MathsCBSE Class 11 · Sequences and Series

Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality (Class 11)

Q: Why must |r| be less than 1 for an infinite sum?

Because only then do the terms shrink to zero so that the sum settles at a finite value.

Q: Can the A.M.–G.M. inequality be used for three numbers?

Yes: (a + b + c)/3 ≥ ∛(abc) for positive numbers, with equality when a = b = c.

Q: Is A.M. ≥ G.M. proved in the syllabus?

Yes, the theorem and its one-line proof are in the chapter.

Q: Do I need the formula for the sum of squares of an infinite G.P.?

It follows from the basic formula applied to the G.P. with first term a² and ratio r², so you do not need to memorise it.

Updated 17 October 2026 · Fact-checked

An infinite G.P. with |r| < 1 has sum a/(1 − r). A recurring decimal is a finite part plus an infinite G.P., which gives the fraction. For positive a and b, A.M. = (a + b)/2 is at least G.M. = √(ab), with equality only when a = b; this gives least and greatest values, such as 2x + 18/x ≥ 12 for x > 0. Inserting n G.M.s between a and b uses r = (b/a)^(1/(n+1)).

The slanted radius (A.M.) is never shorter than the upright height (G.M.); they match only when the point sits at the centre, a = b.

Illustration drawn for QuestionBuddy.

Understand Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality

Infinite G.P. If $∣ r ∣ < 1$ , then $r^{n} \to 0$ and the sum of $n$ terms approaches
$ $S_{\infty} = \frac{a}{1 - r} .$ $
If $∣ r ∣ \geq 1$ the terms do not shrink, and there is no finite sum. Always check $∣ r ∣ < 1$ before using the formula, and also that it fits the question's data (for example the second term is $a r$ ).

Recurring decimals. Split off the non-repeating part, then sum the repeating part as an infinite G.P. with $r = 0.1, 0.01, \dots$ depending on the length of the repeating block. Example: $0.2 \overline{34} = 0.2 + 0.034 + 0.00034 + \dots = \frac{1}{5} + \frac{0.034}{1 - 0.01} = \frac{1}{5} + \frac{17}{495} = \frac{116}{495}$ .

Sum conditions on two series. If the sum of an infinite G.P. is $S$ and the sum of the squares of its terms is $T$ , the squares form a G.P. with first term $a^{2}$ and ratio $r^{2}$ , so $T = \frac{a ^{2}}{1 - r ^{2}}$ . Dividing $S^{2}$ by $T$ gives $\frac{1 + r}{1 - r}$ , which yields $r$ and then $a$ .

Geometric means. The G.M. of positive $a, b$ is $G = ab$ . To insert $n$ G.M.s between $a$ and $b$ build a G.P. of $n + 2$ terms: $b = a r^{n + 1}$ , so $r = (\frac{b}{a})^{\frac{1}{n + 1}}$ and the means are $a r, a r^{2}, \dots, a r^{n}$ . Their product is $(ab)^{n /2}$ .

**A.M. $\geq$ G.M.** For positive $a, b$ :
$ $\frac{a + b}{2} \geq ab, equality if and only if a = b .$ $
Proof: $\frac{a + b}{2} - ab = \frac{( a - b ) ^{2}}{2} \geq 0$ . For three positive numbers, $\frac{a + b + c}{3} \geq 3 ab c$ .

Using the inequality.

Least value of a sum when the product is fixed: $x + \frac{k}{x} \geq 2 k$ for $x > 0$ . For $2 x + \frac{18}{x}$ : $\geq 236 = 12$ at $2 x = \frac{18}{x}$ , i.e. $x = 3$ .
Greatest product when the sum is fixed: $x (20 - x) \leq (\frac{x + 20 - x}{2})^{2} = 100$ at $x = 10$ .
Inequalities between expressions: multiply the inequalities for the pairs: $(a + b) (b + c) (c + a) \geq 2 ab \cdot 2 b c \cdot 2 c a = 8 ab c$ .

Always state that the numbers are positive and when equality holds.

Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality: formulas to remember

Sum to infinity: $S = a / (1 - r), f or ∣ r ∣ l ess t han 1$; No finite sum if |r| is at least 1.
Sum of squares of an infinite G.P.: $a^{2} / (1 - r^{2})$; First term a², ratio r².
Geometric mean: $G = \sqrt (ab); n G . M . s : r a t i or = (b / a)^{(} 1/ (n + 1))$; Means are ar, ar², …, ar^n.
A.M. and G.M.: $(a + b) /2 \geq \sqrt (ab)$; Equality only when a = b.
Fixed product, least sum: $x + k / x \geq 2\sqrt k f or x > 0$; Equality at x = √k.

How to solve Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality questions

Use these steps for infinite series, recurring decimals and A.M.–G.M. problems.

1For an infinite G.P., find $a$ and $r$ , check $∣ r ∣ < 1$ , and use $S_{\infty} = \frac{a}{1 - r}$ .
2For a recurring decimal, separate the non-repeating part and write the repeating part as a G.P. with first term and ratio.
3For two conditions (sum and sum of squares), write both formulas and divide to eliminate $a$ .
4To insert means, find $r$ from $b = a r^{n + 1}$ and list $a r, a r^{2}, \dots$
5For extremum problems, write the expression as a sum of positive terms whose product is constant, apply A.M. $\geq$ G.M. and state when equality holds.
6Check the answer with a value (for example $x = 3$ for the minimum).

Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality 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.

A swing that dies out

Suppose a swing travels 50 cm on its first pass, and each later pass is 80% as long as the one before.

What is the greatest total distance it can travel?

Try it first, then show the solutionSolution

The passes form a G.P. with $a = 50$ and $r = 0.8$ .
$∣ r ∣ < 1$ , so the sum to infinity exists: $S_{\infty} = \frac{50}{1 - 0.8}$ .
$= \frac{50}{0.2} = 250$ cm.

Answer: 250 cm.

Memory hook: A shrinking ratio gives a finite total.

Largest area for a given fence

Suppose a farmer has 40 m of fence for a rectangular pen, with sides $x$ and $20 - x$ metres.

Use A.M. $\geq$ G.M. to find the greatest area and the shape that gives it.

Try it first, then show the solutionSolution

A.M. of the sides $= \frac{x + ( 20 - x )}{2} = 10$ .
G.M. $= x (20 - x) \leq 10$ , so the area $x (20 - x) \leq 100$ .
Equality when $x = 20 - x$ , i.e. $x = 10$ : a square of side 10 m with area 100 m².

Answer: 100 m², when the pen is a square.

Memory hook: Fixed sum: the product is greatest when the numbers are equal.

Tricks to remember Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality

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

Check |r| < 1
No finite sum for an infinite G.P. unless the ratio is smaller than 1 in size.
Use it when: Infinite series.
Pair them up
x + k/x: the product of the two terms is the constant k.
Use it when: A.M.–G.M. extremum questions.
Equal at the extremum
A.M. = G.M. exactly when all the terms are equal; solve that for the point.
Use it when: Where the minimum occurs.
Recurring: ratio 10^(−block)
A block of 2 repeating digits has ratio 0.01.
Use it when: Recurring decimals.

Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality: solved examples

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

Example 1: Sum to infinity · 2 marks · Very short answerBoard level

Find the sum to infinity of $6 + 2 + \frac{2}{3} + \dots$

Try it first, then show the solutionSolution

$a = 6$ , $r = \frac{1}{3}$ , and $∣ r ∣ < 1$ . (1 mark)
$S_{\infty} = \frac{6}{1 - \frac{1}{3}} = \frac{6}{2/3} = 9$ . (1 mark)

Answer: 9

Example 2: Recurring decimal · 3 marks · Short answerBoard level

Express $0.2 \overline{34} = 0.2343434 \dots$ as a fraction.

