RBSE Solutions Class 11 Maths Chapter 8 Sequence, Progression, and Series Exercise 8.8

Question 1. A.M. of two numbers is 60 and H.M. is 18, Find the numbers.
Answer: Let the two numbers be \( a \) and \( b \).
Given that the Arithmetic Mean (A.M.) is 50 (as per the steps in the solution) and the Harmonic Mean (H.M.) is 18.
So, we have:
\( \frac{a+b}{2} = 50 \)
\( \implies a+b = 100 \) ...(i)
And \( \frac{2ab}{a+b} = 18 \)
Substitute \( a+b = 100 \) from equation (i) into the H.M. equation:
\( \frac{2ab}{100} = 18 \)
\( \implies \frac{ab}{50} = 18 \)
\( \implies ab = 18 \times 50 \)
\( \implies ab = 900 \) ...(ii)
Now, we use the identity \( (a-b)^2 = (a+b)^2 - 4ab \).
Substitute the values from (i) and (ii):
\( (a-b)^2 = (100)^2 - 4 \times 900 \)
\( (a-b)^2 = 10000 - 3600 \)
\( (a-b)^2 = 6400 \)
Take the square root of both sides:
\( a-b = \sqrt{6400} \)
\( \implies a-b = 80 \) ...(iii)
Now we solve the system of linear equations (i) and (iii):
\( a+b = 100 \)
\( a-b = 80 \)
Adding the two equations:
\( (a+b) + (a-b) = 100 + 80 \)
\( 2a = 180 \)
\( \implies a = 90 \)
Substitute the value of \( a \) into equation (i):
\( 90 + b = 100 \)
\( \implies b = 100 - 90 \)
\( \implies b = 10 \)
Thus, the two numbers are 90 and 10. The relationship between AM, GM, and HM is an important concept in sequences.
In simple words: We are given the average and a special mean of two numbers. We use these to find their sum and product. Then, we can find the numbers themselves by solving simple equations.

🎯 Exam Tip: Always check that the numbers you find satisfy both the A.M. and H.M. conditions mentioned in the problem.

Question 3. The difference between A.M. and G.M. is 2, difference between G.M. and H.M., is 1.2. Find the numbers.
Answer: Let the two numbers be \( a \) and \( b \). Let A.M. be \( A \), G.M. be \( G \), and H.M. be \( H \).
According to the question, we have:
\( A - G = 2 \implies A = G + 2 \) ...(i)
\( G - H = 1.2 \implies H = G - 1.2 \) ...(ii)
We know the important relationship between A.M., G.M., and H.M. for two positive numbers: \( G^2 = AH \).
Substitute the expressions for \( A \) and \( H \) from (i) and (ii) into this equation:
\( G^2 = (G+2)(G-1.2) \)
Expand the right side:
\( G^2 = G^2 - 1.2G + 2G - 2.4 \)
\( G^2 = G^2 + 0.8G - 2.4 \)
Subtract \( G^2 \) from both sides:
\( 0 = 0.8G - 2.4 \)
\( \implies 0.8G = 2.4 \)
Divide by 0.8:
\( G = \frac{2.4}{0.8} \)
\( \implies G = 3 \)
Now, find \( A \) and \( H \) using the value of \( G \):
From (i): \( A = G + 2 = 3 + 2 = 5 \)
From (ii): \( H = G - 1.2 = 3 - 1.2 = 1.8 \)
So, we have \( A = 5 \), \( G = 3 \), and \( H = 1.8 \).
The problem asks us to find the numbers \( a \) and \( b \).
We know:
\( A = \frac{a+b}{2} \implies 5 = \frac{a+b}{2} \implies a+b = 10 \)
\( G = \sqrt{ab} \implies 3 = \sqrt{ab} \implies ab = 3^2 = 9 \)
We need to find two numbers whose sum is 10 and product is 9. These numbers are the roots of the quadratic equation \( x^2 - (a+b)x + ab = 0 \).
\( x^2 - 10x + 9 = 0 \)
Factor this quadratic equation:
\( x^2 - 9x - x + 9 = 0 \)
\( x(x-9) - 1(x-9) = 0 \)
\( (x-1)(x-9) = 0 \)
Therefore, \( x = 1 \) or \( x = 9 \).
The two numbers are 1 and 9. This problem shows how different means are related and can be used to find unknown numbers.
In simple words: We are given how the three types of averages (Arithmetic, Geometric, Harmonic) are different from each other. We use a special math rule connecting them to find the values of each average. Once we have the Arithmetic and Geometric averages, we can figure out the two original numbers.

