Class 11 FSc Math Chapter 6 Harmonic Progression & Relation Between A.M., G.M., H.M. Solved Board Questions PDF | FBISE & BISE Punjab Study Guide

Theoretical Foundations & Core Concepts

In 11th Class (Inter Part 1) Mathematics under Pakistani Education Boards (FBISE, Punjab, Sindh, KPK), Harmonic Progression (H.P.) represents a specialized sequence intimately connected to Arithmetic Progression (A.P.). Understanding H.P. requires mastery over sequence transformations and the interrelationships between Arithmetic Mean ($A$), Geometric Mean ($G$), and Harmonic Mean ($H$).

1. Definition of Harmonic Progression (H.P.)

A sequence of non-zero real numbers $a_1, a_2, a_3, \dots, a_n$ is said to be in Harmonic Progression if their reciprocals form an Arithmetic Progression (A.P.):

$$\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \dots, \frac{1}{a_n} \quad \text{is in A.P.}$$

Important Note: There is no direct general algebraic formula for the sum of $n$ terms of a Harmonic Progression. All calculations regarding H.P. must be performed by converting terms into their corresponding A.P. equivalents, completing the calculations, and taking reciprocals of the results.

2. General Term of an H.P.

If the first term of the corresponding A.P. is $a$ and its common difference is $d$, then the $n$-th term of the A.P. is given by $a_n^{\text{AP}} = a + (n-1)d$. Consequently, the $n$-th term of the Harmonic Progression ($a_n^{\text{HP}}$) is expressed as:

$$T_n = \frac{1}{a + (n-1)d}$$

3. Harmonic Mean (H.M.) Between Two Numbers

A number $H$ is said to be the Harmonic Mean between two non-zero numbers $a$ and $b$ if $a, H, b$ are in H.P. By definition, $\frac{1}{a}, \frac{1}{H}, \frac{1}{b}$ are in A.P. Thus:

$$\frac{1}{H} - \frac{1}{a} = \frac{1}{b} - \frac{1}{H} \implies \frac{2}{H} = \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} \implies H = \frac{2ab}{a+b}$$

4. Inserting $n$ Harmonic Means Between Two Numbers

To insert $n$ harmonic means $H_1, H_2, \dots, H_n$ between $a$ and $b$, insert $n$ arithmetic means $A_1, A_2, \dots, A_n$ between $\frac{1}{a}$ and $\frac{1}{b}$. The common difference $d$ of the resulting $(n+2)$-term A.P. is:

$$d = \frac{\frac{1}{b} - \frac{1}{a}}{n + 1}$$

Once $A_k = \frac{1}{a} + k \cdot d$ is calculated, the harmonic means are obtained via $H_k = \frac{1}{A_k}$.

5. Relationships Between A.M., G.M., and H.M.

For any two positive unequal real numbers $a$ and $b$, let:

• Arithmetic Mean: $A = \frac{a+b}{2}$
• Geometric Mean: $G = \pm \sqrt{ab}$
• Harmonic Mean: $H = \frac{2ab}{a+b}$

Two fundamental properties govern these means:

1. Geometric Relation: $G^2 = A \times H$ (i.e., $G$ is the geometric mean of $A$ and $H$).
2. Golden Inequality:
   • If $a, b > 0$ and $a \neq b$, then $A > G > H$ (assuming positive $G$).
   • If $a, b < 0$ and $a \neq b$, then $A < G < H$ (assuming negative $G$).

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Formula Summary & Quick Reference

Concept / Property	Mathematical Formula	Conditions / Remarks
$n$-th Term of H.P.	$T_n = \frac{1}{a + (n-1)d}$	$a, d$ are parameters of the reciprocal A.P.
Single Harmonic Mean ($H$)	$H = \frac{2ab}{a+b}$	$a+b \neq 0$, $a \neq 0, b \neq 0$
Arithmetic Mean ($A$)	$A = \frac{a+b}{2}$	For two numbers $a$ and $b$
Geometric Mean ($G$)	$G = \pm \sqrt{ab}$	For real numbers $a$ and $b$ of same sign
Relation Between Means	$G^2 = A \cdot H$	Holds universally for any non-zero $a, b$
Inequality (Positive Real Numbers)	$A > G > H$	$a, b > 0$ and $a \neq b$ (taking $G > 0$)
Inequality (Negative Real Numbers)	$A < G < H$	$a, b < 0$ and $a \neq b$ (taking $G < 0$)

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Step-by-Step Solved Board Exam Questions

Question 1 (Find Term of H.P. - BISE Lahore / FBISE Past Paper)

Problem: If the $4^{\text{th}}$ term of an H.P. is $\frac{1}{13}$ and the $7^{\text{th}}$ term is $\frac{1}{22}$, find the $10^{\text{th}}$ term of the H.P.

