Arithmetic Progression Formulas and Solved Examples

Arithmetic progression is one of the most consistently tested topics in Indian placement aptitude tests: finding the nth term, summing a series, and back-solving for the common difference. Two formulas cover the full range.

This article derives both from scratch, applies them across five worked examples, and maps the question types to the exams where they show up most often.

What Is an Arithmetic Progression?

An arithmetic progression (AP) is a sequence of numbers where consecutive terms differ by a constant amount. That constant is called the common difference, written as d.

The sequence 3, 5, 7, 9, 11 is an AP with d = 2. The sequence 20, 15, 10, 5 is an AP with d = -5. The sequence 7, 7, 7, 7 is a degenerate AP with d = 0 (valid, but rarely tested in placement exams).

Three things are always true in any AP:

• Subtract any term from the one that follows it and you get d. Always.
• Positive d means the sequence grows without bound toward positive infinity.
• Negative d means it shrinks toward negative infinity.

To check if a sequence is an AP, compute the difference between every adjacent pair. If all differences match, it’s an AP. If even one pair diverges, it isn’t. A sequence like 2, 4, 8, 16 fails this test immediately because 4-2=2 but 8-4=4.

The general form: a, a+d, a+2d, a+3d, a+4d, …

The first term is a. The second is a+d. The kth term is a + (k-1)×d. That pattern drives the nth term formula.

The Three Core Formulas

Nth Term Formula

The position-n term of an AP with first term a and common difference d is:

• aₙ = a + (n-1) × d

The Khan Academy arithmetic sequences review derives this the same way: each step adds one d, and you take n-1 steps from the first term to reach the nth.

A common slip is substituting n instead of (n-1). For n=1, the formula must give a + 0×d = a, which is the first term. Use that as a self-check.

Sum of First n Terms

The sum formula comes from a pairing argument. Write the sum twice: once forward, then backward.

• S = a + (a+d) + (a+2d) + … + aₙ
• S = aₙ + (aₙ-d) + (aₙ-2d) + … + a

Each column adds to (a + aₙ). There are n such columns. So 2S = n × (a + aₙ), giving:

• Sₙ = n/2 × (a + aₙ)

Substitute aₙ = a + (n-1)×d to get the fully expanded form:

• Sₙ = n/2 × [2a + (n-1) × d]

Both forms are correct. Use the first when you already know the last term. Use the second when you only have a, d, and n.

Constant Sequence (d = 0)

When d = 0, every term equals a, and the sum reduces to:

• Sₙ = n × a

Aptitude questions rarely test this directly, but it’s a useful boundary check.

Five Solved Examples

These examples mirror the types that appear in placement aptitude sections. Work through each step; the formula application is identical across variants.

Finding a Specific Term

• Problem: Find the 15th term of the AP 3, 9, 15, 21, …
• a = 3, d = 9 - 3 = 6, n = 15
• Step 1: Apply aₙ = a + (n-1) × d
• Step 2: a₁₅ = 3 + (15-1) × 6 = 3 + 84 = 87

Finding the General Term

• Problem: Find the general term of the AP -3, -1/2, 2, …
• a = -3, d = -1/2 - (-3) = 5/2
• Step 1: aₙ = -3 + (n-1) × 5/2
• Step 2: aₙ = -3 + 5n/2 - 5/2 = 5n/2 - 11/2
• Verify: n=1 gives 5/2 - 11/2 = -3 ✓, n=2 gives 10/2 - 11/2 = -1/2 ✓, n=3 gives 15/2 - 11/2 = 2 ✓

Sum of First n Terms

• Problem: Find the sum of the first 10 terms of the AP 1, 11, 21, 31, …
• a = 1, d = 10, n = 10
• Step 1: Sₙ = n/2 × [2a + (n-1) × d]
• Step 2: S₁₀ = 10/2 × [2×1 + 9×10] = 5 × [2 + 90] = 5 × 92 = 460

Back-Solving for the Common Difference

• Problem: The 11th term of an AP is 47 and the first term is 7. Find d.
• a = 7, a₁₁ = 47
• Step 1: 47 = 7 + (11-1) × d → 47 = 7 + 10d
• Step 2: 10d = 40 → d = 4

Sum to a Given Last Term

• Problem: Find the sum of the sequence 10, 15, 20, …, 100.
• a = 10, l = 100, d = 5
• Step 1: Find n: 100 = 10 + (n-1) × 5 → 90 = (n-1) × 5 → n = 19
• Step 2: S₁₉ = 19/2 × (10 + 100) = 19/2 × 110 = 1045

AP in Placement Tests

Arithmetic progression appears in the quantitative aptitude sections of TCS NQT, AMCAT, Infosys InfyTQ, and Wipro NLTH. The question types cluster around three patterns:

Question type	What it asks	Formula used
Find the nth term	Given a, d, find a specific term	aₙ = a + (n-1)×d
Find a missing term	Given some terms, find a or d	Solve the nth term equation
Find the sum	Sum of first n terms, or up to last term l	Sₙ = n/2 × [2a + (n-1)×d] or n/2 × (a + l)

Speed matters. The most reliable shortcut for a “find n” question: once you’ve expressed the last term as a + (n-1)×d, the algebra is one step of linear equation solving. Avoid re-deriving d every time; pull it from the sequence first and write it at the top of your working.

Calendar and date problems use similar modular arithmetic intuitions; that article derives every formula from first principles rather than providing a lookup table. Clock problems in competitive exams follow the same approach. Both are worth pairing with AP practice because placement aptitude sections test all three topics in the same sitting.

For number pattern shortcuts that don’t fit the standard formula framework, see number pattern tricks, which is useful when examiners structure the question around a specific multiplication pattern rather than a general sequence.

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Quantitative aptitude is one layer of the placement filter. The other layer (showing you can work with AI tools) has grown steadily since 2024. The AP sum formula works because pairing terms from opposite ends of the sequence always produces the same constant (a + l), letting you count n/2 such pairs. That kind of constant-stride pattern recognition is also what sequence models in machine learning are trained to identify. If you want to move from recognising patterns to building with them, TinkerLLM is a ₹299 hands-on LLM environment where placement-focused engineers start that experiment.