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Harmonic and Special Series
Harmonic Progression (HP)
A sequence is a Harmonic Progression if the reciprocals of its terms form an Arithmetic Progression. The general form is $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \dots$
$$H_n = \frac{1}{a + (n-1)d}$$
The nth term of an HP is the reciprocal of the nth term of the corresponding AP
$H_n$=nth term of the HP
$a$=First term of the corresponding AP
$d$=Common difference of the corresponding AP
$n$=Position of the term
$a + (n-1)d = 0$
→The term is undefined — HP cannot contain zero
Reciprocal Link: To solve any HP problem, convert terms to their reciprocals to form an AP, solve in the AP domain, then convert back.
No Zero Rule: Since $1/0$ is undefined, zero can never be a term of an HP.
No Direct Formula: There is no direct formula for the sum of an HP — always convert to AP first.
Harmonic Mean (HM)
The Harmonic Mean $H$ between two numbers $a$ and $b$ is defined so that $a, H, b$ are in HP. Equivalently, $\frac{1}{a}, \frac{1}{H}, \frac{1}{b}$ must be in AP.
$$H = \frac{2ab}{a+b}$$
The HM is the reciprocal of the arithmetic mean of the reciprocals
$H$=Harmonic Mean between $a$ and $b$
$a, b$=The two numbers
$a = b$
→$H = a = b$ (all three means coincide)
$a + b = 0$
→$H$ is undefined (division by zero)
Derivation: Since $\frac{1}{a}, \frac{1}{H}, \frac{1}{b}$ are in AP, $\frac{1}{H} - \frac{1}{a} = \frac{1}{b} - \frac{1}{H}$, which gives $\frac{2}{H} = \frac{1}{a} + \frac{1}{b}$.
Verification Shortcut: For HM between 3 and 7: $H = \frac{2(3)(7)}{3+7} = \frac{42}{10} = \frac{21}{5}$. Check: $\frac{1}{3}, \frac{5}{21}, \frac{1}{7}$ has common difference $\frac{5}{21} - \frac{1}{3} = \frac{-2}{21}$ and $\frac{1}{7} - \frac{5}{21} = \frac{-2}{21}$.
To insert $n$ Harmonic Means between two numbers $a$ and $b$, first insert $n$ AMs between $\frac{1}{a}$ and $\frac{1}{b}$, then take their reciprocals.
$$d = \frac{\frac{1}{b} - \frac{1}{a}}{n+1} = \frac{a-b}{ab(n+1)}$$
The common difference of the AP formed by $\frac{1}{a}, A_1, A_2, \dots, A_n, \frac{1}{b}$
$d$=Common difference of the reciprocal AP
$n$=Number of HMs to insert
$a, b$=The boundary terms of the HP
Step-by-Step Method: (1) Take reciprocals $\frac{1}{a}$ and $\frac{1}{b}$. (2) Find $d = \frac{\frac{1}{b} - \frac{1}{a}}{n+1}$. (3) The AMs are $\frac{1}{a} + d, \frac{1}{a} + 2d, \dots$ (4) Take reciprocals of these AMs.
General $k$th HM: The $k$th harmonic mean is $\frac{ab(n+1)}{(n+1-k)b + ka}$.
Relations between AM, GM, and HM
For any two numbers $a$ and $b$, the Arithmetic Mean $A$, Geometric Mean $G$, and Harmonic Mean $H$ satisfy a fundamental identity that links all three.
$$G^2 = A \cdot H$$
The three Pythagorean means are connected — A, G, H are always in GP
$A$=$\frac{a+b}{2}$
$G$=$\pm\sqrt{ab}$
$H$=$\frac{2ab}{a+b}$
GP Connection: Since $G^2 = AH$, the means $A, G, H$ are themselves in Geometric Progression.
Universal: The identity $G^2 = AH$ holds even when $a, b$ are complex numbers.
Finding Numbers: If any two of $A$, $G$, $H$ are known, the two numbers $a$ and $b$ satisfy $a + b = 2A$ and $ab = G^2 = AH$, so $a$ and $b$ are roots of $x^2 - 2Ax + AH = 0$.
For any two distinct positive real numbers, the three means always satisfy a strict ordering.
$$A > G > H \quad (a, b > 0, \; a \neq b)$$
AM is always the largest and HM the smallest for distinct positive reals
$A > G$=Equivalent to $(\sqrt{a} - \sqrt{b})^2 > 0$
$G > H$=Equivalent to $a + b > 2\sqrt{ab}$, same inequality
Proof Sketch for $A > G$: $\frac{a+b}{2} > \sqrt{ab} \Leftrightarrow a + b - 2\sqrt{ab} > 0 \Leftrightarrow (\sqrt{a} - \sqrt{b})^2 > 0$, which is true for distinct positive reals.
