Arithmetic, Geometric And Harmonic Progressions

MathsSequences And SeriesFor JEE aspirants

An arithmetic, geometric or harmonic progression is a sequence built from a single fixed rule: add a constant (AP), multiply by a constant (GP), or take reciprocals of an AP (HP). For any two positive numbers $a$ and $b$ , their arithmetic mean $A = (a + b) /2$ , geometric mean $G = ab$ , and harmonic mean $H = 2 ab / (a + b)$ are always related by the fundamental inequality $A \geq G \geq H$ , with equality only when $a = b$ . This concept forms the backbone of JEE Main and JEE Advanced questions on progressions.

Key Formulas - Quick Reference

AP: $n$ th term $a_{n} = a + (n - 1) d$ ; sum $S_{n} = \frac{n}{2} (2 a + (n - 1) d) = \frac{n}{2} (a + l)$
GP: $n$ th term $a_{n} = a r^{n - 1}$ ; sum $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ ( $r \neq = 1$ ); infinite sum $S_{\infty} = \frac{a}{1 - r}$ if $∣ r ∣ < 1$
HP: $n$ th term $a_{n} = \frac{1}{a + ( n - 1 ) d}$ where $a = 1/ a_{1}$ and $d$ is the common difference of the AP formed by reciprocals
Means of $a$ and $b$ : $A = \frac{a + b}{2}$ , $G = ab$ , $H = \frac{2 ab}{a + b}$
Relations: $G^{2} = A \cdot H$ and $A \geq G \geq H$ (positive reals)
Inserting means: Between $a$ and $b$ , $n$ AMs give common difference $d = \frac{b - a}{n + 1}$ ; $n$ GMs give common ratio $r = (\frac{b}{a})^{1/ (n + 1)}$

1. Sequence, Series, and Progression

A sequence is a succession of numbers formed by a definite rule, such as $1, 3, 5, 7, 9, \dots$ A sequence with a finite number of terms is a finite sequence; one that continues indefinitely is infinite.

A series is the sum of the terms of a sequence. If ${f_{n}}$ is a sequence, the expression $f_{1} + f_{2} + \dots + f_{n}$ is the corresponding series.

A progression is a sequence whose terms follow a specific algebraic condition. The three standard progressions studied in JEE are arithmetic, geometric, and harmonic.

2. Arithmetic Progression (AP)

An arithmetic progression is a sequence in which each term after the first differs from the previous term by a fixed number

d

, called the common difference. The general AP is

a, a + d, a + 2 d, \dots

2.1 $n$ th Term and Sum of $n$ Terms

If $a$ is the first term and $d$ the common difference:

a_{n} = a + (n - 1) d

S_{n} = \frac{n}{2} (2 a + (n - 1) d) = \frac{n}{2} (a + l)

where $l$ is the last term (the $n$ th term of the AP).

2.2 Useful Properties

Adding or subtracting a fixed number from every term of an AP gives another AP with the same $d$ .
Multiplying (or dividing) every term of an AP by a non-zero constant $k$ gives an AP with common difference $k d$ .
If ${a_{n}}$ and ${b_{n}}$ are APs with common differences $d_{1}$ and $d_{2}$ , then ${a_{n} + b_{n}}$ is an AP with common difference $d_{1} + d_{2}$ .
To take three terms in AP conveniently, use $a - d, a, a + d$ . For four terms, use $a - 3 d, a - d, a + d, a + 3 d$ .
If $a_{1}, a_{2}, \dots, a_{n}$ are in AP, then $a_{1} + a_{n} = a_{2} + a_{n - 1} = a_{3} + a_{n - 2} = \dots$ (equidistant terms from the ends sum to the same value).

2.3 Arithmetic Mean (AM)

If three quantities $a, b, c$ are in AP, the middle term is the arithmetic mean of the other two:

b = \frac{a + c}{2}

For $n$ numbers $a_{1}, a_{2}, \dots, a_{n}$ , the arithmetic mean is $A = \frac{1}{n} (a_{1} + a_{2} + \dots + a_{n})$ .