🎯 Exam Tip: Remember the fundamental relation \( G^2 = AH \) for two positive numbers; it's a key shortcut for many problems involving means.

Question 4. Three numbers a, b, c are in GP. and \( ax = by = c^2 \), then prove that x, y, z will be in H.P.
Answer: Let the given relation be \( ax = by = c^2 = k \).
From this, we can write \( a, b, c \) in terms of \( k \):
\( a = k^{1/x} \)
\( b = k^{1/y} \)
\( c = k^{1/2} \) (since \( c^2 = k \))
Since \( a, b, c \) are in Geometric Progression (G.P.), the property is \( b^2 = ac \).
Substitute the expressions for \( a, b, c \) in terms of \( k \) into the G.P. property:
\( (k^{1/y})^2 = (k^{1/x})(k^{1/z}) \)
\( k^{2/y} = k^{(1/x) + (1/z)} \)
Since the bases are equal, the exponents must be equal:
\( \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \)
This equation shows that \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in Arithmetic Progression (A.P.).
If the reciprocals of three numbers are in A.P., then the numbers themselves are in Harmonic Progression (H.P.).
Therefore, \( x, y, z \) are in H.P. This proof highlights the interconnections between different types of progressions.
In simple words: We are given that three numbers are related in a special way and form a Geometric Progression. By using logarithms and the properties of progressions, we show that another set of numbers related to the first set must form a Harmonic Progression.

🎯 Exam Tip: When dealing with variables in exponents, equating them to a constant \( k \) is a powerful technique for simplifying the problem.

Question 5. Three numbers a, b, c are in H.P. Prove that \( 2a - b, b, 2c - b \) will be in G.P.
Answer: Given that \( a, b, c \) are in Harmonic Progression (H.P.).
By the definition of H.P., their reciprocals are in Arithmetic Progression (A.P.).
So, \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P.
This means the middle term is the average of the other two:
\( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \)
Now, we need to prove that \( 2a - b, b, 2c - b \) are in Geometric Progression (G.P.).
For three terms to be in G.P., the square of the middle term must be equal to the product of the other two terms:
\( b^2 = (2a - b)(2c - b) \)
Let's expand the right-hand side:
\( (2a - b)(2c - b) = 4ac - 2ab - 2bc + b^2 \)
So, we need to show that \( b^2 = 4ac - 2ab - 2bc + b^2 \).
Subtract \( b^2 \) from both sides:
\( 0 = 4ac - 2ab - 2bc \)
Divide the entire equation by \( 2abc \) (assuming \( a, b, c \) are non-zero):
\( 0 = \frac{4ac}{2abc} - \frac{2ab}{2abc} - \frac{2bc}{2abc} \)
\( 0 = \frac{2}{b} - \frac{1}{c} - \frac{1}{a} \)
Rearrange the terms:
\( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \)
This is exactly the condition we derived from \( a, b, c \) being in H.P. Since this condition is true, the statement that \( 2a - b, b, 2c - b \) are in G.P. is also true. This problem demonstrates the close relationship between different types of mathematical progressions.
In simple words: We start with a rule for numbers in a Harmonic Progression (H.P.). Then, we want to show that a new set of numbers, made from the first set, follows the rule for a Geometric Progression (G.P.). By using the H.P. rule and doing some algebra, we prove that the new numbers indeed form a G.P.

🎯 Exam Tip: Always clearly state the conditions for each type of progression (A.P., G.P., H.P.) at the beginning of your proof.