Equality Case: $A = G = H$ if and only if $a = b$ (all means coincide for equal numbers).
Negative Numbers: For two distinct negative reals with $G = -\sqrt{ab}$, the ordering reverses: $A < G < H$.
Verification Strategy: Given $a$ and $b$, compute all three means and check the ordering. For $a = 2, b = 8$: $A = 5$, $G = 4$, $H = \frac{16}{5} = 3.2$, confirming $5 > 4 > 3.2$.
Sigma Notation and Summation Properties
The Greek letter $\sum$ (sigma) provides a compact way to express sums. The notation $\sum_{k=m}^{n} a_k$ means $a_m + a_{m+1} + \dots + a_n$.
$$\sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n$$
A compact notation for expressing the sum of a sequence of terms
$k$=Index of summation (dummy variable)
$m, n$=Lower and upper limits of summation
$a_k$=General term expressed as a function of $k$
Linearity: $\sum (ca_k + b_k) = c\sum a_k + \sum b_k$ — constants factor out, sums split.
Constant Sum: $\sum_{k=1}^{n} c = nc$ — a constant summed $n$ times is $nc$.
Dummy Variable: The index letter is arbitrary — $\sum_{k=1}^{n} k^2 = \sum_{j=1}^{n} j^2 = \sum_{i=1}^{n} i^2$.
Three fundamental summation formulas for powers of natural numbers are derived using the telescoping identity $\sum_{k=1}^{n}[k^m - (k-1)^m] = n^m$.
$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$
Closed-form expression for the sum of squares of the first n natural numbers
$\sum k^2$=Sum of squares: $1^2 + 2^2 + \dots + n^2$
$n$=Number of terms
Sum of $n$: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ — the triangular numbers.
Sum of Cubes: $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$ — exactly the square of the sum of $n$.
Telescoping Derivation: Setting $m = 2$ in $\sum[k^2 - (k-1)^2] = n^2$ gives $\sum(2k - 1) = n^2$, which leads to $2\sum k - n = n^2$, so $\sum k = \frac{n(n+1)}{2}$.
Quick Check: For $n = 3$: $\sum k = 6$ vs $\frac{3(4)}{2} = 6$; $\sum k^2 = 14$ vs $\frac{3(4)(7)}{6} = 14$; $\sum k^3 = 36$ vs $6^2 = 36$.
Telescoping Series (Method of Differences)
The Method of Differences simplifies a series by expressing each term $T_r$ as a difference $V_r - V_{r-1}$, causing intermediate terms to cancel.
$$T_r = V_r - V_{r-1} \implies S_n = V_n - V_0$$
The sum collapses to just the first and last values of the decomposition function
$T_r$=General term of the original series
$V_r$=The decomposition function chosen so that $T_r = V_r - V_{r-1}$
$S_n$=Sum of first $n$ terms
Cancellation Mechanism: $(V_1 - V_0) + (V_2 - V_1) + (V_3 - V_2) + \dots + (V_n - V_{n-1}) = V_n - V_0$.
Partial Fractions: The key technique — decompose $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$, so $V_r = -\frac{1}{r+1}$.
Limiting Case: As $n \to \infty$, if $V_n \to L$, then $S_\infty = L - V_0$. For $\sum \frac{1}{r(r+1)}$: $S_\infty = 0 - (-1) = 1$.
Application Strategy
Worked Examples
Problem Statement
If the 4th and 7th terms of an HP are $\frac{2}{13}$ and $\frac{2}{25}$ respectively, find the HP sequence.
HP nth termReciprocal conversion to AP
Strategic Toolkit
$$a_n = a_1 + (n-1)d$$
Step 1
Convert to AP
Take reciprocals of the HP terms to get the corresponding AP terms.
$A_4 = \frac{13}{2}, \quad A_7 = \frac{25}{2}$
Step 2
Find Common Difference
Use the AP formula: $A_7 - A_4 = 3d$.
$3d = \frac{25}{2} - \frac{13}{2} = 6 \implies d = 2$
Step 3
Find First Term
From $A_4 = a_1 + 3d$.
$a_1 = \frac{13}{2} - 3(2) = \frac{13}{2} - 6 = \frac{1}{2}$
Step 4
Write the HP
The AP is $\frac{1}{2}, \frac{5}{2}, \frac{9}{2}, \frac{13}{2}, \dots$ Take reciprocals.
HP: $2, \frac{2}{5}, \frac{2}{9}, \frac{2}{13}, \dots$
Problem Statement
Insert three harmonic means between $\frac{1}{5}$ and $\frac{1}{17}$.
n HMs insertionReciprocal method
Strategic Toolkit
$$d = \frac{\frac{1}{b} - \frac{1}{a}}{n+1}$$
Step 1
Set Up Reciprocal AP
Take reciprocals: we need 3 AMs between 5 and 17.
