Natural Numbers, AP, GP and Means: Placement Aptitude Guide

Arithmetic and geometric progressions appear in the quantitative section of every major placement test (TCS NQT, AMCAT, and Infosys InfyTQ all include them), and the derivable formulas are fewer than most students memorise.

Six cover everything: three natural-number sum formulas, the AP nth-term and sum pair, the GP nth-term and sum pair, and the AM-GM-HM inequality. This article derives each one with a worked example you can verify.

The Three Natural-Number Sum Formulas

Three formulas. Each verifiable in under a minute by substituting small values.

Sum of First n Natural Numbers

Formula: n(n+1)/2

Derive it by writing the sum twice and adding column by column:

• S = 1 + 2 + 3 + … + n
• S = n + (n-1) + (n-2) + … + 1
• Adding: 2S = (n+1) repeated n times = n(n+1)
• Therefore S = n(n+1)/2

Worked example: first 10 natural numbers

• n = 10
• S = 10 x 11 / 2 = 55
• Verify with n = 4: formula gives 4 x 5 / 2 = 10. Check: 1+2+3+4 = 10.

Sum of Squares of First n Natural Numbers

Formula: n(n+1)(2n+1)/6

This appears in quadratic-sequence problems and in Infosys and TCS NQT quantitative rounds.

Worked example: first 4 squares

• n = 4
• S = 4 x 5 x 9 / 6 = 180 / 6 = 30
• Verify: 1 squared + 2 squared + 3 squared + 4 squared = 1 + 4 + 9 + 16 = 30.

Sum of Cubes of First n Natural Numbers

Formula: [n(n+1)/2] squared

The sum of the first n cubes equals the square of the sum of the first n natural numbers. Knowing the first formula gives you the third for free.

Worked example: first 3 cubes

• n = 3
• S = [3 x 4 / 2] squared = [6] squared = 36
• Verify: 1 cubed + 2 cubed + 3 cubed = 1 + 8 + 27 = 36.

A common confusion: the cube formula squares the entire n(n+1)/2 result; the squares formula keeps a separate (2n+1)/3 factor. They are distinct.

Arithmetic Progression: Properties That Matter for Aptitude Tests

An AP is a sequence where consecutive terms differ by a fixed value d. First term: a.

Core AP Formulas

• nth term: Tn = a + (n - 1)d
• Sum of n terms: Sn = n/2 x [2a + (n - 1)d]
• Equivalently: Sn = n/2 x (first term + last term)

Shifting Every Term by a Constant

Add the same constant k to every term of an AP. The result is still an AP with the same common difference d.

Key fact: new sum = old sum + n x k

Example:

• Original series: 1, 3, 5 (d = 2)
• Add 2 to each term: 3, 5, 7 (d still = 2)
• Old sum (3 terms) = 9. New sum = 9 + 3 x 2 = 15. Check: 3+5+7 = 15.

Placement test type: all terms of 5, 8, 11, 14, 17, 20 are increased by 3. New sum?

• Old sum (6 terms) = 6/2 x (5 + 20) = 3 x 25 = 75
• New sum = 75 + 6 x 3 = 93
• Verify new series (8,11,14,17,20,23): 8+11+14+17+20+23 = 93.

Scaling Every Term by a Constant

Multiply every term of an AP by constant k. The result is still an AP, new common difference = k x d.

• New sum = k x old sum

Example:

• Original: 1, 2, 3 (d = 1). Multiply by 3: 3, 6, 9 (d = 3). New sum = 3 x old sum for same n.

Fixed-step arithmetic patterns appear in calendar and date calculations. The calendar problems aptitude guide covers the odd-days method with the same derive-not-memorise approach.

Geometric Progression: Ratio Rules and Sum Formulas

A GP is a sequence where each term is a constant multiple r of the previous one. First term: a.

Core GP Formulas

• nth term: an = a x r to the power (n-1)
• Sum of n terms: Sn = a x (r to the n minus 1) / (r - 1), for r not equal to 1
• Infinite sum: a / (1 - r), when the absolute value of r is less than 1

Constant Multiplication Property

Multiply every term of a GP by constant k. Result: still a GP with the same common ratio r.

Example:

• Original GP: 2, 4, 8, 16 (r = 2). Multiply by 3: 6, 12, 24, 48 (r still = 2).

Product of Two GPs

If two GPs have common ratios r and s, their term-by-term product is a GP with common ratio r x s.

Placement test type: GP 3, 6, 12, 24. What is the 5th term?

• r = 2, a = 3
• 5th term = 3 x 2 to the power 4 = 3 x 16 = 48

For an arithmetic shortcut that exploits number structure in a different way, see Multiplying Any Number by 111.

Pascal’s triangle row sums form a GP: 1, 2, 4, 8, 16, … with common ratio 2. The Pascal’s triangle programming guide covers how this pattern appears in coding rounds.

AM, GM, and HM: The Inequality That Solves One Problem Type

For any set of positive numbers: AM is greater than or equal to GM, which is greater than or equal to HM.

Equality holds only when all numbers in the set are identical.

Definitions for Two Positive Numbers a and b

• Arithmetic Mean: AM = (a + b) / 2
• Geometric Mean: GM = square root of (a x b)
• Harmonic Mean: HM = 2ab / (a + b)

Worked Example: AM, GM, HM for 4 and 9

• a = 4, b = 9
• AM = (4 + 9) / 2 = 6.5
• GM = square root of (4 x 9) = square root of 36 = 6
• HM = 2 x 4 x 9 / (4 + 9) = 72 / 13, approximately 5.54
• Confirmed: AM (6.5) >= GM (6) >= HM (approximately 5.54).

The Product Identity: AM x HM = GM Squared

This compact form solves the common aptitude question type: given two of the three means, find the third.

Placement test type: AM = 5, GM = 4. Find HM.

• HM = GM squared / AM = 16 / 5 = 3.2
• Verify: AM x HM = 5 x 3.2 = 16 = 4 squared = GM squared.

For structured practice on aptitude topics at the same level, see Clock Problems for Competitive Exams.

Everything in One Reference Table

Formula	Expression
Sum of first n naturals	n(n+1)/2
Sum of first n squares	n(n+1)(2n+1)/6
Sum of first n cubes	[n(n+1)/2] squared
AP nth term	a + (n-1)d
AP sum of n terms	n/2 x [2a + (n-1)d]
GP nth term	a x r to the power (n-1)
GP sum of n terms	a x (r to the n minus 1)/(r - 1)
GP infinite sum	a / (1-r), when absolute value of r is less than 1
AM x HM = GM squared	Derives any one mean from the other two

Re-deriving n(n+1)/2 by pairing terms rather than looking it up from a formula sheet is a useful discipline. TinkerLLM applies that same build-from-first-principles approach to language model concepts; the entry point is ₹299.