MathsISC Class 11 · Sequence and Series

AM–GM Relation, Σn, Σn², Σn³ and the Method of Differences

Updated 17 October 2026 · Fact-checked

For positive numbers, AM ≥ GM, with equality only when the numbers are equal. The special sums are Σn = n(n + 1)/2, Σn² = n(n + 1)(2n + 1)/6 and Σn³ = [n(n + 1)/2]². The method of differences finds the nth term of a series whose successive differences form an AP or GP, and then its sum.

The height (GM) can never be taller than the radius (AM); they are equal only when a = b and the height sits at the centre.

Illustration drawn for QuestionBuddy.

Understand AM–GM Inequality, Special Sums and the Method of Differences

AM and GM. For positive a and b: $A M = \frac{a + b}{2}$ , $GM = ab$ and $A M \geq GM$ , since $\frac{a + b}{2} - ab = \frac{( a - b ) ^{2}}{2} \geq 0$ . Equality holds if and only if a = b. Also $A M \cdot H M = G M^{2}$ for two numbers. For n positive numbers the same inequality holds.

Applications. If the sum of two positive numbers is fixed ( $a + b = s$ ), their product is at most $\frac{s ^{2}}{4}$ (when a = b = s/2). If the product is fixed, the sum is at least $2 p$ . For x > 0, $x + \frac{1}{x} \geq 2$ .

Special sums. $\sum_{r = 1}^{n} r = \frac{n ( n + 1 )}{2}$ ; $\sum r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ ; $\sum r^{3} = (\frac{n ( n + 1 )}{2})^{2}$ . Sums of series like $1 \cdot 2 + 2 \cdot 3 + ... + n (n + 1) = \frac{n ( n + 1 ) ( n + 2 )}{3}$ follow by splitting $r (r + 1) = r^{2} + r$ .

Method of differences. For a series $a_{1} + a_{2} + a_{3} + ...$ , if the differences $a_{2} - a_{1}, a_{3} - a_{2}, ...$ form an AP (or GP), find $T_{n}$ by writing $S = a_{1} + a_{2} + ... + a_{n}$ and subtracting a shifted copy of S. Alternatively, fit $T_{n} = A n^{2} + B n + C$ if the second differences are constant. Then find the sum of the series using the special sums.

Steps. (1) Find the general term $T_{n}$ as a polynomial in n. (2) Write $S_{n} = \sum T_{r}$ . (3) Substitute the special sums and simplify. (4) Check with n = 1 and n = 2.

AM–GM Inequality, Special Sums and the Method of Differences: formulas to remember

AM, GM: $$\frac{a + b}{2} ≥ \sqrt{ab}$ (a, b > 0)$; Equality when a = b.
Special sums: $\sum r = \frac{n(n+1)}{2}$ | $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$ | $\sum r^3 = [\frac{n(n+1)}{2}]^2$; r from 1 to n.
Products of consecutive integers: $\sum r(r + 1) = \frac{n(n+1)(n+2)}{3}$; From $r^{2} + r$ .
Constant second differences: $$T_n = An^2 + Bn + C$$; Fit using the first three terms.

AM–GM Inequality, Special Sums and the Method of Differences in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

The largest rectangle with fixed fencing

Suppose a farmer has 40 m of fencing for a rectangular plot.

Use AM ≥ GM to find the dimensions of the largest area.

Try it first, then show the solutionSolution

Length + breadth = 20. By AM ≥ GM, $l b \leq \frac{l + b}{2} = 10$ .
So $l b \leq 100$ , with equality when $l = b = 10$ .
The largest area is 100 m², when the plot is a square.

Answer: 10 m by 10 m: area 100 m².

Memory hook: A fixed sum gives the maximum product at equal parts.

Tricks to remember AM–GM Inequality, Special Sums and the Method of Differences

Short memory aids. Say each one aloud once, then use it on a question the same day.

Equal parts maximise the product
For a fixed sum, the product of positive numbers is greatest when they are equal.
Use it when: Optimisation without calculus.
Split r(r + 1)
Write r(r + 1) = r² + r to use standard sums.
Use it when: Sums of products of consecutive numbers.

AM–GM Inequality, Special Sums and the Method of Differences: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Sum of natural numbers · 1 mark · MCQBoard level

The sum $1 + 2 + 3 + ... + 100$ is:
(a) 5000
(b) 5050
(c) 5100
(d) 10100

Try it first, then show the solutionSolution

$\frac{100 \times 101}{2} = 5050$ .

Answer: (b) 5050

Example 2: Special sums · 3 marks · Short answerBoard level

Find the values of $1^{2} + 2^{2} + ... + 1 0^{2}$ and $1^{3} + 2^{3} + ... + 1 0^{3}$ .

Try it first, then show the solutionSolution

$\sum r^{2} = \frac{10 \times 11 \times 21}{6}$ . (1 mark)
$= 385$ . (0.5 mark)
$\sum r^{3} = (\frac{10 \times 11}{2})^{2} = 5 5^{2} = 3025$ . (1.5 marks)

Answer: 385; 3025.

Example 3: Applying AM ≥ GM · 4 marks · Short answerHarder board level

Prove that $\frac{a + b}{2} \geq ab$ for positive a and b. Two positive numbers have sum 20; find the greatest possible product, and find the least value of $x + \frac{1}{x}$ for x > 0.

