# Geometric Progression : Problem and Solutions

1.        If $\displaystyle a, b, c, d$ are in $\displaystyle G.P.,$ prove that $\displaystyle a+b, b+c, c+d$ are also in $\displaystyle G.P.$

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 $\displaystyle a, b, c, d$ are in $\displaystyle G.P.$ Let $\displaystyle r$ be the common ratio. $\displaystyle \begin{array}{l}\therefore \ a=a,\ b=ar,\ c=a{{r}^{2}},\ d=a{{r}^{3}}\\\\\therefore \ \displaystyle \frac{{b+c}}{{a+b}}=\displaystyle \frac{{ar+a{{r}^{2}}}}{{a+ar}}\\\\\therefore \ \displaystyle \frac{{b+c}}{{a+b}}=\displaystyle \frac{{r\left( {ar+ar} \right)}}{{a+ar}}\\\\\therefore \ \displaystyle \frac{{b+c}}{{a+b}}=r\\\\\text{Again},\\\\\ \ \ \displaystyle \frac{{c+d}}{{b+c}}=\displaystyle \frac{{a{{r}^{2}}+a{{r}^{3}}}}{{ar+a{{r}^{2}}}}\\\\\therefore \ \displaystyle \frac{{c+d}}{{b+c}}=\displaystyle \frac{{r\left( {ar+a{{r}^{2}}} \right)}}{{ar+a{{r}^{2}}}}\\\\\therefore \ \displaystyle \frac{{c+d}}{{b+c}}=r\\\\\therefore \displaystyle \frac{{b+c}}{{a+b}}=\displaystyle \frac{{c+d}}{{b+c}}\end{array}$ $\displaystyle \therefore \ a+b, b+c, c+d$ are also in $\displaystyle G.P.$

2.         Find three numbers in $\displaystyle G.P.$ whose sum is $\displaystyle 13$ and the sum of whose squares is $\displaystyle 91$.

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 Let the three numbers in $\displaystyle G.P.$ be $\displaystyle a, ar, ar^2$ By the problem, $\displaystyle a+ar+ar^2=13$ $\displaystyle a(1+r+r^2)=13---(1)$ $\displaystyle a^2+(ar)^2+(ar^2)^2=91$ $\displaystyle \therefore \ a^2+a^2r^2+a^2r^4=91$ $\displaystyle \therefore \ a^2(1+r^2+r^4)=91---(2)$ Squaring both sides of equation $\displaystyle (1),$ $\displaystyle a^2(1+r+r^2)^2=169$ $\displaystyle \therefore a^2(1+2r+3{{r}^{2}}+2{{r}^{3}}+{{r}^{4}})=169$ $\displaystyle \begin{array}{l}\therefore {{a}^{2}}\left[ {1+2r+3{{r}^{2}}+2{{r}^{3}}+{{r}^{4}}} \right]=169\\\\\therefore {{a}^{2}}\left[ {1+{{r}^{2}}+{{r}^{4}}+2r+2{{r}^{2}}+2{{r}^{3}}} \right]=169\\\\\therefore {{a}^{2}}\left[ {\left( {1+{{r}^{2}}+{{r}^{4}}} \right)+2r\left( {1+r+{{r}^{2}}} \right)} \right]=169\\\\\therefore {{a}^{2}} {\left( {1+{{r}^{2}}+{{r}^{4}}} \right)+2r\cdot {{a}^{2}}\left( {1+r+{{r}^{2}}} \right)}=169\\\\\therefore 91+2ar\cdot 13=169\\\\\therefore 26ar=78\\\\\therefore ar=3\\\\\therefore a=\displaystyle \frac{3}{r}\end{array}$ Substituting $\displaystyle a=\frac{3}{r}$ in equation $\displaystyle (1),$ $\displaystyle \begin{array}{l}\ \ \frac{3}{r}(1+r+{{r}^{2}})=13\\\\\therefore 3+3r+3{{r}^{2}}=13r\\\\\therefore 3{{r}^{2}}-10r+3=0\\\\\therefore (3r-1)(r-3)=0\end{array}$ $\displaystyle \therefore r=\frac{1}{3}$ (or) $\displaystyle r=3.$ When $\displaystyle r=\frac{1}{3}, a= 9.$ Therefore the numbers are 9, 3 and 1. When $\displaystyle r=3, a= 1.$ Therefore the numbers are 1, 3 and 9.

