\displaystyle \textbf{Note:}
\displaystyle \text{If }n\text{ geometric means are inserted between two positive numbers }a\text{ and }b,\text{ then}
\displaystyle r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}},
\displaystyle \text{where }r\text{ is the common ratio of the resulting G.P.}
\displaystyle \text{The geometric mean between }a\text{ and }b\text{ is }\sqrt{ab}.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Insert }6\text{ geometric means between }27\text{ and }\frac{1}{81}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }G_1,G_2,G_3,G_4,G_5,G_6\text{ be the six geometric means inserted between }27\text{ and }\frac{1}{81}.
\displaystyle \text{Here }a=27,\quad b=\frac{1}{81},\quad n=6.
\displaystyle \therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}=\left(\frac{\frac{1}{81}}{27}\right)^{\frac{1}{7}}=\left(\frac{1}{3^7}\right)^{\frac{1}{7}}=\frac{1}{3}.
\displaystyle \therefore G_1=ar=27\left(\frac{1}{3}\right)=9.
\displaystyle G_2=ar^2=27\left(\frac{1}{3}\right)^2=3.
\displaystyle G_3=ar^3=27\left(\frac{1}{3}\right)^3=1.
\displaystyle G_4=ar^4=27\left(\frac{1}{3}\right)^4=\frac{1}{3}.
\displaystyle G_5=ar^5=27\left(\frac{1}{3}\right)^5=\frac{1}{9}.
\displaystyle G_6=ar^6=27\left(\frac{1}{3}\right)^6=\frac{1}{27}.
\displaystyle \therefore \text{The six geometric means are }9,\ 3,\ 1,\ \frac{1}{3},\ \frac{1}{9},\ \frac{1}{27}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Insert }5\text{ geometric means between }16\text{ and }\frac{1}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }G_1,G_2,G_3,G_4,G_5\text{ be the five geometric means inserted between }16\text{ and }\frac{1}{4}.
\displaystyle \text{Here }a=16,\quad b=\frac{1}{4},\quad n=5.
\displaystyle \therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}=\left(\frac{\frac{1}{4}}{16}\right)^{\frac{1}{6}}=\left(\frac{1}{2^6}\right)^{\frac{1}{6}}=\frac{1}{2}.
\displaystyle \therefore G_1=ar=16\left(\frac{1}{2}\right)=8.
\displaystyle G_2=ar^2=16\left(\frac{1}{2}\right)^2=4.
\displaystyle G_3=ar^3=16\left(\frac{1}{2}\right)^3=2.
\displaystyle G_4=ar^4=16\left(\frac{1}{2}\right)^4=1.
\displaystyle G_5=ar^5=16\left(\frac{1}{2}\right)^5=\frac{1}{2}.
\displaystyle \therefore \text{The five geometric means are }8,\ 4,\ 2,\ 1,\ \frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Insert }5\text{ geometric means between }\frac{32}{9}\text{ and }\frac{81}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }G_1,G_2,G_3,G_4,G_5\text{ be the five geometric means inserted between }\frac{32}{9}\text{ and }\frac{81}{2}.
\displaystyle \text{Here }a=\frac{32}{9},\quad b=\frac{81}{2},\quad n=5.
\displaystyle \therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}=\left(\frac{\frac{81}{2}}{\frac{32}{9}}\right)^{\frac{1}{6}}=\left(\frac{3^6}{2^6}\right)^{\frac{1}{6}}=\frac{3}{2}.
\displaystyle \therefore G_1=ar=\frac{32}{9}\times\frac{3}{2}=\frac{16}{3}.
\displaystyle G_2=ar^2=\frac{32}{9}\times\left(\frac{3}{2}\right)^2=8.
\displaystyle G_3=ar^3=\frac{32}{9}\times\left(\frac{3}{2}\right)^3=12.
\displaystyle G_4=ar^4=\frac{32}{9}\times\left(\frac{3}{2}\right)^4=18.
\displaystyle G_5=ar^5=\frac{32}{9}\times\left(\frac{3}{2}\right)^5=27.
\displaystyle \therefore \text{The five geometric means are }\frac{16}{3},\ 8,\ 12,\ 18,\ 27.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the geometric mean of each of the following pairs of numbers:}
\displaystyle \text{i) }2\text{ and }8\qquad\text{ii) }a^3b\text{ and }ab^3\qquad\text{iii) }-8\text{ and }-2
\displaystyle \text{Answer:}
\displaystyle \text{i) Geometric mean}=\sqrt{2\times8}=\sqrt{16}=4.
