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

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.