3.        The product of three consecutive terms of a $\displaystyle G.P.$ is $\displaystyle 8.$ The sum of product of these terms taken in pairs is $\displaystyle 14.$ Find the numbers.

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 Let the three consecutive terms in $\displaystyle G.P.$ be $\displaystyle \frac{a}{r}, a, ar$ By the problem, $\displaystyle \frac{a}{r}\cdot a\cdot ar=8$ $\displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{a}{r}\cdot a\cdot ar=8\\\\\therefore \ {{a}^{3}}=8\\\\\therefore \ a=2\\\\\ \ \text{Again},\\\\\ \ \displaystyle \frac{a}{r}\cdot a+a\cdot ar+\displaystyle \frac{a}{r}\cdot a=14\\\\\therefore \displaystyle \frac{{{{a}^{2}}}}{r}+{{a}^{2}}r+{{a}^{2}}=14\\\\\therefore {{a}^{2}}\left( {\displaystyle \frac{1}{r}+r+1} \right)=14\\\\\therefore 4\left( {\displaystyle \frac{1}{r}+r+1} \right)=14\\\\\therefore \displaystyle \frac{1}{r}+r+1=\displaystyle \frac{7}{2}\\\\\therefore 2+2{{r}^{2}}+2r=7r\\\\\therefore 2{{r}^{2}}-5r+2=0\\\\\therefore (2r-1)(r-2)=0\\\\\therefore \ r=\displaystyle \frac{1}{2}\ (\text{or})\ r=2\end{array}$ Therefore the numbers are 4, 2 , 1 or 1, 2, 4.

4.         The sum of two positive numbers is $\displaystyle 6$ times their geometric mean. Show that the numbers are in the ratio $\displaystyle \left( {3+2\sqrt{2}} \right):\left( {3-2\sqrt{2}} \right)$.

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 Let the two positive numbers be $\displaystyle a$ and $\displaystyle b.$ $\displaystyle G.M.$ between $\displaystyle a$ and $\displaystyle b.$ = $\displaystyle \sqrt{ab}$ By the problem, $\displaystyle \begin{array}{l}\ \ \ a+b=6\sqrt{{ab}}\\\\\therefore \ {{a}^{2}}+2ab+{{b}^{2}}=36ab\\\\\therefore \ {{a}^{2}}+{{b}^{2}}=34ab\\\\\therefore \ \displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{ab}}=34\\\\\therefore \ \displaystyle \frac{a}{b}+\displaystyle \frac{b}{a}=34\\\\\text{Let}\ \displaystyle \frac{a}{b}=x.\\\\\therefore \displaystyle \frac{b}{a}=\displaystyle \frac{1}{x}\\\\\therefore x+\displaystyle \frac{1}{x}=34\\\\\therefore {{x}^{2}}-34x+1=0\\\\\therefore {{x}^{2}}-34x=-1\\\\\therefore {{x}^{2}}-34x+289=288\\\\\therefore {{(x-17)}^{2}}=288\\\\\therefore x-17=12\sqrt{2}\\\\\therefore x=17+12\sqrt{2}\\\\\therefore x=9+12\sqrt{2}+8\\\\\therefore x={{3}^{2}}+2\left( 3 \right)\left( {2\sqrt{2}} \right)+{{\left( {2\sqrt{2}} \right)}^{2}}\\\\\therefore x={{\left( {3+2\sqrt{2}} \right)}^{2}}\\\\\therefore x=\displaystyle \frac{{{{{\left( {3+2\sqrt{2}} \right)}}^{2}}}}{{9-8}}\\\\\therefore x=\displaystyle \frac{{{{{\left( {3+2\sqrt{2}} \right)}}^{2}}}}{{{{3}^{2}}-{{{\left( {2\sqrt{2}} \right)}}^{2}}}}\\\\\therefore x=\displaystyle \frac{{\left( {3+2\sqrt{2}} \right)\left( {3+2\sqrt{2}} \right)}}{{\left( {3+2\sqrt{2}} \right)\left( {3-2\sqrt{2}} \right)}}\\\\\therefore x=\displaystyle \frac{{3+2\sqrt{2}}}{{3-2\sqrt{2}}}\\\\\therefore \displaystyle \frac{a}{b}=\displaystyle \frac{{3+2\sqrt{2}}}{{3-2\sqrt{2}}}\end{array}$

