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

Last Updated : 07 Apr, 2021
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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 an = arn-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 an = arn-1, where r is the common ratio.

=> a7 = 1/4

=> 16(r7-1) = 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 an = arn-1, where r is the common ratio.

=> a7 = 81/2

=> (32/9)(r7-1) = 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 an = arn-1.

So, the last term of the GP ,i.e., (n+2)th term will be, an+2 = arn+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 = ar4-1, 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.



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