Inserting $n$ AMs between $a$ and $b$ : if $A_{1}, A_{2}, \dots, A_{n}$ are such that $a, A_{1}, A_{2}, \dots, A_{n}, b$ is an AP, the common difference is $d = \frac{b - a}{n + 1}$ , and

A_{k} = a + \frac{k ( b - a )}{n + 1} = \frac{( n + 1 - k ) a + k b}{n + 1}

Solved Example 1

If the 1st and 2nd terms of an AP are

1

and

- 3

respectively, find the

n

th term and the sum of the first

n

terms.

Solution:

Here $a = 1$ and $a + d = - 3$ , so $d = - 4$ .

$n$ th term: $a_{n} = 1 + (n - 1) (- 4) = 5 - 4 n$ .

Sum: $S_{n} = \frac{n}{2} (a + a_{n}) = \frac{n}{2} (1 + 5 - 4 n) = n (3 - 2 n)$ .

Solved Example 2

If 6 arithmetic means are inserted between

1

and

9/2

, find the 4th arithmetic mean.

Solution:

Let the six AMs be $A_{1}, A_{2}, \dots, A_{6}$ . Then $1, A_{1}, A_{2}, \dots, A_{6}, 9/2$ are in AP with 8 terms.

$\frac{9}{2} = 1 + 7 d \Rightarrow d = \frac{1}{2}$ .

Therefore $A_{4} = 1 + 4 \cdot \frac{1}{2} = 3$ .

Solved Example 3

Prove that

2, 3, 5

cannot be any three terms (not necessarily consecutive) of an AP.

Solution:

Suppose $2, 3, 5$ are the $n$ th, $m$ th, and $p$ th terms of an AP with first term $a$ and common difference $d$ .

Then $2 = a + (n - 1) d, 3 = a + (m - 1) d, 5 = a + (p - 1) d$ .

Subtracting: $\frac{2 - 3}{3 - 5} = \frac{n - m}{m - p}$ .

The left side is irrational; the right side is rational. This is a contradiction, so no such AP exists.

Solved Example 4

The interior angles of a polygon are in arithmetic progression. The smallest angle is

12 0^{\circ}

and the common difference is

5^{\circ}

. Find the number of sides.

Solution:

Let the polygon have $n$ sides. Sum of interior angles $= (n - 2) \cdot 18 0^{\circ}$ .

Sum of the AP $= \frac{n}{2} [2 (120) + (n - 1) (5)]$ .

Equating: $\frac{n}{2} [240 + 5 (n - 1)] = (n - 2) (180)$

$\Rightarrow n [48 + (n - 1)] = 72 (n - 2) \Rightarrow n^{2} - 25 n + 144 = 0 \Rightarrow (n - 9) (n - 16) = 0$ .

So $n = 9$ or $n = 16$ . For $n = 16$ , the largest angle $= 120 + 15 \cdot 5 = 19 5^{\circ} > 18 0^{\circ}$ , impossible. Hence $n = 9$ .

Solved Example 5

Let

S_{n}

denote the sum of the first

n

terms of an AP. If

S_{n} = n^{2} P

and

S_{m} = m^{2} P

where

m, n, P

are positive integers and

m \neq = n

, find

S_{P}

Solution:

Let first term be $a$ , common difference $d$ . Then

$S_{n} = \frac{n}{2} (2 a + (n - 1) d) = n^{2} P \Rightarrow 2 a + (n - 1) d = 2 n P$

$S_{m} = \frac{m}{2} (2 a + (m - 1) d) = m^{2} P \Rightarrow 2 a + (m - 1) d = 2 m P$

Subtracting: $(n - m) d = 2 (n - m) P \Rightarrow d = 2 P$ , hence $a = P$ .

Therefore $S_{P} = \frac{P}{2} (2 P + (P - 1) (2 P)) = \frac{P}{2} \cdot 2 P^{2} = P^{3}$ .