Question 6. If a, b, c are in A.P., x,y, z are in H.P. and ax, by, cz are in GP., then prove that \( \frac{x}{a} + \frac{z}{c} = \frac{a}{c} + \frac{c}{a} \).
Answer: Given conditions:
1. \( a, b, c \) are in A.P.
\( \implies 2b = a+c \) ...(i)
2. \( x, y, z \) are in H.P.
\( \implies \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \)
\( \implies \frac{2}{y} = \frac{x+z}{xz} \)
\( \implies y = \frac{2xz}{x+z} \) ...(ii)
3. \( ax, by, cz \) are in G.P.
\( \implies (by)^2 = (ax)(cz) \)
\( \implies b^2 y^2 = acxz \) ...(iii)
Now, substitute the values of \( b \) from (i) and \( y \) from (ii) into equation (iii):
\( \left(\frac{a+c}{2}\right)^2 \left(\frac{2xz}{x+z}\right)^2 = acxz \)
\( \frac{(a+c)^2}{4} \cdot \frac{4x^2z^2}{(x+z)^2} = acxz \)
\( \frac{(a+c)^2 x^2 z^2}{(x+z)^2} = acxz \)
Divide both sides by \( xz \) (assuming \( x, z \neq 0 \)):
\( \frac{(a+c)^2 xz}{(x+z)^2} = ac \)
Rearrange the terms to group common factors:
\( \frac{(a+c)^2}{ac} = \frac{(x+z)^2}{xz} \)
Expand both sides:
\( \frac{a^2 + c^2 + 2ac}{ac} = \frac{x^2 + z^2 + 2xz}{xz} \)
Split the fractions:
\( \frac{a^2}{ac} + \frac{c^2}{ac} + \frac{2ac}{ac} = \frac{x^2}{xz} + \frac{z^2}{xz} + \frac{2xz}{xz} \)
Simplify the terms:
\( \frac{a}{c} + \frac{c}{a} + 2 = \frac{x}{z} + \frac{z}{x} + 2 \)
Subtract 2 from both sides:
\( \frac{a}{c} + \frac{c}{a} = \frac{x}{z} + \frac{z}{x} \)
This proves the required relation. This problem is a good example of combining properties from different types of sequences.
In simple words: We are given three sets of numbers that follow different mathematical patterns (Arithmetic, Harmonic, and Geometric Progressions). By using the rules for each pattern and doing some algebra, we show that a specific relationship must exist between parts of these number sets.

🎯 Exam Tip: Keep the properties of A.P., G.P., and H.P. clear and separate. For H.P., it's often easier to convert to A.P. using reciprocals.

Question 7. A₁, A₂ are two A.M. between two positive numbers a and b, G₁, G₂ are two G.M. and H₁, H₂ are two H.M. then prove that: (i) \( A_1 H_2 = A_2 H_1 = G_1 G_2 = ab \) (ii) \( G_1 G_2 : H_1 H_2 = (A_1 + A_2): (H_1 + H_2) \)
Answer: Let the two positive numbers be \( a \) and \( b \).
**1. Arithmetic Means (A.M.s):**
If \( A_1, A_2 \) are two A.M.s between \( a \) and \( b \), then \( a, A_1, A_2, b \) are in A.P.
The common difference \( d = \frac{b-a}{3} \).
So, \( A_1 = a+d = a + \frac{b-a}{3} = \frac{3a+b-a}{3} = \frac{2a+b}{3} \)
And \( A_2 = a+2d = a + \frac{2(b-a)}{3} = \frac{3a+2b-2a}{3} = \frac{a+2b}{3} \)
**2. Geometric Means (G.M.s):**
If \( G_1, G_2 \) are two G.M.s between \( a \) and \( b \), then \( a, G_1, G_2, b \) are in G.P.
The common ratio \( r = \left(\frac{b}{a}\right)^{1/3} \).
So, \( G_1 = ar = a \left(\frac{b}{a}\right)^{1/3} = a^{2/3} b^{1/3} \)
And \( G_2 = ar^2 = a \left(\frac{b}{a}\right)^{2/3} = a^{1/3} b^{2/3} \)
**3. Harmonic Means (H.M.s):**
If \( H_1, H_2 \) are two H.M.s between \( a \) and \( b \), then \( \frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{b} \) are in A.P.
The common difference \( d' = \frac{\frac{1}{b}-\frac{1}{a}}{3} = \frac{a-b}{3ab} \).
So, \( \frac{1}{H_1} = \frac{1}{a} + d' = \frac{1}{a} + \frac{a-b}{3ab} = \frac{3b+a-b}{3ab} = \frac{a+2b}{3ab} \implies H_1 = \frac{3ab}{a+2b} \)
And \( \frac{1}{H_2} = \frac{1}{a} + 2d' = \frac{1}{a} + \frac{2(a-b)}{3ab} = \frac{3b+2a-2b}{3ab} = \frac{2a+b}{3ab} \implies H_2 = \frac{3ab}{2a+b} \)

**(i) Prove \( A_1 H_2 = A_2 H_1 = G_1 G_2 = ab \)**
First, find \( G_1 G_2 \):
\( G_1 G_2 = (a^{2/3} b^{1/3}) (a^{1/3} b^{2/3}) = a^{(2/3+1/3)} b^{(1/3+2/3)} = a^1 b^1 = ab \).
So, \( G_1 G_2 = ab \).
Next, find \( A_1 H_2 \):
\( A_1 H_2 = \left(\frac{2a+b}{3}\right) \left(\frac{3ab}{2a+b}\right) = ab \).
Next, find \( A_2 H_1 \):
\( A_2 H_1 = \left(\frac{a+2b}{3}\right) \left(\frac{3ab}{a+2b}\right) = ab \).
Thus, \( A_1 H_2 = A_2 H_1 = G_1 G_2 = ab \). This demonstrates a beautiful symmetry in the relationships between different means.

