Note:

• If there are $\displaystyle n$ Geometric Means inserted between $\displaystyle a \text{ and } b$, then
• $\displaystyle r = \Big( \frac{b}{a} \Big)^{\frac{1}{n+1}}$
• Geometric mean between $\displaystyle a \text{ and } b$ is $\displaystyle \sqrt{ab}$

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{i) } 2 \text{ and } 8 \hspace{1.0cm} \text{ii) } a^3b \text{ and } ab^3 \hspace{1.0cm} \text{iii) } -8 \text{ and } - 2$

$\displaystyle \text{i) } 2 \text{ and } 8$

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

$\displaystyle \text{ii) } a^3b \text{ and } ab^3$

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

$\displaystyle \text{iii) } -8 \text{ and } - 2$

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

$\displaystyle \\$

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

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

$\displaystyle \\$

$\displaystyle \text{Question 6: Find the two numbers whose A.M. is } 25 \text{ and GM is } 20$.

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

$\displaystyle \text{Also GM } = 20 \Rightarrow \sqrt{ab} = 20 \Rightarrow ab = 400 .$ .. … … … … ii)

Solving i) and ii) we get

$\displaystyle a( 50 -a ) = 400$

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

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

$\displaystyle \Rightarrow a = 40$ or $\displaystyle a = 10$

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

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

$\displaystyle \text{Hence the two numbers are } 10 \text{ and } 40$

$\displaystyle \\$

$\displaystyle \text{Question 7: Construct a quadratic in } x \text{ such that A.M. of its roots is } A \text{ and G.M. is } G$.

Let the two roots of the quadratic equation be $\displaystyle a \text{ and } b$

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

$\displaystyle \text{Also } G^2 = ab .$ .. … … … … ii)

We know the quadratic equation having roots $\displaystyle a \text{ and } b$ is given by $\displaystyle x^2 - ( a+b) x + ab = 0$

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

$\displaystyle \\$

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

$\displaystyle \text{Let the two numbers be } a \text{ and } b$

$\displaystyle \text{Given } a+b = 6 \sqrt{ab}$

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

Applying componendo and dividendo

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

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

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

Applying componendo and dividendo once again

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

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

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

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

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

$\displaystyle \\$

Question 9: If AM and GM of roots of a quadratic equation are $\displaystyle 8 \text{ and } 5$ respectively, then obtain the quadratic equation.

Let the two roots of the quadratic equation be $\displaystyle a \text{ and } b$

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

$\displaystyle \text{Also GM } = 5 \Rightarrow \sqrt{ab} = 5 \Rightarrow ab = 25$

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

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

$\displaystyle \\$

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

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

$\displaystyle \text{Also GM } = 8 \Rightarrow \sqrt{ab} = 8 \Rightarrow ab = 64 .$ .. … … … … ii)

Solving i) and ii) we get

$\displaystyle a( 20-a) = 6$

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

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

$\displaystyle \Rightarrow a = 16 \ or \ a = 4$

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

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

$\displaystyle \text{Hence the two numbers are } 4 \text{ and } 16$.

$\displaystyle \\$

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

$\displaystyle \text{Let } G_1, G_2, G_3, \ldots , G_n$ be the $\displaystyle n$ GMs inserted between $\displaystyle a \text{ and } b$

$\displaystyle \text{Let } r$ be the common ratio.

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

$\displaystyle G_1 = ar$

$\displaystyle G_2 = ar^2$

$\displaystyle \ldots$

$\displaystyle G_n = ar^n$

$\displaystyle \text{Let } G$ be the GM between $\displaystyle a \text{ and } b$

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

$\displaystyle \text{Now } \text{Given } G_1 \times G_2 \times \ldots \times G_n$

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

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

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

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

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

$\displaystyle = ( \sqrt{ab})^n$

$\displaystyle = (G)^n$. Hence proved.

$\displaystyle \\$

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

$\displaystyle \text{Let the two numbers be } a \text{ and } b$

Given A.M. = 2 G.M.

$\displaystyle \text{Given } \frac{a+b}{2} = 2 \sqrt{ab}$

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

Applying componendo and dividendo

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

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

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

Applying componendo and dividendo once again

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Let the two numbers be } a \text{ and } b$

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

$\displaystyle \text{Let } G_1 \text{ and } G_2$ be two GM between $\displaystyle a \text{ and } b$

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

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

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

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

$\displaystyle \text{LHS } = \frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1}$

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

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