Try it first, then show the solutionSolution

$\frac{a + b}{2} - ab = \frac{( a - b ) ^{2}}{2} \geq 0$ , equal only if a = b. (1.5 marks)
With $a + b = 20$ : $ab \leq (\frac{20}{2})^{2} = 100$ at a = b = 10. (1.5 marks)
$x + \frac{1}{x} \geq 2 x \cdot \frac{1}{x} = 2$ with equality at x = 1. (1 mark)

Answer: Proved; maximum product 100; least value 2.

Example 4: Sum of a product series · 4 marks · Short answerHarder board level

Find the sum of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + 20 \cdot 21$ .

Try it first, then show the solutionSolution

$T_{r} = r (r + 1) = r^{2} + r$ . (1 mark)
$S_{n} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2} = \frac{n ( n + 1 ) ( n + 2 )}{3}$ . (1.5 marks)
For n = 20: $\frac{20 \times 21 \times 22}{3} = 3080$ . (1.5 marks)

Answer: 3080

Example 5: Method of differences · 5 marks · Long answerHOTS / competency

Find the nth term and the sum of the first n terms of the series $3 + 7 + 13 + 21 + ...$ . Find the sum to 10 terms.

Try it first, then show the solutionSolution

The differences are 4, 6, 8, ... in AP, so the second difference is 2. (1 mark)
Fit $T_{n} = n^{2} + n + 1$ : check $T_{1} = 3$ , $T_{2} = 7$ , $T_{3} = 13$ , $T_{4} = 21$ . (1.5 marks)
$S_{n} = \sum (r^{2} + r + 1) = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2} + n = \frac{n ( n + 1 ) ( n + 2 )}{3} + n = \frac{n ( n ^{2} + 3 n + 5 )}{3}$ . (1.5 marks)
For n = 10: $385 + 55 + 10 = 450$ . (1 mark)

Answer: Tₙ = n² + n + 1; Sₙ = n(n² + 3n + 5)/3; S₁₀ = 450.

Where students lose marks in AM–GM Inequality, Special Sums and the Method of Differences

The slips examiners see most often, and what to write instead.

Mistake: Using AM ≥ GM for negative numbers
Applying the inequality to negative terms.
Do this: It needs positive numbers.
Usually costs: 2 marks.
Mistake: Mixing Σn² and (Σn)²
Writing Σn² as [n(n+1)/2]².
Do this: That is Σn³; Σn² = n(n+1)(2n+1)/6.
Usually costs: 1 to 2 marks.
Mistake: No check of the nth term
Not verifying the fitted formula.
Do this: Verify with the first three terms.
Usually costs: 1 mark.

Exam tips

State that the numbers are positive before using AM ≥ GM.
Write the general term first, then use the standard sums.
Verify the formula for Tₙ by substituting n = 1, 2, 3.
Use the sums as a check on small n.

Practice questions on AM–GM Inequality, Special Sums and the Method of Differences

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If the AM of two positive numbers is 10 and their GM is 8, their product is:
(a) 64
(b) 80
(c) 100
(d) 20
Hint
GM² = ab.
Answer
(a) 64
Q2 · 2 marks · Short answerBoard level
Find the sum of the first 15 odd natural numbers.
Hint
Sum of first n odd numbers is n².
Answer
$1 5^{2} = 225$ .
Q3 · 3 marks · Short answerHarder board level
Find the sum $1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + ...$ to 10 terms.
Hint
$T_{r} = (2 r - 1) (2 r + 1) = 4 r^{2} - 1$ .
Answer
$S_{10} = 4 \times 385 - 10 = 1530$ .
Q4 · 3 marks · Short answerHarder board level
The AM of two numbers is 25 and their GM is 24. Find the numbers.
Hint
a + b = 50, ab = 576.
Answer
The numbers satisfy $t^{2} - 50 t + 576 = 0$ , so $t = 32$ and $18$ ; check: AM $= 25$ , GM $= 576 = 24$ .
Q5 · 3 marks · Short answerHarder board level
Find the sum to n terms of $2 + 5 + 10 + 17 + ...$ and evaluate for n = 10.
Hint
$T_{r} = r^{2} + 1$ .
Answer
$S_{n} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + n$ ; for n = 10: $385 + 10 = 395$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a company cuts a rectangular sheet from a strip with perimeter 60 cm, and wants maximum area.
(i) State the sum of length and breadth. (1 mark)
(ii) State the maximum area using AM ≥ GM. (1 mark)
(iii) State the dimensions. (1 mark)
(iv) If the perimeter is 100 cm, find the maximum area. (1 mark)
Hint
l + b = half the perimeter.
Answer
(i) 30 cm. (ii) $l b \leq 1 5^{2} = 225$ cm². (iii) 15 cm by 15 cm. (iv) $l + b = 50$ , area at most $2 5^{2} = 625$ cm².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Σn?Show answer
n(n + 1)/2.
2.Σn³?Show answer
[n(n + 1)/2]².
3.Equality in AM ≥ GM?Show answer
When the numbers are equal.
4.Second differences constant means?Show answer
Quadratic nth term.

AM–GM Inequality, Special Sums and the Method of Differences: frequently asked questions

What is the relation AM · HM = GM²?

For two positive numbers, the product of their arithmetic and harmonic means equals the square of their geometric mean.

How can I find Tₙ if the differences form a GP?

Use the method of subtracting shifted sums, or the form T_n = A r^n + C (a Bn term appears only when the differences are a GP plus an AP).

Why is the sum of cubes the square of the sum?

It is the identity 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2.