5.         The sum of an infinite $\displaystyle G.P$ is $\displaystyle 57$ and the sum of their cubes is $\displaystyle 9747,$ find the fourth term.

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 Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$. $\displaystyle \therefore$ Given $\displaystyle G.P$ is $\displaystyle a, ar, ar^2,...$ Let sum to infinity be $\displaystyle S$. $\displaystyle \therefore S =57$ Since $\displaystyle S=\frac{a}{{1-r}},$ $\displaystyle \frac{a}{{1-r}}=57---(1)$ When each terms are cubed, given $\displaystyle G.P$ becomes ... $\displaystyle a^3, a^3r^3, a^3r^6,...$ It is also a $\displaystyle G.P$ with the first term $\displaystyle a^3$ and the common ratio $\displaystyle r^3.$ Let the sum to infinity of that $\displaystyle G.P$ be $\displaystyle S^{*}$ $\displaystyle \therefore S^{*} =9747$ Since $\displaystyle {{S}^{*}}=\frac{{{{a}^{3}}}}{{1-{{r}^{3}}}},$ $\displaystyle \frac{{{{a}^{3}}}}{{1-{{r}^{3}}}}=9747---(2)$ Cubing equation $\displaystyle (1)$ on both sides, $\displaystyle \ \ \ \frac{{{{a}^{3}}}}{{{{{\left( {1-r} \right)}}^{3}}}}={{57}^{3}}---(3)$ Dividing equation $\displaystyle (3)$ by equation $\displaystyle (2),$ $\displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{\displaystyle \frac{{{{a}^{3}}}}{{{{{\left( {1-r} \right)}}^{3}}}}}}{{\displaystyle \frac{{{{a}^{3}}}}{{1-{{r}^{3}}}}}}=\displaystyle \frac{{{{{57}}^{3}}}}{{9747}}\\\\\therefore \ \ \displaystyle \frac{{{{a}^{3}}}}{{{{{\left( {1-r} \right)}}^{3}}}}\times \displaystyle \frac{{1-{{r}^{3}}}}{{{{a}^{3}}}}=19\\\\\therefore \ \ \displaystyle \frac{{\left( {1-r} \right)\left( {1+r+{{r}^{2}}} \right)}}{{\left( {1-r} \right)\left( {1-2r+{{r}^{2}}} \right)}}=19\\\\\therefore \ \ \displaystyle \frac{{1+r+{{r}^{2}}}}{{1-2r+{{r}^{2}}}}=19\\\\\therefore \ \ 1+r+{{r}^{2}}=19-38r+19{{r}^{2}}\\\\\therefore \ \ 18{{r}^{2}}-39r+18=0\\\\\therefore \ \ 6{{r}^{2}}-13r+6=0\\\\\therefore \ \ (3r-2)(2r-3)=0\\\\\therefore \ \ r=\displaystyle \frac{2}{3}\ (\text{or})\ r=\displaystyle \frac{3}{2}\end{array}$ Since the sum to infinity exists, |$\displaystyle r$|<1. $\displaystyle \begin{array}{l}\therefore \ \ r=\displaystyle \frac{2}{3}\\\\\therefore \ \ \displaystyle \frac{a}{{1-\displaystyle \frac{2}{3}}}=57\\\\\therefore \ \ a=19\\\\\therefore \ \ {{u}_{4}}=a{{r}^{3}}\\\\\therefore \ \ {{u}_{4}}=19\times \displaystyle \frac{8}{{27}}\\\\\therefore \ \ {{u}_{4}}=\displaystyle \frac{{152}}{{27}}\end{array}$

6.         The sum of an infinite $\displaystyle G.P$ is $\displaystyle x$ and the sum of their squares is $\displaystyle y,$ find the common ratio of that $\displaystyle G.P$ in terms of $\displaystyle x$ and $\displaystyle y$.