3. Geometric Progression (GP)

A geometric progression is a sequence whose first term is non-zero and each succeeding term is

r

times the preceding term, where

r \neq = 0

is the common ratio. The general GP is

a, a r, a r^{2}, \dots

3.1 $n$ th Term and Sum of $n$ Terms

a_{n} = a r^{n - 1}

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} = \frac{a ( 1 - r ^{n} )}{1 - r}, r \neq = 1

If $r = 1$ , then $S_{n} = na$ .

Infinite GP: if $∣ r ∣ < 1$ , the sum of the infinite series $a + a r + a r^{2} + \dots$ converges:

S_{\infty} = \frac{a}{1 - r}

3.2 Useful Properties

Multiplying (or dividing) every term of a GP by a non-zero constant gives a GP with the same common ratio.
Raising every term of a GP (common ratio $r$ ) to power $k$ gives a GP with common ratio $r^{k}$ .
The termwise product of two GPs (common ratios $r_{1}, r_{2}$ ) is a GP with common ratio $r_{1} r_{2}$ .
To take three terms in GP conveniently, use $a / r, a, a r$ . For four terms, use $a / r^{3}, a / r, a r, a r^{3}$ .
If $a_{1}, a_{2}, \dots, a_{n}$ are in GP, then $a_{1} a_{n} = a_{2} a_{n - 1} = a_{3} a_{n - 2} = \dots$
If $a_{1}, a_{2}, a_{3}, \dots$ is a GP of positive terms, then $lo g a_{1}, lo g a_{2}, lo g a_{3}, \dots$ is an AP (and conversely).

3.3 Geometric Mean (GM)

If $a, b, c$ are in GP, the middle term is the geometric mean: $b = a c$ .

For $n$ positive numbers, $G = (a_{1} a_{2} \dots a_{n})^{1/ n}$ .

Inserting $n$ GMs between $a$ and $b$ : if $a, G_{1}, G_{2}, \dots, G_{n}, b$ is a GP, then

r = (\frac{b}{a})^{1/ (n + 1)}, G_{k} = a (\frac{b}{a})^{k / (n + 1)}

Solved Example 6

The third term of a GP is

7

. Find the product of the first five terms.

Solution:

Let the terms be $a / r^{2}, a / r, a, a r, a r^{2}$ (with middle term as $a$ ). Given $a = 7$ .

Product $= \frac{a}{r ^{2}} \cdot \frac{a}{r} \cdot a \cdot a r \cdot a r^{2} = a^{5} = 7^{5} = 16807$ .

Solved Example 7

a^{x} = b^{y} = c^{z}

and

x, y, z

are in GP, show that

lo g_{b} a = lo g_{c} b

Solution:

Let $a^{x} = b^{y} = c^{z} = k$ . Taking logarithms: $x lo g a = y lo g b = z lo g c = lo g k$ .

So $x = \frac{lo g k}{lo g a}, y = \frac{lo g k}{lo g b}, z = \frac{lo g k}{lo g c}$ .

Since $x, y, z$ are in GP: $y^{2} = x z \Rightarrow \frac{( lo g k ) ^{2}}{( lo g b ) ^{2}} = \frac{( lo g k ) ^{2}}{lo g a \cdot lo g c}$ .

$\Rightarrow (lo g b)^{2} = lo g a \cdot lo g c \Rightarrow \frac{lo g a}{lo g b} = \frac{lo g b}{lo g c} \Rightarrow lo g_{b} a = lo g_{c} b$ .

Solved Example 8

The sum of three numbers in GP is

21

and the sum of their squares is

189

. Find the numbers.

Solution:

Let the numbers be $a, a r, a r^{2}$ .

