Note:

1. If there are n Geometric Means inserted between a and b , then

r = \Big( \frac{b}{a} \Big)^{\frac{1}{n+1}}

2. Geometric mean between a and b is \sqrt{ab}

Question 1: Insert 6 geometric means between 27 and \frac{1}{81} .

Answer:

Let G_1, G_2, G_3, G_4, G_5, G_6 be the  6 GMs inserted between a = 27 and b = \frac{1}{81} .

n = 6 ( As 6 geometric means are being inserted).

\therefore r = \Big( \frac{1/81}{27} \Big)^{\frac{1}{6+1}} = \Big( \frac{1}{3^7} \Big)^{\frac{1}{7}} = \frac{1}{3} . Therefore,

G_1 = ar^{ } = 27 \Big( \frac{1}{3} \Big)^{ } = 9

G_2 = ar^{2 } = 27 \Big( \frac{1}{3} \Big)^{2 } = 3

G_3 = ar^{3 } = 27 \Big( \frac{1}{3} \Big)^{3 } = 1

G_4 = ar^{4 } = 27 \Big( \frac{1}{3} \Big)^{4 } = \frac{1}{3}

G_5 = ar^{5 } = 27 \Big( \frac{1}{3} \Big)^{5 } = \frac{1}{9}

G_6 = ar^{6 } = 27 \Big( \frac{1}{3} \Big)^{ 6} = \frac{1}{27}

Hence the 6 geometric means between a = 27 and b = \frac{1}{81} are 9, 3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}

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Question 2: Insert 5 geometric means between 16 and \frac{1}{4} .

Answer:

Let G_1, G_2, G_3, G_4, G_5 be the  5 GMs inserted between a = 16 and b = \frac{1}{4} .

n = 5 ( As 5 geometric means are being inserted).

\therefore r = \Big( \frac{1/4}{16} \Big)^{\frac{1}{5+1}} = \Big( \frac{1}{3^6} \Big)^{\frac{1}{6}} = \frac{1}{2} . Therefore,

G_1 = ar^{ } = 16 \Big( \frac{1}{2} \Big)^{ } = 8

G_2 = ar^{2 } = 16 \Big( \frac{1}{2} \Big)^{2 } = 4

G_3 = ar^{3 } = 16 \Big( \frac{1}{2} \Big)^{3 } = 2

G_4 = ar^{4 } = 16 \Big( \frac{1}{2} \Big)^{4 } = 1

G_5 = ar^{5 } = 16 \Big( \frac{1}{2} \Big)^{5 } = \frac{1}{2}

Hence the 5 geometric means between a = 16 and b = \frac{1}{4} are 8, 4, 2, 1, \frac{1}{2}

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Question 3: Insert 5 geometric means between \frac{32}{9} and \frac{81}{2} .

Answer:

Let G_1, G_2, G_3, G_4, G_5 be the  5 GMs inserted between \frac{32}{9} and \frac{81}{2} .

n = 5 ( As 5 geometric means are being inserted).

\therefore r = \Big( \frac{81/2}{32/9} \Big)^{\frac{1}{5+1}} = \Big( \frac{3^6}{2^6} \Big)^{\frac{1}{6}} = \frac{3}{2} . Therefore,

G_1 = ar^{ } = \Big( \frac{32}{9} \Big) \Big( \frac{3}{2} \Big)^{ } = \frac{16}{3}

G_2 = ar^{2 } = \Big( \frac{32}{9} \Big) \Big( \frac{3}{2} \Big)^{2 } = 8

G_3 = ar^{3 } = \Big( \frac{32}{9} \Big) \Big( \frac{3}{2} \Big)^{3 } = 12

G_4 = ar^{4 } = \Big( \frac{32}{9} \Big) \Big( \frac{3}{2} \Big)^{4 } = 18

G_5 = ar^{5 } = \Big( \frac{32}{9} \Big) \Big( \frac{3}{2} \Big)^{5 } = 27

Hence the 5 geometric means between \frac{32}{9} and \frac{81}{2}  are \frac{16}{3} 8, 12, 18, 27

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

i) 2 \text{ and } 8         ii) a^3b \text{ and } ab^3         iii) -8 \text{ and }  - 2

Answer:

i)       2 \text{ and } 8

Geometric mean between  2 \text{ and } 8  = \sqrt{2 \times 8} = \sqrt{16} = 4

ii)      a^3b \text{ and } ab^3

Geometric mean between  a^3b \text{ and } ab^3  = \sqrt{a^3b \times ab^3} = \sqrt{a^4b^4} = a^2b^2

iii)     -8 \text{ and }  - 2

Geometric mean between  -8 \text{ and } -2  = \sqrt{(-8) \times (-2)} = \sqrt{16} = 4 

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Question 5: If a is the G.M. of 2 and \frac{1}{4} , find a

Answer:

a = \sqrt{2 \times \frac{1}{4} } = \frac{1}{\sqrt{2}}

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

Answer:

Given A.M. = 25 \Rightarrow \frac{a+b}{2} = 25 = a+b = 50      … … … … … i)

Also GM = 20 \Rightarrow \sqrt{ab} = 20 \Rightarrow ab = 400      … … … … … ii)