Solution:

Step 1: Convert H.P. terms to corresponding A.P. terms.
Let the corresponding Arithmetic Progression have first term $a$ and common difference $d$.
Given $T_4^{\text{HP}} = \frac{1}{13} \implies a_4^{\text{AP}} = 13$
Given $T_7^{\text{HP}} = \frac{1}{22} \implies a_7^{\text{AP}} = 22$

Step 2: Form system of linear equations for the A.P.
$$a + 3d = 13 \quad \text{--- (Equation 1)}$$ $$a + 6d = 22 \quad \text{--- (Equation 2)}$$

Step 3: Solve for $d$ and $a$.
Subtract Equation 1 from Equation 2: $$(a + 6d) - (a + 3d) = 22 - 13$$ $$3d = 9 \implies d = 3$$

Substitute $d = 3$ into Equation 1: $$a + 3(3) = 13 \implies a + 9 = 13 \implies a = 4$$

Step 4: Find the $10^{\text{th}}$ term of the A.P. and invert for H.P.
$$a_{10}^{\text{AP}} = a + 9d = 4 + 9(3) = 4 + 27 = 31$$ Therefore, the $10^{\text{th}}$ term of the Harmonic Progression is: $$T_{10}^{\text{HP}} = \frac{1}{a_{10}^{\text{AP}}} = \frac{1}{31}$$

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Question 2 (Inserting Harmonic Means - BISE Rawalpindi / FBISE)

Problem: Insert 4 Harmonic Means between $\frac{1}{3}$ and $\frac{1}{23}$.

Solution:

Step 1: Transform to an A.P. problem.
Let $H_1, H_2, H_3, H_4$ be the required harmonic means between $\frac{1}{3}$ and $\frac{1}{23}$.
Then $A_1, A_2, A_3, A_4$ (reciprocals of $H_1, H_2, H_3, H_4$) are 4 Arithmetic Means between $3$ and $23$.
The sequence $3, A_1, A_2, A_3, A_4, 23$ forms an A.P. with total terms $N = 6$.

Step 2: Calculate the common difference $d$.
Here $a = 3$ and $a_6 = 23$. $$a_6 = a + 5d \implies 23 = 3 + 5d \implies 5d = 20 \implies d = 4$$

Step 3: Calculate the arithmetic means $A_1, A_2, A_3, A_4$.
$$A_1 = a + d = 3 + 4 = 7$$ $$A_2 = a + 2d = 3 + 8 = 11$$ $$A_3 = a + 3d = 3 + 12 = 15$$ $$A_4 = a + 4d = 3 + 16 = 19$$

Step 4: Convert arithmetic means back to harmonic means.
$$H_1 = \frac{1}{A_1} = \frac{1}{7}, \quad H_2 = \frac{1}{A_2} = \frac{1}{11}, \quad H_3 = \frac{1}{A_3} = \frac{1}{15}, \quad H_4 = \frac{1}{A_4} = \frac{1}{19}$$ Hence, the 4 Harmonic Means are $\frac{1}{7}, \frac{1}{11}, \frac{1}{15}, \frac{1}{19}$.

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Question 3 (Theoretical Proof of Means - Board Long Question)

Problem: For any two distinct positive real numbers $a$ and $b$, prove that:
(i) $G^2 = A \cdot H$
(ii) $A > G > H$ (where $G > 0$).