$a (1 + r + r^{2}) = 21 \dots (1)$

$a^{2} (1 + r^{2} + r^{4}) = 189 \dots (2)$

Squaring (1) and dividing (2) by it:

\frac{1 + r ^{2} + r ^{4}}{( 1 + r + r ^{2} ) ^{2}} = \frac{189}{441} = \frac{3}{7}

Using $1 + r^{2} + r^{4} = (1 + r + r^{2}) (1 - r + r^{2})$ :

\frac{1 - r + r ^{2}}{1 + r + r ^{2}} = \frac{3}{7} \Rightarrow 7 (1 - r + r^{2}) = 3 (1 + r + r^{2})

$\Rightarrow 4 r^{2} - 10 r + 4 = 0 \Rightarrow 2 r^{2} - 5 r + 2 = 0 \Rightarrow r = 2$ or $r = 1/2$ .

With $r = 2$ : $a \cdot 7 = 21 \Rightarrow a = 3$ , giving $3, 6, 12$ . With $r = 1/2$ : numbers are $12, 6, 3$ .

Solved Example 9

Find the sum of the infinite series

(1 + \frac{1}{2 ^{2}}) + (\frac{1}{2} + \frac{1}{2 ^{4}}) + (\frac{1}{2 ^{2}} + \frac{1}{2 ^{6}}) + \dots

Solution:

Split into two infinite GPs:

S = (1 + \frac{1}{2} + \frac{1}{2 ^{2}} + \dots) + (\frac{1}{2 ^{2}} + \frac{1}{2 ^{4}} + \frac{1}{2 ^{6}} + \dots)

First GP: $a = 1, r = 1/2$ , sum $= \frac{1}{1 - 1/2} = 2$ .

Second GP: $a = 1/4, r = 1/4$ , sum $= \frac{1/4}{1 - 1/4} = \frac{1}{3}$ .

Total $S = 2 + \frac{1}{3} = \frac{7}{3}$ .

4. Harmonic Progression (HP)

A sequence

a_{1}, a_{2}, a_{3}, \dots

(with all

a_{i} \neq = 0

) is a harmonic progression if the sequence of reciprocals

\frac{1}{a _{1}}, \frac{1}{a _{2}}, \frac{1}{a _{3}}, \dots

is an arithmetic progression.

4.1 $n$ th Term of HP

If $1/ a_{1} = A$ and the common difference of the reciprocal AP is $D$ , then the $n$ th term of the HP is

a_{n} = \frac{1}{A + ( n - 1 ) D}

Note: There is no closed formula for the sum of

n

terms of an HP. Problems on HP are almost always solved by converting to the corresponding AP of reciprocals.

4.2 Harmonic Mean (HM)

If $a, H, b$ are in HP, the harmonic mean $H$ of $a$ and $b$ satisfies

\frac{1}{H} = \frac{1}{2} (\frac{1}{a} + \frac{1}{b}) \Rightarrow H = \frac{2 ab}{a + b}

For $n$ positive numbers, the harmonic mean $H$ satisfies $\frac{1}{H} = \frac{1}{n} (\frac{1}{a _{1}} + \frac{1}{a _{2}} + \dots + \frac{1}{a _{n}})$ .

Inserting $n$ HMs between $a$ and $b$ : if $a, H_{1}, H_{2}, \dots, H_{n}, b$ is an HP, then the reciprocals form an AP, and

\frac{1}{H _{k}} = \frac{1}{a} + \frac{k ( a - b )}{( n + 1 ) ab} = \frac{( n + 1 - k ) b + k a}{ab ( n + 1 )}

Solved Example 10

Find the 4th and the 8th terms of the HP

6, 4, 3, \dots

Solution:

Reciprocals: $\frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \dots$ Check: $\frac{1}{4} - \frac{1}{6} = \frac{1}{12}$ and $\frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ . So this is an AP with $A = 1/6, D = 1/12$ .

4th term of the AP $= \frac{1}{6} + 3 \cdot \frac{1}{12} = \frac{5}{12}$ ; 8th term $= \frac{1}{6} + 7 \cdot \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$ .

Therefore the 4th term of the HP $= \frac{12}{5}$ and the 8th term $= \frac{4}{3}$ .