$5, A_1, A_2, A_3, 17$ are in AP
Step 2
Find Common Difference
With $n+2 = 5$ total terms, $a_5 = a_1 + 4d$.
$17 = 5 + 4d \implies d = 3$
Step 3
Compute AMs
Add $d$ successively to find the arithmetic means.
$A_1 = 8, \quad A_2 = 11, \quad A_3 = 14$
Step 4
Take Reciprocals
The harmonic means are the reciprocals of the AMs.
HMs: $\frac{1}{8}, \frac{1}{11}, \frac{1}{14}$
Problem Statement
For $a = 9$ and $b = 4$, find A, G, H and verify $G^2 = AH$.
AM-GM-HM relationshipG² = AH identity
Strategic Toolkit
$$A = \frac{a+b}{2}, \; G = \sqrt{ab}, \; H = \frac{2ab}{a+b}$$
Step 1
Calculate AM
Arithmetic Mean of 9 and 4.
$A = \frac{9+4}{2} = \frac{13}{2}$
Step 2
Calculate GM
Geometric Mean of 9 and 4.
$G = \sqrt{9 \times 4} = \sqrt{36} = 6$
Step 3
Calculate HM
Harmonic Mean of 9 and 4.
$H = \frac{2(9)(4)}{9+4} = \frac{72}{13}$
Step 4
Verify Identity
Check $G^2 = AH$.
$G^2 = 36$; $AH = \frac{13}{2} \times \frac{72}{13} = \frac{72}{2} = 36$ ✓. Also $\frac{13}{2} > 6 > \frac{72}{13} \approx 5.54$, so $A > G > H$ ✓
Problem Statement
Evaluate $\sum_{k=1}^{10} k(k+1)$.
Sigma notationSum of squaresSum of n
Strategic Toolkit
$$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$$
Step 1
Expand the General Term
Write $k(k+1) = k^2 + k$.
$\sum_{k=1}^{10} k(k+1) = \sum_{k=1}^{10} k^2 + \sum_{k=1}^{10} k$
Step 2
Apply Formulas
Use the standard summation formulas with $n = 10$.
$\sum k^2 = \frac{10(11)(21)}{6} = 385$; $\sum k = \frac{10(11)}{2} = 55$
Step 3
Combine
Add the two results.
$S = 385 + 55 = 440$
Memorize formulas
HP nth Term
General term of a Harmonic Progression.
Formula
$H_n = \frac{1}{a + (n-1)d}$
Harmonic Mean
HM between two numbers a and b.
Formula
$H = \frac{2ab}{a+b}$
Means Identity
Product of AM and HM equals square of GM.
Formula
$G^2 = A \cdot H$
Sum of n
Sum of first n natural numbers.
Formula
$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
Sum of Squares
Sum of squares of first n natural numbers.
Formula
$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
Sum of Cubes
Sum of cubes equals square of sum of n.
Formula
$\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$
Review
1
If the 4th and 7th terms of an HP are $\frac{2}{13}$ and $\frac{2}{25}$ respectively, what is the first term of the HP?
Ans
$2$
Review
Tap to reveal
Convert the HP to an AP of reciprocals
If the given terms are in harmonic progression, their reciprocals are in arithmetic progression. So the 4th and 7th terms become $\frac{13}{2}$ and $\frac{25}{2}$ in an AP.
Backtrack to the first term
The common difference is $d = \frac{25/2 - 13/2}{3} = 2$. Hence the first AP term is $\frac{13}{2} - 3(2) = \frac{1}{2}$, so the first HP term is its reciprocal.
$$a_1 = \frac{1}{1/2} = 2$$
2
If the (positive) GM and HM between two numbers are 4 and $\frac{16}{5}$ respectively, find the two numbers.
Ans
$2$ and $8$
Review
Tap to reveal
Use the GM-HM-AM identity
For two numbers, $G^2 = AH$. With $G = 4$ and $H = \frac{16}{5}$, the arithmetic mean is forced by the identity.
Recover the pair from sum and product
Compute $A = \frac{G^2}{H} = \frac{16}{16/5} = 5$, so $a+b = 10$ and $ab = 16$. The numbers solve $x^2 - 10x + 16 = 0$.
$$(x-2)(x-8) = 0 \Rightarrow \{a,b\} = \{2,8\}$$
3
Three harmonic means are inserted between $\frac{1}{5}$ and $\frac{1}{17}$. What is the second harmonic mean?
Ans
$\frac{1}{11}$
Review
Tap to reveal
Work with reciprocals first
Harmonic means become arithmetic means after taking reciprocals. So insert three AMs between $5$ and $17$ instead.
