Class 11 RD Sharma Solutions - Chapter 20 Geometric Progressions- Exercise 20.6

Chapter 20 of RD Sharma's Class 11 Mathematics textbook focuses on the Geometric Progressions (GPs) an essential concept in sequences and series. Exercise 20.6 delves into the problems related to the properties and applications of the geometric progressions offering the students practice in solving real-world problems using these principles. This exercise aims to the enhance understanding and application of GPs in various mathematical contexts.

Geometric Progressions

A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by the constant known as the common ratio. The general form of a GP is a, ar,ar₂,ar₃,… where a is the first term and r is the common ratio. The GPs are widely used in mathematical modeling, financial calculations, and various scientific fields due to their unique properties and ease of manipulation.

Question 1: Insert 6 geometric means between 27 and 1/81.

Solution:

Let the six geometric means be A₁, A₂, A₃, A₄, A₅, A₆.

Now, these 6 terms are to be added between 27 and 1/81.

So the G.P. becomes, 27, A₁, A₂, A₃, A₄, A₅, A₆,1/81 with first term(a) = 27, number of terms(n) = 8 and 8th term(a₈) = 1/81.

We know nth term of a G.P. is given by a_n = ar^n-1, where r is the common ratio.

=> a₈ = 1/81

=> 27(r)^8-1 = 1/81

=> r⁷ = 1/(81×27)

=> r⁷ = (1/3)⁷

=> r = 1/3

So, geometric means are:

A₁ = ar = 27(1/3) = 9

A₂ = ar² = 27(1/3)² = 3

A₃ = ar³ = 27(1/3)³ = 1

A₄ = ar⁴ = 27(1/3)⁴ = 1/3

A₅ = ar⁵ = 27(1/3)⁵ = 1/9

A₆ = ar⁶ = 27(1/3)⁶ = 1/27

Therefore, the 6 geometric means between 27 and 1/81 are 9, 3, 1, 1/3, 1/9, 1/27. 

Question 2. Insert 5 geometric means between 16 and 1/4.

Solution:

Let the five geometric means be A₁, A₂, A₃, A₄, A₅.

Now, these 5 terms are to be added between 16 and 1/4.

So the G.P. becomes, 16, A₁, A₂, A₃, A₄, A₅,1/4 with first term(a) = 16, number of terms(n) = 7 and 7th term(a₇) = 1/4.

We know nth term of a G.P. is given by a_n = ar^n-1, where r is the common ratio.

=> a₇ = 1/4

=> 16(r^7-1) = 1/4

=> r⁶ = 1/64

=> r⁶ = (1/2)⁶

=> r = 1/2

So, geometric means are:

A₁ = ar = 16(1/2) = 8

A₂ = ar² = 16(1/2)² = 4

A₃ = ar³ = 16(1/2)³ = 2

A₄ = ar⁴ = 16(1/2)⁴ = 1

A₅ = ar⁵ = 16(1/2)⁵ = 1/2

Therefore, the 5 geometric means between 16 and 1/4 are 8, 4, 2, 1, 1/2.

Question 3. Insert 5 geometric means between 32/9 and 81/2.

Solution:

Let the five geometric means be A₁, A₂, A₃, A₄, A₅.

Now, these 5 terms are to be added between 32/9 and 81/2.

So the G.P. becomes, 32/9, A₁, A₂, A₃, A₄, A₅, 81/2 with first term(a) = 32/9, number of terms(n) = 7 and 7th term(a₇) = 81/2.

We know nth term of a G.P. is given by a_n = ar^n-1, where r is the common ratio.

=> a₇ = 81/2

=> (32/9)(r^7-1) = 81/2

=> r⁶ = (81×9)/(2×32)

=> r⁶ = (3/2)⁶

=> r = 3/2

So, geometric means are:

A₁ = ar = (32/9)×(3/2) = 16/3

A₂ = ar² = (32/9)×(3/2)² = 8

A₃ = ar³ = (32/9)×(3/2)³ = 12

A₄ = ar⁴ = (32/9)×(3/2)⁴ = 18

A₅ = ar⁵ = (32/9)×(3/2)⁵ = 27

Therefore, the 5 geometric means between 32/9 and 81/2 are 16/3, 8, 12, 18, 27.

Question 4. Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a³b and ab³

(iii) –8 and –2

Solution:

We know geometric mean between two numbers, a and b is given by \sqrt{ab}    .

