + (2n − 1) = n² on instant recall for series sums.
⚠️ Common mistakes & traps
CAT is designed so that careless errors here cost you marks. Internalise each trap before the exam.
- Using a + nd for the nth term instead of a + (n − 1)d (off-by-one on the count of d added).
- Forgetting the "+1" in n = (last − first)/d + 1, which undercounts the terms by one.
- Writing unknown terms as a, a + d, a + 2d for sum-product problems instead of the symmetric form, which forces messy quadratics.
- Taking the common difference of four symmetric terms as d instead of 2d (the spacing in a − 3d, a − d, a + d, a + 3d is 2d).
- Dropping the negative root of n in a sum equation, or keeping a non-integer n — the number of terms must be a positive integer.
📈 CAT exam insight & PYQ analysis
🎴 Flashcards — instant recall
Tap a card to reveal the answer. Drill these until they are automatic.
📌 Quick revision
Chapter test
🏆 Vidaara CAT success checklist
You have truly mastered Arithmetic Progression when you can tick every box below.
- Recall every formula in this chapter without looking them up
- Solve each topic’s practice set with at least 80% accuracy
- Use the chapter shortcuts to cut your solving time in half
- Spot and avoid every common trap listed above
- Score 80%+ on the timed chapter test
📋 Chapter mastery scorecard
Track where you stand. Aim for the target before moving to the next chapter.
| Skill checkpoint | Target |
|---|---|
| Concept theory & formulas understood | 100% |
| Topic practice sets attempted (3 topics) | 3/3 |
| Best topic-test score | — → 80%+ |
| Chapter test score | — → 80%+ |
| Flashcards drilled to instant recall | 12 cards |
Formula Reference Sheet
Term & general form
| nth term | a_n = a + (n − 1)d |
|---|---|
| Common difference | d = a_n − a_(n−1) |
| Number of terms | n = (last − first)/d + 1 |
| mth from the end | l − (m − 1)d (l = last term) |
| Neighbour-average property | a_n = (a_(n−1) + a_(n+1)) / 2 |
Sum & shortcuts
| Sum of n terms | S_n = n/2 [2a + (n − 1)d] |
|---|---|
| Sum via first & last | S_n = n/2 (first + last) |
| Sum from average | S_n = n × (average term) |
| First n naturals | 1 + 2 + … + n = n(n + 1)/2 |
| First n odd numbers | 1 + 3 + … + (2n − 1) = n² |