Solving i) and ii) we get

a( 50 -a ) = 400

\Rightarrow a^2 - 50a + 400 = 0

\Rightarrow ( a - 40)(a - 10) = 0

\Rightarrow a = 40 or  a = 10

When a = 40, b = \frac{400}{40} = 10

When a = 10, b = \frac{400}{10} = 40

Hence the two numbers are 10 and 40

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

Answer:

Let the two roots of the quadratic equation be a and b

\therefore A = \frac{a+b}{2} \Rightarrow a + b = 2 A    … … … … … i)

Also G^2 = ab      … … … … … ii)

We know the quadratic equation having roots a and b is given by x^2 - ( a+b) x + ab = 0

\Rightarrow x^2 - 2A x + G^2= 0

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

Answer:

Let the two numbers be a and b

Given  a+b = 6 \sqrt{ab}

\Rightarrow \frac{a+b}{2\sqrt{ab}} = \frac{3}{1}

Applying componendo and dividendo

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

\Rightarrow \Big( \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} \Big)^2 = \frac{4}{2}

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

Applying componendo and dividendo once again

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

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

\Rightarrow \frac{a}{b} =\Big( \frac{\sqrt{2}+1}{\sqrt{2}-1}\Big)^2

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

\Rightarrow \frac{a}{b} =\frac{3+2\sqrt{2}}{3-2\sqrt{2}} . Hence proved.

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

Answer:

Let the two roots of the quadratic equation be a and b

Given AM = 8 \Rightarrow \frac{a+b}{2} = 8 \Rightarrow a + b = 16     … … … … … i)

Also GM = 5 \Rightarrow \sqrt{ab} = 5 \Rightarrow ab = 25

Therefore the quadratic equation will be x^2 - ( a+b) x + ab = 0

\Rightarrow x^2 - 16x + 25 = 0

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Question 10: If AM and GM of two positive numbers a \ and \ b are 10 \ and \  8 respectively, find the numbers.

Answer:

Given AM = 10 \Rightarrow \frac{a+b}{2} =10 \Rightarrow a + b = 20      … … … … … i)

Also GM = 8 \Rightarrow \sqrt{ab} = 8 \Rightarrow ab = 64      … … … … … ii)

Solving i) and ii) we get

a( 20-a) = 6

\Rightarrow a^2 - 20a + 64 = 0

\Rightarrow ( a-16)(a-4) = 0

\Rightarrow a = 16 \ or \ a = 4

When a = 16, b = \frac{64}{16} = 4

When a = 4, b = \frac{64}{4} = 16

Hence the two numbers are 4 and 16 .

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Question 11: Prove that the product of n geometric means between two quantities is equal to the n^{th} power of a geometric mean of those two quantities.

Answer:

Let G_1, G_2, G_3, \ldots , G_n be the  n GMs inserted between a  and b 

Let r  be the common ratio.

\therefore r = \Big( \frac{b}{a} \Big)^{\frac{1}{n+1}}

G_1 = ar

G_2 = ar^2

\ldots

G_n = ar^n

Let G be the GM between a and b

\therefore G = \sqrt{ab} \Rightarrow G^2 = ab

Now Given G_1 \times G_2 \times \ldots \times G_n

= ar \times ar^2 \times \ldots \times ar^n

= a^n \cdot r^{1+2+3+\ldots+ n}

= a^n \Big[ \Big(  \frac{b}{a} \Big)^{\frac{1}{n+1}}  \Big]^{\frac{n(n+10)}{2}}

= a^n \Big( \frac{b}{a} \Big)^{\frac{n}{2}}

= a^{\frac{n}{2}} b^{\frac{b}{2}}

= ( \sqrt{ab})^n

= (G)^n . Hence proved.

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

Answer:

Let the two numbers be a and b

Given A.M. = 2 G.M.

Given  \frac{a+b}{2} = 2 \sqrt{ab}

\Rightarrow \frac{a+b}{2\sqrt{ab}} = \frac{2}{1}

Applying componendo and dividendo

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

\Rightarrow \Big( \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} \Big)^2 = \frac{3}{1}

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

Applying componendo and dividendo once again

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

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

\Rightarrow \frac{a}{b} =\Big( \frac{\sqrt{3}+1}{\sqrt{3}-1}\Big)^2

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

\Rightarrow \frac{a}{b} =\frac{2+\sqrt{3}}{2-\sqrt{3}} . Hence proved.

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Question 13: If one A.M., A and two geometric means G_1 and G_2 inserted between any two positive numbers, show that \frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1} = 2A

Answer:

Let the two numbers be a and b

\therefore A = \frac{a+b}{2}

Let G_1 and G_2 be two GM between a and b

\therefore r = ( \frac{b}{a} )^{\frac{1}{3}}

\therefore G_1 = a ( \frac{b}{a} )^{\frac{1}{3}} = a^{2/3} b^{1/3}

G_2 = a ( \frac{b}{a} )^{\frac{2}{3}} = a^{1/3} b^{2/3}

To prove : \frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1} = 2A

LHS = \frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1}

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

= a + b = 2 ( \frac{a+b}{2}) = 2A= RHS