(i) 2 and 8

Here, a = 2 and b = 8

So, G.M. = \sqrt{2(8)}

= \sqrt{16}

= 4

(ii) a³b and ab³

Here, a = a³b and b = ab³

So, G.M. = \sqrt{a^3b×ab^3}

= \sqrt{a^4b^4}

= a²b²

(iii) –8 and –2

Here, a = –8 and b = –2

G.M. = \sqrt{(-8)×(-2)}

= \sqrt{16}

= 4

Question 5. If a is the G.M. of 2 and 1/4 find a.

Solution:

We know geometric mean between two numbers, a and b is given by \sqrt{ab}.

According to the question,

a = \sqrt{2×(1/4)}

= \sqrt{1/2}

= \frac{1}{\sqrt{2}}

Therefore, the value of a is \frac{1}{\sqrt{2}}.

Question 6. Find the two numbers whose A.M. is 25 and GM is 20.

Solution:

We know geometric mean between two numbers, a and b is given by \sqrt{ab} and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

=> \sqrt{ab} = 20 ……. (1)

And

=> (a+b)/2 = 25

=> a+b = 50

=> b = 50–a ……. (2)

From (1) and (2), we get,

=> \sqrt{a(50-a)} = 20

Squaring both sides, we get,

=> a(50 –a) = 400

=> a² – 50a + 400 = 0

=> a²– 40a–10a+400 = 0

=> a(a– 40) – 10(a– 40) = 0

=> (a– 40) (a– 10) = 0

=> a = 40 or a = 10

Putting these in (2) we get,

When a = 40, then b = 10 and 

When a = 10, then b = 40.

Therefore, the numbers are 10 and 40.

Question 7. Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Solution:

Suppose the roots of the quadratic equation are a and b.

We know geometric mean between two numbers, a and b is given by \sqrt{ab} and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

A.M. of roots = (a+b)/2 = A

a + b = 2A ….. (1) 

And G.M. of roots = \sqrt{ab} = G

ab = G²  … (2)

Now, we know a quadratic equation in x with roots a and b is given by,

x² – (a+b)x + (ab) = 0

From (1) and (2), we get,

x² – 2Ax + G² = 0 

Therefore, the required quadratic equation is x² – 2Ax + G² = 0.

Question 8. The sum of two numbers is 6 times their geometric means. Show that the numbers are in the ratio (3+2\sqrt{2}):(3-2\sqrt{2})

Solution:

Let the two numbers be a and b. We know geometric mean between two numbers, a and b is given by \sqrt[]{ab}.

According to the question,

=> a+b = 6\sqrt{(ab)}

=> \frac{a+b}{2\sqrt{(ab)}} = \frac{3}{1}

Applying Componendo and Dividendo on both sides, we get,

=> \frac{a+b+2\sqrt{(ab)}}{a+b-2\sqrt{(ab)}} = \frac{3+1}{3-1}

=> \left[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right]^2 = \frac{2}{1}

=> \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{2}}{1}

By again applying Componendo and Dividendo on both sides, we get,

=> \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

=> \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

=> \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}

Squaring both sides, we get

=> \frac{a}{b} = \left[\frac{\sqrt{2}+1}{\sqrt{2}-1}\right]^2

=> \frac{a}{b} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}

=> \frac{a}{b} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}

Hence proved.

Question 9. If AM and GM of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.

Solution:

Suppose the roots of the quadratic equation are a and b.

We know geometric mean between two numbers, a and b is given by \sqrt{ab} and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

A.M. of roots = (a+b)/2 = 8

a+b = 16 ….. (1)

And G.M. of roots = \sqrt{ab} = 5

ab = 25  .… (2)

Now, let the quadratic equation with roots a and b is given by,

x² – (a+b)x + (ab) = 0

From (1) and (2), we get,

x² – 16x + 25 = 0

Therefore, the required quadratic equation is x² – 16x + 25 = 0.

Question 10. If AM and GM of the two positive numbers a and b are 10 and 8 respectively. Find the numbers.

Solution:

We know geometric mean between two numbers, a and b is given by \sqrt{ab} and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

=> \sqrt{ab} = 8 ……. (1)

And,

=> (a+b)/2 = 10

=> a+b = 20

=> b = 20–a ……. (2)

From (1) and (2), we get,

=> \sqrt{a(20–a)} = 8

Squaring both sides, we get,

=> a(20–a) = 64

=> a²–20a+64 = 0

=> a²–16a–4a+64 = 0

=> a(a–16) – 4(a–16) = 0

=> (a–4) (a–16) = 0

=> a = 4 or a = 16

Putting a = 4 in (2), we get b = 16. And,

Putting a = 16 in (2), we get b = 4.