Solved Example 11

a, x, y, z, b

are in AP with

x + y + z = 15

, and

a, x, y, z, b

are in HP with

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}

, find

a

and

b

Solution:

When $a, x, y, z, b$ are in AP: $x + y + z = 3 \cdot \frac{a + b}{2}$ (three AMs between $a$ and $b$ ).

$\Rightarrow 15 = \frac{3 ( a + b )}{2} \Rightarrow a + b = 10 \dots (1)$

When $a, x, y, z, b$ are in HP: $\frac{1}{a}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{b}$ are in AP.

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2} (\frac{1}{a} + \frac{1}{b}) \Rightarrow \frac{5}{3} = \frac{3 ( a + b )}{2 ab} = \frac{30}{2 ab}$

$\Rightarrow ab = 9 \dots (2)$

Then $(a - b)^{2} = (a + b)^{2} - 4 ab = 100 - 36 = 64 \Rightarrow a - b = \pm 8$ .

Combining with (1): ${a, b} = {9, 1}$ .

5. Relationship Between AM, GM, and HM

5.1 Fundamental Inequality: $A \geq G \geq H$

For any $n$ positive real numbers $a_{1}, a_{2}, \dots, a_{n}$ :

A = \frac{a _{1} + a _{2} + \dots + a _{n}}{n} \geq G = (a_{1} a_{2} \dots a_{n})^{1/ n} \geq H = \frac{n}{\frac{1}{a _{1}} + \frac{1}{a _{2}} + \dots + \frac{1}{a _{n}}}

Equality at any place holds if and only if $a_{1} = a_{2} = \dots = a_{n}$ .

For just two positive numbers $a$ and $b$ , the three means satisfy the beautiful identity

G^{2} = A \cdot H

5.2 Weighted Means

Given positive reals $a_{1}, \dots, a_{n}$ and positive rational weights $m_{1}, \dots, m_{n}$ , define

A^{*} = \frac{m _{1} a _{1} + m _{2} a _{2} + \dots + m _{n} a _{n}}{m _{1} + m _{2} + \dots + m _{n}}

G^{*} = (a_{1}^{m_{1}} \cdot a_{2}^{m_{2}} \dots a_{n}^{m_{n}})^{1/ (m_{1} + m_{2} + \dots + m_{n})}

H^{*} = \frac{m _{1} + m _{2} + \dots + m _{n}}{\frac{m _{1}}{a _{1}} + \frac{m _{2}}{a _{2}} + \dots + \frac{m _{n}}{a _{n}}}

The same inequality holds: $A^{*} \geq G^{*} \geq H^{*}$ , with equality iff all $a_{i}$ are equal.

5.3 Arithmetic Mean of $m$ th Powers

Let $a_{1}, a_{2}, \dots, a_{n}$ be positive reals (not all equal), and let $m$ be a real number. Then:

$\frac{a _{1}^{m} + a _{2}^{m} + \dots + a _{n}^{m}}{n} > (\frac{a _{1} + a _{2} + \dots + a _{n}}{n})^{m}$ when $m \in R ∖ [0, 1]$
$\frac{a _{1}^{m} + a _{2}^{m} + \dots + a _{n}^{m}}{n} < (\frac{a _{1} + a _{2} + \dots + a _{n}}{n})^{m}$ when $m \in (0, 1)$
Equality when $m \in {0, 1}$

Solved Example 12

Show that

x + \frac{1}{x} \geq 2

x > 0

, and

x + \frac{1}{x} \leq - 2

x < 0

Solution:

Case 1: $x > 0$ . Apply AM $\geq$ GM to $x$ and $1/ x$ :

\frac{x + 1/ x}{2} \geq x \cdot \frac{1}{x} = 1 \Rightarrow x + \frac{1}{x} \geq 2

Case 2: $x < 0$ . Let $y = - x$ , so $y > 0$ . Then $y + 1/ y \geq 2$ , i.e. $- x - \frac{1}{x} \geq 2$ , giving $x + \frac{1}{x} \leq - 2$ .