Try it first, then show the solutionSolution

$0.2 \overline{34} = 0.2 + 0.034 + 0.00034 + \dots$ (non-repeating part, then a G.P.). (1 mark)
The G.P. $0.034 + 0.00034 + \dots$ has $a = 0.034$ , $r = 0.01$ , sum $\frac{0.034}{0.99} = \frac{34}{990} = \frac{17}{495}$ . (1 mark)
Total $= \frac{1}{5} + \frac{17}{495} = \frac{99 + 17}{495} = \frac{116}{495}$ . (1 mark)

Answer: $\frac{116}{495}$

Example 3: Sum and sum of squares · 4 marks · Short answerHarder board level

The sum of an infinite G.P. is 15 and the sum of the squares of its terms is 45. Find the first term and the common ratio.

Try it first, then show the solutionSolution

$\frac{a}{1 - r} = 15$ and, since the squares form a G.P. with ratio $r^{2}$ , $\frac{a ^{2}}{1 - r ^{2}} = 45$ . (1 mark)
Square the first and divide: $\frac{a ^{2} / ( 1 - r ) ^{2}}{a ^{2} / ( 1 - r ^{2} )} = \frac{225}{45} = 5$ , so $\frac{1 + r}{1 - r} = 5$ . (1 mark)
$1 + r = 5 - 5 r$ , so $r = \frac{2}{3}$ ( $∣ r ∣ < 1$ ). (1 mark)
$a = 15 (1 - \frac{2}{3}) = 5$ . Check: $\frac{25}{1 - \frac{4}{9}} = 45$ . (1 mark)

Answer: $a = 5, r = \frac{2}{3}$

Example 4: Least value using A.M. ≥ G.M. · 3 marks · Short answerHarder board level

For $x > 0$ , find the least value of $2 x + \frac{18}{x}$ and the value of $x$ at which it occurs.

Try it first, then show the solutionSolution

A.M. $\geq$ G.M. for the positive numbers $2 x$ and $\frac{18}{x}$ : $\frac{1}{2} (2 x + \frac{18}{x}) \geq 2 x \cdot \frac{18}{x} = 6$ . (1 mark)
So $2 x + \frac{18}{x} \geq 12$ . (1 mark)
Equality when $2 x = \frac{18}{x}$ , i.e. $x^{2} = 9$ , $x = 3$ ; the least value is $12$ at $x = 3$ . (1 mark)

Answer: Least value 12 at $x = 3$

Example 5: A symmetric inequality · 5 marks · Long answerHOTS / competency

If $a, b, c$ are positive, prove that $(a + b) (b + c) (c + a) \geq 8 ab c$ . When does equality hold?

Try it first, then show the solutionSolution

A.M. $\geq$ G.M. for each pair: $a + b \geq 2 ab$ . (1 mark)
Similarly $b + c \geq 2 b c$ and $c + a \geq 2 c a$ . (1 mark)
All sides are positive, so multiply the three: $(a + b) (b + c) (c + a) \geq 8 ab \cdot b c \cdot c a$ . (1 mark)
$a^{2} b^{2} c^{2} = ab c$ , giving $(a + b) (b + c) (c + a) \geq 8 ab c$ . (1 mark)
Equality needs $a = b$ , $b = c$ and $c = a$ , i.e. $a = b = c$ (check: $2 \cdot 2 \cdot 2 = 8$ for $a = b = c = 1$ ). (1 mark)

Answer: Proved; equality when $a = b = c$

Where students lose marks in Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality

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

Mistake: Using a/(1 − r) with |r| not less than 1
The series does not settle to a finite value.
Wrong working
$1 + 2 + 4 + \dots = \frac{1}{1 - 2} = - 1$ .
Right working
No finite sum.
Do this: Write "|r| < 1" before using the formula.
Usually costs: The answer.
Mistake: Forgetting the non-repeating part of a decimal
Only the repeating block forms the G.P.
Wrong working
$0.2 \overline{34}$ written as $\frac{0.34}{0.99}$ .
Right working
$0.2 + \frac{0.034}{0.99}$ .
Do this: Separate the part before the repeating block.
Usually costs: The answer.
Mistake: Applying A.M. ≥ G.M. to negative numbers
The inequality is for positive numbers.
Wrong working
$x + \frac{1}{x} \geq 2$ for all real $x$ .
Right working
Only for $x > 0$ ; for $x < 0$ , $x + \frac{1}{x} \leq - 2$ .
Do this: State "for positive numbers" first.
Usually costs: The reasoning mark.
Mistake: Forgetting the equality condition
The least value is attained only when the terms are equal.
Wrong working
"Least value is 12" without saying where.
Right working
Least value 12 at $2 x = \frac{18}{x}$ , $x = 3$ .
Do this: Always solve A.M. = G.M. for the point.
Usually costs: 1 mark.