\displaystyle \text{ii) Geometric mean}=\sqrt{a^3b\times ab^3}=\sqrt{a^4b^4}=a^2b^2.
\displaystyle \text{iii) Geometric mean}=\sqrt{(-8)\times(-2)}=\sqrt{16}=4.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }a\text{ is the geometric mean of }2\text{ and }\frac{1}{4},\text{ find }a.
\displaystyle \text{Answer:}
\displaystyle a=\sqrt{2\times\frac{1}{4}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the two numbers whose A.M. is }25\text{ and G.M. is }20.
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Given, A.M.}=25
\displaystyle \Rightarrow \frac{a+b}{2}=25
\displaystyle \Rightarrow a+b=50\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, G.M.}=20
\displaystyle \Rightarrow \sqrt{ab}=20
\displaystyle \Rightarrow ab=400\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{From (i), }b=50-a.
\displaystyle \text{Substituting in (ii),}
\displaystyle a(50-a)=400
\displaystyle \Rightarrow a^2-50a+400=0
\displaystyle \Rightarrow (a-40)(a-10)=0
\displaystyle \Rightarrow a=40\text{ or }a=10
\displaystyle \text{When }a=40,\ b=50-40=10.
\displaystyle \text{When }a=10,\ b=50-10=40.
\displaystyle \therefore \text{The two numbers are }10\text{ and }40.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Construct a quadratic equation in }x\text{ such that the A.M. of its roots is }A
\displaystyle \text{and the G.M. of its roots is }G.
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots of the quadratic equation be }a\text{ and }b.
\displaystyle \text{Given, }A=\frac{a+b}{2}
\displaystyle \Rightarrow a+b=2A\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }G=\sqrt{ab}
\displaystyle \Rightarrow ab=G^2\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{The quadratic equation whose roots are }a\text{ and }b\text{ is}
\displaystyle x^2-(a+b)x+ab=0.
\displaystyle \text{Using (i) and (ii),}
\displaystyle x^2-2Ax+G^2=0.
\displaystyle \therefore \text{The required quadratic equation is }x^2-2Ax+G^2=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The sum of two numbers is }6\text{ times their geometric mean. Show that the numbers}
\displaystyle \text{are in the ratio }(3+2\sqrt{2}):(3-2\sqrt{2}).
\displaystyle \text{Answer:}
\displaystyle \text{Let the two positive numbers be }a\text{ and }b,\text{ where }a>b.
\displaystyle \text{Given, }a+b=6\sqrt{ab}
\displaystyle \Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{3}{1}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{3+1}{3-1}
\displaystyle \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=2
\displaystyle \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\sqrt{2}
\displaystyle \text{Applying componendo and dividendo 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{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}
\displaystyle \Rightarrow \frac{a}{b}=\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2
\displaystyle =\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}
\displaystyle =\frac{3+2\sqrt{2}}{3-2\sqrt{2}}
\displaystyle \therefore a:b=(3+2\sqrt{2}):(3-2\sqrt{2}).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the A.M. and G.M. of the roots of a quadratic equation are }8\text{ and }5
\displaystyle \text{respectively, obtain the quadratic equation.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots of the quadratic equation be }a\text{ and }b.
\displaystyle \text{Given, A.M.}=8
\displaystyle \Rightarrow \frac{a+b}{2}=8
\displaystyle \Rightarrow a+b=16\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, G.M.}=5
\displaystyle \Rightarrow \sqrt{ab}=5
\displaystyle \Rightarrow ab=25\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{The quadratic equation whose roots are }a\text{ and }b\text{ is}
\displaystyle x^2-(a+b)x+ab=0.
\displaystyle \text{Using (i) and (ii),}
\displaystyle x^2-16x+25=0.
\displaystyle \therefore \text{The required quadratic equation is }x^2-16x+25=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the A.M. and G.M. of two positive numbers }a\text{ and }b\text{ are }10\text{ and }8
\displaystyle \text{respectively, find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, A.M.}=10
\displaystyle \Rightarrow \frac{a+b}{2}=10
\displaystyle \Rightarrow a+b=20\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, G.M.}=8
\displaystyle \Rightarrow \sqrt{ab}=8
\displaystyle \Rightarrow ab=64\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{From (i), }b=20-a.