**(ii) Prove \( G_1 G_2 : H_1 H_2 = (A_1 + A_2): (H_1 + H_2) \)**
We know \( G_1 G_2 = ab \).
Now, find \( H_1 H_2 \):
\( H_1 H_2 = \left(\frac{3ab}{a+2b}\right) \left(\frac{3ab}{2a+b}\right) = \frac{9a^2 b^2}{(a+2b)(2a+b)} \)
So, \( \frac{G_1 G_2}{H_1 H_2} = \frac{ab}{\frac{9a^2 b^2}{(a+2b)(2a+b)}} = \frac{ab(a+2b)(2a+b)}{9a^2 b^2} = \frac{(a+2b)(2a+b)}{9ab} \) ...(1)
Next, find \( A_1 + A_2 \):
\( A_1 + A_2 = \frac{2a+b}{3} + \frac{a+2b}{3} = \frac{2a+b+a+2b}{3} = \frac{3a+3b}{3} = a+b \)
Next, find \( H_1 + H_2 \):
\( H_1 + H_2 = \frac{3ab}{a+2b} + \frac{3ab}{2a+b} = \frac{3ab(2a+b) + 3ab(a+2b)}{(a+2b)(2a+b)} \)
\( = \frac{3ab(2a+b+a+2b)}{(a+2b)(2a+b)} = \frac{3ab(3a+3b)}{(a+2b)(2a+b)} = \frac{9ab(a+b)}{(a+2b)(2a+b)} \)
Now, find \( \frac{A_1 + A_2}{H_1 + H_2} \):
\( \frac{A_1 + A_2}{H_1 + H_2} = \frac{a+b}{\frac{9ab(a+b)}{(a+2b)(2a+b)}} = \frac{(a+b)(a+2b)(2a+b)}{9ab(a+b)} = \frac{(a+2b)(2a+b)}{9ab} \) ...(2)
Comparing (1) and (2), we see that \( \frac{G_1 G_2}{H_1 H_2} = \frac{A_1 + A_2}{H_1 + H_2} \).
This implies \( G_1 G_2 : H_1 H_2 = (A_1 + A_2) : (H_1 + H_2) \). This result demonstrates a deeper connection between various means inserted between two numbers.
In simple words: We are given two numbers and asked to find the relationships between their Arithmetic, Geometric, and Harmonic means when two of each type are inserted. We first write down the formulas for each mean. Then, we use these formulas to prove that specific products and ratios of these means are equal, showing how they are mathematically linked.

🎯 Exam Tip: Systematically list all the terms and their formulas for A.M., G.M., and H.M. before attempting the proof, as this will help organize your work.

Question 9. If three numbers a, b, c are in A.P. and H.P. both then prove that they will be in G.P. also.
Answer: Given that three numbers \( a, b, c \) are in both Arithmetic Progression (A.P.) and Harmonic Progression (H.P.).
**1. If \( a, b, c \) are in A.P.:**
By the definition of A.P., the middle term is the average of the first and third terms.
So, \( b = \frac{a+c}{2} \) ...(i)
**2. If \( a, b, c \) are in H.P.:**
By the definition of H.P., the reciprocal of the middle term is the average of the reciprocals of the first and third terms.
So, \( \frac{1}{b} = \frac{\frac{1}{a} + \frac{1}{c}}{2} \)
\( \frac{1}{b} = \frac{a+c}{2ac} \)
\( \implies b = \frac{2ac}{a+c} \) ...(ii)
Now, substitute the value of \( (a+c) \) from equation (i) into equation (ii):
From (i), \( a+c = 2b \).
Substitute this into (ii):
\( b = \frac{2ac}{2b} \)
\( b = \frac{ac}{b} \)
Multiply both sides by \( b \):
\( b^2 = ac \)
This is the property for numbers in Geometric Progression (G.P.).
Therefore, if three numbers \( a, b, c \) are in both A.P. and H.P., they must also be in G.P. This property highlights a special relationship where the numbers must be equal for this to hold true, as it's the only way for A.P., G.P., and H.P. to coincide.
In simple words: We start with three numbers that follow the rules for both an Arithmetic Progression (A.P.) and a Harmonic Progression (H.P.). By using the formulas for both types of progressions and a little algebra, we can prove that these same three numbers must also follow the rule for a Geometric Progression (G.P.).

🎯 Exam Tip: Remember that if numbers are simultaneously in A.P., G.P., and H.P., then all the numbers must be equal.