Locate the middle inserted value
With five AP terms total, $d = (17 - 5)/4 = 3$, giving $5, 8, 11, 14, 17$. The reciprocals of the inserted terms are the harmonic means.
$$\frac{1}{8},\ \frac{1}{11},\ \frac{1}{14} \Rightarrow \text{second HM} = \frac{1}{11}$$
4
For $a = 2i$ and $b = 4i$, compute $G^2$ and $AH$ to verify the identity $G^2 = AH$.
Ans
$G^2 = AH = -8$
Review
Tap to reveal
Compute the arithmetic and geometric parts
The arithmetic mean is $A = \frac{2i + 4i}{2} = 3i$, and the geometric mean squared is just the product $ab$.
Check the harmonic-mean identity
Here $G^2 = (2i)(4i) = -8$. Also $H = \frac{2ab}{a+b} = \frac{2(2i)(4i)}{6i} = \frac{8i}{3}$, so multiplying by $A$ reproduces the same value.
$$AH = (3i)\left(\frac{8i}{3}\right) = -8 = G^2$$
5
If $a^2, b^2, c^2$ are in AP, then $a+b, c+a, b+c$ are in:
Ans
HP
Review
Tap to reveal
Translate the AP condition
If $a^2, b^2, c^2$ are in AP, then $2b^2 = a^2 + c^2$, equivalently $b^2 - a^2 = c^2 - b^2$.
Invert the sums to spot the harmonic pattern
After factoring differences of squares, the condition becomes an AP relation among the reciprocals of $a+b$, $c+a$, and $b+c$.
$$\frac{1}{a+b},\ \frac{1}{c+a},\ \frac{1}{b+c} \text{ are in AP } \Rightarrow a+b,\ c+a,\ b+c \text{ are in HP}$$
6
Evaluate $\sum_{k=1}^{20} k(k+1)(k+2)$ given that $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$ and $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Ans
$53130$
Review
Tap to reveal
Expand the cubic product
Rewrite each term as $k(k+1)(k+2) = k^3 + 3k^2 + 2k$, so the whole sum splits into standard power sums.
Substitute the known formulas
Use the given identities for $\sum k^3$, $\sum k^2$, and $\sum k$ at $n=20$ and add the pieces.
$$44100 + 3(2870) + 2(210) = 53130$$
7
The first term of an HP is $-\frac{1}{3}$ and the fifth term is $\frac{1}{5}$. What is the 9th term?
Ans
$\frac{1}{13}$
Review
Tap to reveal
Convert the given HP terms to an AP
Reciprocals of the HP terms form an AP, so $A_1 = -3$ and $A_5 = 5$.
Extend the AP to the ninth term
The common difference is $d = (5 - (-3))/4 = 2$. Then $A_9 = -3 + 8(2) = 13$, so the 9th HP term is the reciprocal of 13.
$$T_9 = \frac{1}{13}$$
8
The sum $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{99 \cdot 101}$ equals:
Ans
$\frac{50}{101}$
Review
Tap to reveal
Split each term into a telescoping difference
Use partial fractions: $\frac{1}{(2r-1)(2r+1)} = \frac{1}{2}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)$. That makes the series telescope.
Keep only the endpoints
All interior terms cancel, leaving only the first and last boundary terms of the telescope.
$$\frac{1}{2}\left(1 - \frac{1}{101}\right) = \frac{50}{101}$$
9
If the GM and AM between two positive numbers are 6 and 5 respectively, find the numbers.
Ans
No such real numbers exist
Review
Tap to reveal
Check the AM-GM order
For positive numbers, the arithmetic mean must be at least the geometric mean. Here the stated values give the reverse order.
Conclude impossibility
Since $AM = 5$ and $GM = 6$, the condition $AM \ge GM$ fails, so no real positive pair can satisfy the data.
$$5 < 6 \Rightarrow \text{no such real numbers exist}$$
10
Find $n$ so that $\frac{a^{n+1} + b^{n+1}}{a^n + b^n}$ is the HM between $a$ and $b$.
Ans
$n = -1$
Review
Tap to reveal
Set the given ratio equal to the harmonic mean
The harmonic mean of $a$ and $b$ is $\frac{2ab}{a+b}$, so equate the expression to this value and clear denominators.
Factor the resulting identity
After expansion and rearrangement, the condition becomes $(a-b)(a^{n+1}-b^{n+1}) = 0$. Since the numbers are distinct, the exponent must force equality of the powers.
$$a^{n+1} = b^{n+1} \Rightarrow n+1=0 \Rightarrow n=-1$$
Question 1 of 10
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1
If the 4th and 7th terms of an HP are $\frac{2}{13}$ and $\frac{2}{25}$ respectively, what is the first term of the HP?