Solved Example 13

a_{i}

are all positive reals, prove that

(1 + a_{1} + a_{1}^{2}) (1 + a_{2} + a_{2}^{2}) \dots (1 + a_{n} + a_{n}^{2}) \geq 3^{n} a_{1} a_{2} \dots a_{n}

Solution:

Apply AM $\geq$ GM to the three positive quantities $1, a_{i}, a_{i}^{2}$ for each $i$ :

\frac{1 + a _{i} + a _{i}^{2}}{3} \geq (1 \cdot a_{i} \cdot a_{i}^{2})^{1/3} = a_{i}

$\Rightarrow 1 + a_{i} + a_{i}^{2} \geq 3 a_{i}$ for every $i$ .

Multiplying these $n$ inequalities (all sides positive):

i = 1 \prod n (1 + a_{i} + a_{i}^{2}) \geq 3^{n} i = 1 \prod n a_{i} = 3^{n} a_{1} a_{2} \dots a_{n}

Solved Example 14

x, y, z

are positive reals with

x + y + z = a

, prove that

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{a}

Solution:

Apply AM $\geq$ HM to $x, y, z$ :

\frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x + y + z} = \frac{9}{a}$ .

Solved Example 15

a, b, c

are positive reals with

a + b + c = 1

, prove that

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}

Solution:

Add $1$ to each term on the left. It is equivalent to prove

(\frac{a}{b + c} + 1) + (\frac{b}{c + a} + 1) + (\frac{c}{a + b} + 1) \geq \frac{3}{2} + 3

i.e. $(a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b}) \geq \frac{9}{2}$ , i.e. $\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} \geq \frac{9}{2}$ .

Apply AM $\geq$ HM to $b + c, c + a, a + b$ :

\frac{( b + c ) + ( c + a ) + ( a + b )}{3} \geq \frac{3}{\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b}}

$\Rightarrow \frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b} \geq \frac{9}{2 ( a + b + c )} = \frac{9}{2}$ . $□$

Common Mistakes to Avoid

Watch out

Applying AM $\geq$ GM $\geq$ HM to negative or zero values. The inequality only holds for strictly positive reals. Always verify positivity first.
Forgetting the equality condition. Equality in $A \geq G \geq H$ requires all terms equal, not just some. Skipping this loses marks in "prove and state when equality holds" style questions.
Adding formulas for HP directly. There is no closed sum formula for HP. Always convert to the reciprocal AP first.
Choosing bad symmetric substitution. Using $a, a r, a r^{2}$ for three GP terms leads to messier algebra than $a / r, a, a r$ when the product is fixed. Pick symmetry to match the given condition.
Missing the sign case in $x + 1/ x$ . The inequality flips for $x < 0$ (giving $\leq - 2$ ). Always split into positive and negative cases if the domain is not specified.
Confusing AM insertion count. If $n$ AMs are inserted between $a$ and $b$ , the resulting AP has $n + 2$ terms and common difference $\frac{b - a}{n + 1}$ , not $\frac{b - a}{n}$ .

Frequently Asked Questions

Q1. What is the difference between AP, GP, and HP with examples?

An AP grows by adding a fixed number: $2, 5, 8, 11, \dots$ (common difference $3$ ). A GP grows by multiplying by a fixed number: $2, 6, 18, 54, \dots$ (common ratio $3$ ). An HP is a sequence whose reciprocals form an AP: $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \dots$ is an HP because its reciprocals $2, 5, 8, 11, \dots$ form an AP.

Q2. What is the relation between AM, GM, and HM for two numbers?

For two positive reals $a$ and $b$ , the three means satisfy $A \geq G \geq H$ , with equality only when $a = b$ . They also satisfy the identity $G^{2} = A \cdot H$ , so the GM is the geometric mean of the AM and HM themselves.

Q3. How many arithmetic means are inserted between two numbers if the resulting AP has $k$ terms?

