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.$
Show/Hide Solution
| $ \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$.
Show/Hide Solution
| 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.
Show/Hide Solution
| 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)$.
Show/Hide Solution
| 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.
Show/Hide Solution
| 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$.
Show/Hide Solution
| 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.$
Show/Hide Solution
| $ \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<b<c$, $ \displaystyle {d=-\frac{1}{{\sqrt{2}}}}$ $ \displaystyle \begin{array}{l}\therefore \ \ a=\displaystyle \frac{1}{2}-\displaystyle \frac{1}{{\sqrt{2}}}\\\\\ \ \ \ b=\displaystyle \frac{1}{2}\\\\\ \ \ \ c=\displaystyle \frac{1}{2}+\displaystyle \frac{1}{{\sqrt{2}}}\end{array}$ |
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}$ |
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!