# 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 A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}.Now, these 6 terms are to be added between 27 and 1/81.

So the G.P. becomes, 27, A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6},1/81 with first term(a) = 27, number of terms(n) = 8 and 8th term(a_{8}) = 1/81.We know nth term of a G.P. is given by a

_{n}= ar^{n-1}, where r is the common ratio.=> a

_{8}= 1/81=> 27(r)

^{8-1}= 1/81=> r

^{7}= 1/(81Ã—27)=> r

^{7}= (1/3)^{7}=> r = 1/3

So, geometric means are:

A

_{1}= ar = 27(1/3) = 9A

_{2}= ar^{2}= 27(1/3)^{2}= 3A

_{3}= ar^{3}= 27(1/3)^{3}= 1A

_{4}= ar^{4}= 27(1/3)^{4}= 1/3A

_{5}= ar^{5}= 27(1/3)^{5}= 1/9A

_{6}= ar^{6}= 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 A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}.Now, these 5 terms are to be added between 16 and 1/4.

So the G.P. becomes, 16, A

_{1}, A_{2}, A_{3}, A_{4}, A_{5},1/4 with first term(a) = 16, number of terms(n) = 7 and 7th term(a_{7}) = 1/4.We know nth term of a G.P. is given by a

_{n}= ar^{n-1}, where r is the common ratio.=> a

_{7}= 1/4=> 16(r

^{7-1}) = 1/4=> r

^{6}= 1/64=> r

^{6}= (1/2)^{6}=> r = 1/2

So, geometric means are:

A

_{1}= ar = 16(1/2) = 8A

_{2}= ar^{2}= 16(1/2)^{2}= 4A

_{3}= ar^{3}= 16(1/2)^{3}= 2A

_{4}= ar^{4}= 16(1/2)^{4}= 1A

_{5}= ar^{5}= 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 A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}.Now, these 5 terms are to be added between 32/9 and 81/2.

So the G.P. becomes, 32/9, A

_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 81/2 with first term(a) = 32/9, number of terms(n) = 7 and 7th term(a_{7}) = 81/2.We know nth term of a G.P. is given by a

_{n}= ar^{n-1}, where r is the common ratio.=> a

_{7}= 81/2=> (32/9)(r

^{7-1}) = 81/2=> r

^{6}= (81Ã—9)/(2Ã—32)=> r

^{6}= (3/2)^{6}=> r = 3/2

So, geometric means are:

A

_{1}= ar = (32/9)Ã—(3/2) = 16/3A

_{2}= ar^{2}= (32/9)Ã—(3/2)^{2}= 8A

_{3}= ar^{3}= (32/9)Ã—(3/2)^{3}= 12A

_{4}= ar^{4}= (32/9)Ã—(3/2)^{4}= 18A

_{5 }= ar^{5}= (32/9)Ã—(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) a ^{3}b and ab^{3}**

**(iii) â€“8 and â€“2**

**Solution:**

We know geometric mean between two numbers, a and b is given by .

(i)2 and 8Here, a = 2 and b = 8

So, G.M. =

=

= 4

(ii)a^{3}b and ab^{3}Here, a = a

^{3}b and b = ab^{3}So, G.M. =

=

= a

^{2}b^{2}

(iii)â€“8 and â€“2Here, a = â€“8 and b = â€“2

G.M. =

=

= 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 .

According to the question,

a =

=

=

Therefore, the value of a is.

**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 and arithmetic mean between two numbers, a and b is given by (a+b)/2.

According to the question,

=> = 20 â€¦â€¦. (1)

And

=> (a+b)/2 = 25

=> a+b = 50

=> b = 50â€“a â€¦â€¦. (2)

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

=> = 20

Squaring both sides, we get,

=> a(50 â€“a) = 400

=> a

^{2}â€“ 50a + 400 = 0=> a

^{2}â€“ 40aâ€“10a+400 = 0=> a(aâ€“ 40) â€“ 10(aâ€“ 40) = 0

=> (aâ€“ 40) (aâ€“ 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 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 â€¦.. (1)

And G.M. of roots = = G

ab = G

^{2}â€¦ (2)Now, we know a quadratic equation in x with roots a and b is given by,

x

^{2}â€“ (a+b)x + (ab) = 0From (1) and (2), we get,

x

^{2}â€“ 2Ax + G^{2}= 0

Therefore, the required quadratic equation is x^{2}â€“ 2Ax + G^{2}= 0.

**Question 8. The sum of two numbers is 6 times their geometric means. Show that the numbers are in the ratio **

**Solution:**

Let the two numbers be a and b. We know geometric mean between two numbers, a and b is given by .

According to the question,

=> a+b = 6

=> =

Applying Componendo and Dividendo on both sides, we get,

=> =

=> =

=> =

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

=> =

=> =

=> =

Squaring both sides, we get

=> =

=> =

=> =

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 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 â€¦.. (1)

And G.M. of roots = = 5

ab = 25 .â€¦ (2)

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

x

^{2}â€“ (a+b)x + (ab) = 0From (1) and (2), we get,

x

^{2}â€“ 16x + 25 = 0

Therefore, the required quadratic equation is x^{2}â€“ 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:**

According to the question,

=> = 8 â€¦â€¦. (1)

And,

=> (a+b)/2 = 10

=> a+b = 20

=> b = 20â€“a â€¦â€¦. (2)

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

=> = 8

Squaring both sides, we get,

=> a(20â€“a) = 64

=> a

^{2}â€“20a+64 = 0=> a

^{2}â€“16aâ€“4a+64 = 0=> a(aâ€“16) â€“ 4(aâ€“16) = 0

=> (aâ€“4) (aâ€“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}= ar^{n+1 }We know geometric mean between two numbers, a and b is given by .

The GM of a and ar

^{n+1}will be, G_{1}= =Hence, L.H.S. =

^{ }=Now, R.H.S. = Product of n geometric means between these two quantities, G

_{2}= ar Ã— ar^{2}Ã— . . . . Ã— ar^{n}=

=

=

=

= 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 = **

**Solution:**

According to the question,

AM = 2(GM)

=> =

=> =

Applying Componendo and Dividendo on both sides, we get,

=> =

=> =

=> =

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

=> =

=> =

=> =

Squaring both sides, we get

=> =

=> =

=> =

=> =

Hence, proved.

**Question 13. If one AM, A**,** and two geometric means G**_{1} and G_{2} are inserted between any two positive numbers, show that

_{1}and G

_{2}are inserted between any two positive numbers, show that

**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,G

_{1},G_{2},b.Now, we know b = ar

^{4-1}, where r is the common ratio.=> r

^{3}==> r =

So, G

_{1}= ar = =G

_{2}= ar^{2}= =Now, L.H.S. =

=

= ab

^{0}+ a^{0}b= a+b

= 2A [From (1)]

= R.H.S.

Hence, proved.

## Please

Loginto comment...