Solution:

Part (i): Proof of $G^2 = A \cdot H$
By standard definitions: $$A = \frac{a+b}{2}, \quad G = \sqrt{ab}, \quad H = \frac{2ab}{a+b}$$ Multiply $A$ and $H$: $$A \cdot H = \left(\frac{a+b}{2}\right) \left(\frac{2ab}{a+b}\right) = \frac{2ab(a+b)}{2(a+b)} = ab$$ Since $G = \sqrt{ab}$, we have $G^2 = ab$. Thus: $$G^2 = A \cdot H \quad \text{(Hence Proved)}$$

Part (ii): Proof of $A > G > H$ for $a \neq b, a, b > 0$
1. To show $A > G$: $$A - G = \frac{a+b}{2} - \sqrt{ab} = \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2}$$ Since $a \neq b$, $(\sqrt{a} - \sqrt{b})^2 > 0$. Thus: $$A - G > 0 \implies A > G \quad \text{--- (Statement 1)}$$ 2. To show $G > H$: $$G - H = \sqrt{ab} - \frac{2ab}{a+b} = \sqrt{ab}\left(1 - \frac{2\sqrt{ab}}{a+b}\right) = \sqrt{ab}\left(\frac{a+b-2\sqrt{ab}}{a+b}\right) = \frac{\sqrt{ab}(\sqrt{a}-\sqrt{b})^2}{a+b}$$ Since $a, b > 0$ and $a \neq b$, all factors in the numerator and denominator are positive. $$G - H > 0 \implies G > H \quad \text{--- (Statement 2)}$$ Combining Statement 1 and Statement 2: $$A > G > H \quad \text{(Hence Proved)}$$

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Question 4 (Finding Numbers from Given Means - BISE Multan / Sargodha)

Problem: Find two positive real numbers $a$ and $b$ if their Arithmetic Mean is $5$ and their Harmonic Mean is $\frac{16}{5}$.

Solution:

Step 1: Set up equations using standard definitions.
Given $A = 5$: $$\frac{a+b}{2} = 5 \implies a + b = 10 \quad \text{--- (Equation 1)}$$ Given $H = \frac{16}{5}$: $$\frac{2ab}{a+b} = \frac{16}{5} \quad \text{--- (Equation 2)}$$

Step 2: Substitute Equation 1 into Equation 2.
$$\frac{2ab}{10} = \frac{16}{5} \implies \frac{ab}{5} = \frac{16}{5} \implies ab = 16 \quad \text{--- (Equation 3)}$$

Step 3: Form a quadratic equation for $a$ and $b$.
From Equation 1, $b = 10 - a$. Substitute into Equation 3: $$a(10 - a) = 16 \implies 10a - a^2 = 16 \implies a^2 - 10a + 16 = 0$$

Step 4: Factorize to find values.
$$a^2 - 8a - 2a + 16 = 0 \implies a(a - 8) - 2(a - 8) = 0$$ $$(a - 8)(a - 2) = 0 \implies a = 8 \quad \text{or} \quad a = 2$$ If $a = 8$, then $b = 10 - 8 = 2$.
If $a = 2$, then $b = 10 - 2 = 8$.
Thus, the required numbers are $2$ and $8$.

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Interactive Practice Quiz (Clickable MCQs)

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Frequently Asked Questions (FAQs)

1. How do I calculate the sum of $n$ terms of a Harmonic Progression in board exams?

Unlike Arithmetic Progressions (A.P.) and Geometric Progressions (G.P.), there is no simple closed-form algebraic formula for the sum of $n$ terms of a Harmonic Progression ($S_n$). In FBISE and Board examination questions, you are never asked to evaluate $\sum H_n$ directly via a formula. If such a sum appears, convert each term to its value or express the sum in terms of standard mathematical functions.

2. What is the distinction between $A \ge G \ge H$ and $A > G > H$?

The non-strict inequality $A \ge G \ge H$ holds for any non-negative numbers where the numbers *can* be equal. If $a = b$, then $A = G = H$. However, if $a$ and $b$ are strictly distinct positive numbers ($a \neq b, a, b > 0$), the strict Golden Inequality $A > G > H$ applies. Pay close attention to the conditions specified in board questions.

3. Why must we convert H.P. terms into A.P. terms before solving?

Harmonic Progression does not possess a constant common difference or common ratio between consecutive terms. Because its defining characteristic is that the reciprocals of its terms form an A.P., converting to A.P. allows us to use well-established linear formulas ($a_n = a + (n-1)d$). Once the required term or mean in A.P. is computed, taking the reciprocal yields the correct H.P. solution.