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 Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$. $\displaystyle \therefore$ Given $\displaystyle G.P$ is $\displaystyle a, ar, ar^2,...$ By the problem, the sum to infinity of the progression is $\displaystyle x$. $\displaystyle \therefore \ \ \frac{a}{{1-r}}=x$ $\displaystyle \therefore \ \ a=x(1-r)$ When each terms are squared, given $\displaystyle G.P$ becomes, $\displaystyle a^2, a^2r^2, a^2r^4,...$ It is also a $\displaystyle G.P$ with the first term $\displaystyle a^2$ and the common ratio $\displaystyle r^2.$ By the problem, the sum to infinity of above $\displaystyle G.P$ is $\displaystyle y$. $\displaystyle \begin{array}{l}\therefore \ \ \displaystyle \frac{{{{a}^{2}}}}{{1-{{r}^{2}}}}=y\\\\\therefore \ \ \displaystyle \frac{a}{{1-r}}\times \displaystyle \frac{a}{{1+r}}=y\\\\\therefore \ \ x\times \displaystyle \frac{a}{{1+r}}=y\\\\\therefore \ \ x\times \displaystyle \frac{{x(1-r)}}{{1+r}}=y\\\\\therefore \ \ \displaystyle \frac{{1+r}}{{1-r}}=\displaystyle \frac{{{{x}^{2}}}}{y}\\\\\therefore \ \ \displaystyle \frac{{\left( {1+r} \right)+\left( {1-r} \right)}}{{\left( {1+r} \right)-\left( {1-r} \right)}}=\displaystyle \frac{{{{x}^{2}}+y}}{{{{x}^{2}}-y}}\\\ \ \ \ \left[ {\because \ \text{componendo and dividendo}} \right]\\\\\therefore \ \ \displaystyle \frac{2}{{2r}}=\displaystyle \frac{{{{x}^{2}}+y}}{{{{x}^{2}}-y}}\\\\\therefore \ \ r=\displaystyle \frac{{{{x}^{2}}-y}}{{{{x}^{2}}+y}}\end{array}$

7.         Given that $\displaystyle a, b, c$ are in $\displaystyle A.P.$ and $\displaystyle a^2, b^2, c^2$ are in $\displaystyle G.P.$ If $\displaystyle a<b<c$ and $\displaystyle a+b+c=\frac{3}{2},$ Find the value of $\displaystyle a, b$ and $\displaystyle c.$