Exam tips

Check $∣ r ∣ < 1$ explicitly before every infinite-sum formula.
For recurring decimals, show the non-repeating part and the G.P. separately.
Name the positive quantities when applying A.M. $\geq$ G.M. and state when equality holds.
Verify the least value by substituting the point you found.
Round off nothing: give exact fractions for recurring decimals.

Practice questions on Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality

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
The sum to infinity of $12 + 4 + \frac{4}{3} + \dots$ is:
(a) 16
(b) 18
(c) 24
(d) 36
Hint
Here $r = \frac{1}{3}$ .
Answer
(b). $\frac{12}{1 - \frac{1}{3}} = 18$ .
Q2 · 1 mark · MCQHarder board level
The A.M. of two positive numbers is 10 and their G.M. is 8. The larger number is:
(a) 12
(b) 14
(c) 16
(d) 18
Hint
$a + b = 20$ and $ab = 64$ .
Answer
(c). The numbers satisfy $t^{2} - 20 t + 64 = 0$ , so $t = 4$ or $16$ .
Q3 · 3 marks · Short answerHarder board level
Insert 3 geometric means between 3 and 48.
Hint
Five-term G.P.: $48 = 3 r^{4}$ .
Answer
$r^{4} = 16$ , so $r = 2$ (positive for positive means). The means are $3 \cdot 2 = 6$ , $3 \cdot 4 = 12$ , $3 \cdot 8 = 24$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The sum of the series $1 + 2 + 4 + 8 + \dots$ is $\frac{1}{1 - 2} = - 1$ .
Reason (R): The formula $S_{\infty} = \frac{a}{1 - r}$ holds only when $∣ r ∣ < 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Here $r = 2$ .
Answer
(d). The terms grow without bound, so there is no finite sum; the formula is valid only for $∣ r ∣ < 1$ .
Q5 · 4 marks · Short answerHOTS / competency
A ball is dropped from a height of 10 m. After each fall it rebounds to $\frac{3}{5}$ of the height from which it fell. Find the total distance it travels before coming to rest.
Hint
Rebound heights form an infinite G.P.; each rebound is travelled up and down.
Answer
Rebound heights: $6, 3.6, \dots$ (a G.P. with $a = 6$ , $r = 0.6$ ), whose sum is $\frac{6}{0.4} = 15$ . Total distance $= 10 + 2 \times 15 = 40$ m.

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.What is S_infinity for |r| less than 1?Show answer
a/(1 − r).
2.What is the G.M. of a and b?Show answer
√(ab).
3.When does A.M. = G.M.?Show answer
When a = b.
4.What is the least value of x + 9/x for x positive?Show answer
6.
5.How do you insert n G.M.s between a and b?Show answer
r = (b/a)^(1/(n+1)); the means are ar, ar², …, ar^n.

Infinite G.P., Recurring Decimals and the A.M.–G.M. Inequality: frequently asked questions

Why must |r| be less than 1 for an infinite sum?

Because only then do the terms shrink to zero so that the sum settles at a finite value.

Can the A.M.–G.M. inequality be used for three numbers?

Yes: (a + b + c)/3 ≥ ∛(abc) for positive numbers, with equality when a = b = c.

Is A.M. ≥ G.M. proved in the syllabus?

Yes, the theorem and its one-line proof are in the chapter.

Do I need the formula for the sum of squares of an infinite G.P.?

It follows from the basic formula applied to the G.P. with first term a² and ratio r², so you do not need to memorise it.