Therefore, the numbers are 4 and 16.

Question 11. Prove that the product of n geometric means between two quantities is equal to the nth power of geometric mean of those two quantities.

Solution:

Suppose we have a GP with first term a, common ratio r and number of terms n.

We have to add these n terms of GP between two quantities such that the GP remains maintained. So total number of terms become (n+2).

We know nth term of a G.P. is given by a_n = ar^n-1.

So, the last term of the GP ,i.e., (n+2)^th term will be, a_n+2 = ar^n+2-1 = arⁿ⁺¹  

We know geometric mean between two numbers, a and b is given by \sqrt{ab}.

The GM of a and arⁿ⁺¹ will be, G₁ = \sqrt{a(ar^{n+1})} = \left(a^2r^{n+1}\right)^{\frac{1}{2}}      

Hence, L.H.S. = G^n_1 = \left(a^2r^{n+1}\right)^{\frac{n}{2}}

Now, R.H.S. = Product of n geometric means between these two quantities, G₂ = ar × ar² × . . . . × arⁿ

= a^nr^{1+2+....+n}

= a^nr^{\frac{n(n+1)}{2}}

= \left[a^2r^{n+1}\right]^{\frac{n}{2}}

= G^n_1

= L.H.S.

Hence, Proved.

Question 12: If the AM of two positive numbers a and b (a>b) is twice their geometric mean. Prove that a:b = (2+\sqrt{3}):(2-\sqrt{3})

Solution:

We know geometric mean between two numbers, a and b is given by \sqrt{ab} and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

AM = 2(GM)

=> \frac{a+b}{2} = 2\sqrt{ab}

=> \frac{a+b}{2\sqrt{(ab)}}      = \frac{2}{1}

Applying Componendo and Dividendo on both sides, we get,

=> \frac{a+b+2\sqrt{(ab)}}{a+b-2\sqrt{(ab)}} = \frac{2+1}{2-1}

=> \left[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right]^2 = \frac{3}{1}

=> \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}= \frac{\sqrt{3}}{1}

By again applying Componendo and Dividendo on both sides, we get,

=> \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}= \frac{\sqrt{3}+1}{\sqrt{3}-1}

=> \frac{2\sqrt{a}}{2\sqrt{b}}= \frac{\sqrt{3}+1}{\sqrt{3}-1}

=> \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}

Squaring both sides, we get

=> \frac{a}{b} = \left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]^2

=> \frac{a}{b} = \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}

=> \frac{a}{b} = \frac{4+2\sqrt{3}}{4-2\sqrt{3}}

=> \frac{a}{b} = \frac{2+\sqrt{3}}{2-\sqrt{3}}

Hence, proved.

Question 13. If one AM, A, and two geometric means G₁ and G₂ are inserted between any two positive numbers, show that \frac{G^2_1}{G_2}+\frac{G^2_2}{G_1} = 2A

Solution:

Let the two positive numbers be a and b.

Now value of one AM between a and b, A = (a+b)/2.

So, 2A = a+b . . . . (1) 

If we add two geometric means between a and b, the GP becomes a,G₁,G₂,b.

Now, we know b = ar^4-1, where r is the common ratio.

=> r³ = \frac{b}{a}

=> r = \left[\frac{b}{a}\right]^{\frac{1}{3}}

So, G₁ = ar = a\left[\frac{b}{a}\right]^{\frac{1}{3}} = a^{\frac{2}{3}}b^{\frac{1}{3}}

G₂ = ar² = a\left[\frac{b}{a}\right]^{\frac{2}{3}} = a^{\frac{1}{3}}b^{\frac{2}{3}}

Now, L.H.S. = \frac{G^2_1}{G_2}+\frac{G^2_2}{G_1}

= \frac{a^{\frac{4}{3}}b^{\frac{2}{3}}}{a^{\frac{1}{3}}b^{\frac{2}{3}}}+\frac{a^{\frac{2}{3}}b^{\frac{4}{3}}}{a^{\frac{2}{3}}b^{\frac{1}{3}}}

= ab⁰ + a⁰b

= a+b

= 2A [From (1)]

= R.H.S.

Hence, proved.

Read More: Geometric Series

Conclusion

Exercise 20.6 in Chapter 20 provides the students with a deeper insight into the geometric progressions focusing on the applying the concepts to solve various types of problems. The Mastery of GPs is crucial for the understanding more complex mathematical concepts and for the practical applications in the fields like finance and engineering. By working through these problems students can solidify their grasp of the geometric sequences and their applications.