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 $\displaystyle a, b, c$ are in $\displaystyle A.P.$ (given) Let the common difference be $\displaystyle d$ where $\displaystyle d\ne 0$. $\displaystyle \therefore \ a=b-d\ \operatorname{and}\ c=b+d$ By the problem, $\displaystyle \begin{array}{l}\ \ \ \ a+b+c=\displaystyle \frac{3}{2}\\\\\therefore \ \ b-d+b+b+d=\displaystyle \frac{3}{2}\\\\\therefore \ \ 3b=\displaystyle \frac{3}{2}\\\\\therefore \ \ b=\displaystyle \frac{1}{2}\\\\\therefore \ a=b-\displaystyle \frac{1}{2}\ \operatorname{and}\ c=b+\displaystyle \frac{1}{2}\end{array}$ $\displaystyle a^2, b^2, c^2$ are in $\displaystyle G.P.$ (given) $\displaystyle \therefore \ {{\left( {\frac{1}{2}-d} \right)}^{2}},{\displaystyle \frac{1}{4}},{{\left( {\frac{1}{2}+d} \right)}^{2}}$ are in $\displaystyle G.P.$ $\displaystyle \begin{array}{l}\therefore \displaystyle \frac{{\displaystyle \frac{1}{4}}}{{{{{\left( {\displaystyle \frac{1}{2}-d} \right)}}^{2}}}}=\displaystyle \frac{{{{{\left( {\displaystyle \frac{1}{2}+d} \right)}}^{2}}}}{{\displaystyle \frac{1}{4}}}\\\\\therefore {{\left( {\displaystyle \frac{1}{4}-{{d}^{2}}} \right)}^{2}}=\displaystyle \frac{1}{{16}}\\\\\therefore \displaystyle \frac{1}{4}-{{d}^{2}}=\displaystyle \frac{1}{4}\ (\text{or})\displaystyle \frac{1}{4}-{{d}^{2}}=-\displaystyle \frac{1}{4}\\\\\therefore {{d}^{2}}=0\ (\text{or})\ {{d}^{2}}=\displaystyle \frac{1}{2}\end{array}$ But $\displaystyle {{d}^{2}}=0$ is impossible, $\displaystyle \begin{array}{l}\therefore \ {{d}^{2}}=\displaystyle \frac{1}{2}\\\\\therefore \ d=\pm \displaystyle \frac{1}{{\sqrt{2}}}\end{array}$ Since $\displaystyle a 8. Find the sum to infinity of the series$ \displaystyle 1+\frac{2}{3}+\frac{4}{{{{3}^{2}}}}+\frac{{10}}{{{{3}^{3}}}}+\frac{{14}}{{{{3}^{4}}}}+...$Show/Hide Solution  Let$ \displaystyle S=1+\frac{2}{3}+\frac{4}{{{{3}^{2}}}}+\frac{{10}}{{{{3}^{3}}}}+\frac{{14}}{{{{3}^{4}}}}+... ---(1) \displaystyle \therefore \frac{1}{3}S=\frac{1}{3}+\frac{2}{{{{3}^{2}}}}+\frac{4}{{{{3}^{3}}}}+\frac{{10}}{{{{3}^{4}}}}+\frac{{14}}{{{{3}^{6}}}}+...\ ---(2)$By equation$ \displaystyle (1)$- equation$ \displaystyle (2) \displaystyle \begin{array}{l}\ \ \ \ S-\displaystyle \frac{1}{3}S=1+\displaystyle \frac{1}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}+\displaystyle \frac{4}{{{{3}^{3}}}}+\displaystyle \frac{4}{{{{3}^{4}}}}+\displaystyle \frac{4}{{{{3}^{6}}}}+...\\\\\therefore \ \ S\left( {1-\displaystyle \frac{1}{3}} \right)=\displaystyle \frac{4}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}\left( {1+\displaystyle \frac{1}{3}+\displaystyle \frac{1}{{{{3}^{2}}}}+\displaystyle \frac{1}{{{{3}^{3}}}}+...} \right)\\\\\therefore \ \ \displaystyle \frac{2}{3}S=\displaystyle \frac{4}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}\left( {\displaystyle \frac{1}{{1-\displaystyle \frac{1}{3}}}} \right)\\\\\therefore \ \ \displaystyle \frac{2}{3}S=\displaystyle \frac{4}{3}+\displaystyle \frac{4}{{{{3}^{2}}}}\times \displaystyle \frac{3}{2}\\\\\therefore \ \ \displaystyle \frac{2}{3}S=2\\\\\therefore \,\ \ S=3\end{array}$9. If the$ \displaystyle (p+q)^{\text{th}}$term of a$ \displaystyle G.P$is$ \displaystyle m$and the$ \displaystyle (p-q)^{\text{th}}$term is$ \displaystyle n,$find the$ \displaystyle p^{\text{th}}$term in terms of$ \displaystyle m$and$ \displaystyle n$. Show/Hide Solution  Let the first term of the$ \displaystyle G.P$be$ \displaystyle a$and the common ratio be$ \displaystyle r$. By the problem,$ \displaystyle \begin{array}{l}\,\ \ \ {{u}_{{p+q}}}=m\\\\\therefore \ \ a{{r}^{{p+q-1}}}=m\\\\\,\ \ \ {{u}_{{p-q}}}=n\\\\\therefore \ \ a{{r}^{{p-q-1}}}=n\\\\\therefore \ a{{r}^{{p+q-1}}}\times a{{r}^{{p-q-1}}}=mn\\\\\therefore \ {{a}^{2}}{{r}^{{2p-2}}}=mn\\\\\therefore \ {{(a{{r}^{{p-1}}})}^{2}}=mn\\\\\therefore \ a{{r}^{{p-1}}}=\sqrt{{mn}}\\\\\therefore \ {{u}_{p}}=\sqrt{{mn}}\end{array}$10. If$ \displaystyle x>1$and$ \displaystyle {{\log }_{2}}x$,$ \displaystyle {{\log }_{3}}x$,$ \displaystyle {{\log }_{x}}16$are in$ \displaystyle G.P.$, find the value of$ \displaystyle x$. Show/Hide Solution $ \displaystyle {{\log }_{3}}x$,$ \displaystyle {{\log }_{x}}16$are in$ \displaystyle G.P.$.$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \displaystyle \frac{{{{{\log }}_{3}}x}}{{{{{\log }}_{2}}x}}=\displaystyle \frac{{{{{\log }}_{x}}16}}{{{{{\log }}_{3}}x}}\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}={{\log }_{2}}x\cdot {{\log }_{x}}16\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=\displaystyle \frac{{{{{\log }}_{x}}16}}{{{{{\log }}_{x}}2}}\ \left[ {\because {{{\log }}_{b}}a=\displaystyle \frac{1}{{{{{\log }}_{a}}b}}} \right]\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}={{\log }_{2}}16\ \left[ {\because \displaystyle \frac{{{{{\log }}_{b}}x}}{{{{{\log }}_{b}}y}}={{{\log }}_{y}}x} \right]\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}={{\log }_{2}}{{2}^{4}}\\\\\therefore \ \ \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\\therefore \ \ \ \ {{\log }_{3}}x=\pm 2\\\\\therefore \ \ \ x={{3}^{2}}\ (\text{or})\ x={{3}^{{-2}}}\\\\\therefore \ \ \ x=9\ (\text{or})\ x=\displaystyle \frac{1}{9}\end{array}$Since$ \displaystyle x>1$,$ \displaystyle x=\frac{1}{9}$is impossible.$ \displaystyle \therefore \ \ \ x=9$11. How many terms of the sesies$ \displaystyle \sqrt{3}+ 3+ 3\sqrt{3}+ ...$gives the sum$ \displaystyle 39+13\sqrt{3}$. Show/Hide Solution  Given series :$ \displaystyle \sqrt{3}+ 3+ 3\sqrt{3}+ ... \displaystyle \left. \begin{array}{l}\displaystyle \frac{3}{{\sqrt{3}}}=\sqrt{3}\\\\\displaystyle \frac{{3\sqrt{3}}}{3}=\sqrt{3}\end{array} \right\}\text{yields}\ \text{common ratio} \displaystyle \therefore\ \ \ \ $Given terms are in$ \displaystyle G.P.$with$ \displaystyle a=\sqrt{3}$and$ \displaystyle d=\sqrt{3}$. By the problem,$ \displaystyle \begin{array}{l}\ \ \ \ {{S}_{n}}=39+13\sqrt{3}\\\\\therefore \ \ \displaystyle \frac{{a\left( {{{r}^{n}}-1} \right)}}{{r-1}}=39+13\sqrt{3}\\\\\therefore \ \ \displaystyle \frac{{\sqrt{3}\left( {{{{\left( {\sqrt{3}} \right)}}^{n}}-1} \right)}}{{\sqrt{3}-1}}=13\sqrt{3}\left( {\sqrt{3}+1} \right)\\\\\therefore \ \ \sqrt{3}\left( {{{{\left( {\sqrt{3}} \right)}}^{n}}-1} \right)=13\sqrt{3}\left( {\sqrt{3}+1} \right)\left( {\sqrt{3}-1} \right)\\\\\therefore \ \ {{\left( {\sqrt{3}} \right)}^{n}}-1=13\left( {3-1} \right)\\\\\therefore \ \ {{\left( {\sqrt{3}} \right)}^{n}}=27\\\\\therefore \ \ {{\left( {\sqrt{3}} \right)}^{n}}=\ {{\left( {\sqrt{3}} \right)}^{6}}\\\\\therefore \ \ n=6\end{array}\$