If the resulting AP has $k$ terms starting at $a$ and ending at $b$ , then $k - 2$ arithmetic means are inserted. The common difference of the AP is $\frac{b - a}{k - 1} = \frac{b - a}{n + 1}$ where $n = k - 2$ is the number of inserted AMs.

Q4. Why is there no formula for the sum of an HP?

The reciprocals of HP terms form an AP, but summing the reciprocals of an AP does not simplify to a closed form. HP problems are always converted to the reciprocal AP and solved there. In cases where a specific HP has telescoping structure (e.g. $\frac{1}{n ( n + 1 )}$ ), summation is possible via partial fractions, but this is not a general formula.

Q5. When does an infinite GP have a finite sum?

An infinite GP $a + a r + a r^{2} + \dots$ has a finite sum $\frac{a}{1 - r}$ if and only if $∣ r ∣ < 1$ . If $∣ r ∣ \geq 1$ , the series diverges (either grows without bound or oscillates). This convergence condition is a common JEE question in decimal-to-fraction and repeating-pattern problems.

Q6. Is AM $\geq$ GM applicable to any real numbers?

No. AM $\geq$ GM applies only to positive reals. For example, taking $a = - 1, b = - 4$ gives AM $= - 2.5$ but GM $= (- 1) (- 4) = 2$ , so AM $<$ GM. Always check positivity before applying the inequality in a proof.

Q7. How is $G^{2} = A \cdot H$ derived for two numbers?

For positive reals $a$ and $b$ : $A = \frac{a + b}{2}$ , $G = ab$ , $H = \frac{2 ab}{a + b}$ . So $A \cdot H = \frac{a + b}{2} \cdot \frac{2 ab}{a + b} = ab = G^{2}$ . This identity means $G$ is itself the geometric mean of $A$ and $H$ .

Q8. What is the standard trick for choosing terms in AP or GP?

For AP: three terms as $a - d, a, a + d$ ; four terms as $a - 3 d, a - d, a + d, a + 3 d$ . For GP: three terms as $a / r, a, a r$ ; four terms as $a / r^{3}, a / r, a r, a r^{3}$ . Symmetric choices make the given sum or product conditions cleaner because middle terms simplify.

Previous year questions on Arithmetic, Geometric And Harmonic Progressions

43 questions from past papers, each with a step-by-step solution.

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1. Sequence, Series, and Progression

2. Arithmetic Progression (AP)

2.1 nth Term and Sum of n Terms

2.2 Useful Properties

2.3 Arithmetic Mean (AM)

3. Geometric Progression (GP)

3.1 nth Term and Sum of n Terms

3.2 Useful Properties

3.3 Geometric Mean (GM)

4. Harmonic Progression (HP)

4.1 nth Term of HP

4.2 Harmonic Mean (HM)

5. Relationship Between AM, GM, and HM

5.1 Fundamental Inequality: A≥G≥H

5.2 Weighted Means

5.3 Arithmetic Mean of mth Powers

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between AP, GP, and HP with examples?

Q2. What is the relation between AM, GM, and HM for two numbers?

Q3. How many arithmetic means are inserted between two numbers if the resulting AP has k terms?

Q4. Why is there no formula for the sum of an HP?

Q5. When does an infinite GP have a finite sum?

Q6. Is AM ≥ GM applicable to any real numbers?

Q7. How is G2=A⋅H derived for two numbers?

Q8. What is the standard trick for choosing terms in AP or GP?

Previous year questions on Arithmetic, Geometric And Harmonic Progressions

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2.1 $n$ th Term and Sum of $n$ Terms

3.1 $n$ th Term and Sum of $n$ Terms

4.1 $n$ th Term of HP

5.1 Fundamental Inequality: $A \geq G \geq H$

5.3 Arithmetic Mean of $m$ th Powers

Q3. How many arithmetic means are inserted between two numbers if the resulting AP has $k$ terms?

Q6. Is AM $\geq$ GM applicable to any real numbers?

Q7. How is $G^{2} = A \cdot H$ derived for two numbers?