\displaystyle \text{Substituting in (ii),}
\displaystyle a(20-a)=64
\displaystyle \Rightarrow a^2-20a+64=0
\displaystyle \Rightarrow (a-16)(a-4)=0
\displaystyle \Rightarrow a=16\text{ or }a=4
\displaystyle \text{When }a=16,\ b=20-16=4.
\displaystyle \text{When }a=4,\ b=20-4=16.
\displaystyle \therefore \text{The two numbers are }4\text{ and }16.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove that the product of }n\text{ geometric means between two positive quantities}
\displaystyle \text{is equal to the }n^{\mathrm{th}}\text{ power of the geometric mean of those quantities.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }G_1,G_2,G_3,\ldots,G_n\text{ be the }n\text{ geometric means inserted between }a\text{ and }b.
\displaystyle \text{Let }r\text{ be the common ratio of the resulting G.P.}
\displaystyle \therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}
\displaystyle G_1=ar,\quad G_2=ar^2,\quad\ldots,\quad G_n=ar^n
\displaystyle \text{Let }G\text{ be the geometric mean between }a\text{ and }b.
\displaystyle \therefore G=\sqrt{ab}
\displaystyle \text{Now,}
\displaystyle G_1G_2\cdots G_n=(ar)(ar^2)\cdots(ar^n)
\displaystyle =a^nr^{1+2+\cdots+n}
\displaystyle =a^nr^{\frac{n(n+1)}{2}}
\displaystyle =a^n\left[\left(\frac{b}{a}\right)^{\frac{1}{n+1}}\right]^{\frac{n(n+1)}{2}}
\displaystyle =a^n\left(\frac{b}{a}\right)^{\frac{n}{2}}
\displaystyle =a^{\frac{n}{2}}b^{\frac{n}{2}}
\displaystyle =(ab)^{\frac{n}{2}}
\displaystyle =(\sqrt{ab})^n
\displaystyle =G^n
\displaystyle \therefore \text{The product of the }n\text{ geometric means is }G^n.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the A.M. of two positive numbers }a\text{ and }b,\text{ where }a>b,\text{ is twice their}
\displaystyle \text{geometric mean, prove that }a:b=(2+\sqrt{3}):(2-\sqrt{3}).
\displaystyle \text{Answer:}
\displaystyle \text{Given, A.M.}=2\text{ G.M.}
\displaystyle \therefore \frac{a+b}{2}=2\sqrt{ab}
\displaystyle \Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{2}{1}
\displaystyle \text{Applying componendo and dividendo,}
\displaystyle \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{2+1}{2-1}
\displaystyle \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=3
\displaystyle \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\sqrt{3}\qquad[a>b>0]
\displaystyle \text{Applying componendo and dividendo 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{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}
\displaystyle \Rightarrow \frac{a}{b}=\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2
\displaystyle =\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)^2}
\displaystyle =\frac{4+2\sqrt{3}}{4-2\sqrt{3}}
\displaystyle =\frac{2+\sqrt{3}}{2-\sqrt{3}}
\displaystyle \therefore a:b=(2+\sqrt{3}):(2-\sqrt{3}).
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If one arithmetic mean }A\text{ and two geometric means }G_1\text{ and }G_2\text{ are inserted}
\displaystyle \text{between two positive numbers, show that }\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=2A.
\displaystyle \text{Answer:}
\displaystyle \text{Let the two positive numbers be }a\text{ and }b.
\displaystyle \therefore A=\frac{a+b}{2}
\displaystyle \text{Let }G_1\text{ and }G_2\text{ be the two geometric means inserted between }a\text{ and }b.
\displaystyle \therefore a,G_1,G_2,b\text{ are in G.P.}
\displaystyle \text{Let }r\text{ be the common ratio. Then}
\displaystyle r=\left(\frac{b}{a}\right)^{\frac{1}{3}}
\displaystyle \therefore G_1=ar=a\left(\frac{b}{a}\right)^{\frac{1}{3}}=a^{\frac{2}{3}}b^{\frac{1}{3}}
\displaystyle G_2=ar^2=a\left(\frac{b}{a}\right)^{\frac{2}{3}}=a^{\frac{1}{3}}b^{\frac{2}{3}}
\displaystyle \text{LHS}=\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}
\displaystyle =\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}}}
\displaystyle =a+b
\displaystyle =2\left(\frac{a+b}{2}\right)
\displaystyle =2A
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

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