Geometric Progression : Problems and Solutions -Part (2)

Definition: Geometric Progression

A geometric progression is a sequence in which the ratio of each term to the one before it, is a constant.

ကိန်းစဉ်တစ်ခု၏ နီးစပ် (ကပ်လျက်) ကိန်းနှစ်လုံး (နောက်ကိန်းနှင့် ရှေ့ကိန်း)၏ အချိုးသည် ကိန်းသေဖြစ်လျှင် ၎င်းကိန်းစဉ်ကို geometric progression ဟုခေါ်သည်။

This ratio is called the common ratio and is denoted by $r$.

If $u_{1}, u_{2}, u_{3}, \ldots, u_{n-1}, u_{n}$ is an G.P., then $\dfrac{u_{2}}{u_{1}}=\dfrac{u_{3}}{u_{2}}=\ldots=\dfrac{u_{n}}{u_{n-1}}=$ constant.

$\dfrac{u_{n}}{u_{n-1}}=r$ and $u_{n}=u_{n-1} \cdot r$ where $r$ is called the common ratio.

If the first term is $a$, the the sequence may be $a, ar, ar^2, ar^3, \ldots$.

Hence the $n^{\text{th}}$ term is,

$\begin{array}{|l|}\hline u_n = ar^{n-1}\\ \hline \end{array}$

Definition: Geometric Mean (G.M.)

In a finite geometric progression, the terms between the first term and the last term are called the geometric means.

အဆုံးရှိ geometric progression တစ်ခု၏ ရှေ့ဆုံးကိန်းနှင့် နောက်ဆုံးကိန်းကြားရှိ ကိန်းများအားလုံးကို geonetric means (G.M) ဟုခေါ်သည်။

If $u_{1}, u_{2}, u_{3}, \ldots u_{n-1}, u_{n}$ is an G.P., then

$u_{2}, u_{3}, \ldots, u_{n-1}$ are called geometric means.

The geometric mean between two numbers $x$ and $y$ is given by

$\begin{array}{|l|}\hline \text{G.M } = \sqrt{xy}\\ \hline \end{array}$

Exercises

1. If $x, y, z$ are in G.P., prove that $\log x, \log y, \log z$ are in A.P.





$x, y, z$ ane in G.P. $\begin{aligned} &\\ \therefore \dfrac{y}{x}&=\dfrac{z}{y} \\\\ \log \left(\dfrac{y}{x}\right)&=\log \left(\dfrac{z}{y}\right) \\\\ \log y-\log x&=\log z-\log y\\\\ \end{aligned}$ $\therefore \quad \log x, \log y, \log z$ are in A.P.



2. If the $n^{\text {th }}$ term of the G.P. $5,10,20, \ldots$ is equal to the $n^{\text {th }}$ term of the G.P. $1280,640,320, \ldots$, find the value of $n$.





$\begin{aligned} 5,10,20, \ldots \text { is a G.P. }& \\\\ a &=5 \\\\ r &=\dfrac{10}{5}=2 . \\\\ u_{n} &=a r^{n-1} \\\\ &=5(2)^{n-1} \\\\ 1280,640,320, \ldots \text { is a G.P. } \\\\ a &=1280 \\\\ r &=\dfrac{640}{1280}=\dfrac{1}{2}\\\\ u_{n} &=a r^{n-1} \\\\ &=1280\left(\dfrac{1}{2}\right)^{n-1} \\\\ &=1280(2)^{1-n}\\\\ \text{By the problem,}&\\\\ 5(2)^{n-1} &=1280(2)^{1-n} \\\\ \therefore \dfrac{2^{n-1}}{2^{1-n}} &=256 \\\\ 2^{2 n-2} &=2^{8} \\\\ 2 n-2 &=8 \\\\ 2 n &=10 \\\\ n &=5 \end{aligned}$



3. Find three numbers in G.P. whose sum is 19 and whose product is $216 .$





Let the three numbers in GP be $a$, ar, $a r^{2}\\\\$. By the problem, $\begin{aligned} &\\ a+a r+a r^{2}&=19 \\\\ a\left(1+r+r^{2}\right)&=19\dots(1)\\\\ a\cdot ar\cdot a r^{2} &=216 \\\\ a^{3} r^{3} &=6^{3} \\\\ (a r)^{3} &=6^{3}\\\\ ar &=6\\\\ a&=\dfrac{6}{r}\dots(2)\\\\ \end{aligned}$ Substituting $a=\dfrac{6}{r}$ in equation (1), $\begin{aligned} &\\ \dfrac{6}{r}\left(1+r+r^{2}\right)&=19\\\\ 6+6 r+6 r^{2}&=19 r\\\\ 6 r^{2}-13 r+6&=0\\\\ (3 r-2)(2 r-3)&=0\\\\ r=\dfrac{2}{3} \text{ or } r=\dfrac{3}{2}\\\\ \text{When } r=\dfrac{3}{2},&\\\\ a=\dfrac{6}{3 / 2}&=4\\\\ \therefore\ 1^{\text{st}} \text { number } &= 4\\\\ 2^{\text{nd}} \text { number } &= 6\\\\ 3^{\text{rd}} \text { number } &= 9\\\\ \text{When } r=\dfrac{2}{3},&\\\\ a=\dfrac{6}{2 / 3}&=9\\\\ \therefore\ 1^{\text{st}} \text { number } &= 9\\\\ 2^{\text{nd}} \text { number } &= 6\\\\ 3^{\text{rd}} \text { number } &= 4\\\\ \end{aligned}$



4. If in a GP the $(p+q)^{\text {th }}$ term is $a$ and the $(p-q)^{\text {th }}$ term is $b$, prove that the $p^{\text {th }}$ term is $\sqrt{a b}$.





Let the first term and the common ratio of given G.P. be $A$ and $R$ respectively. By the problem, $\begin{aligned} &\\ u_{p+q}&=a \\\\ A R^{p+q-1}&=a\ldots(1) \\\\ u_{p-q}&=b\\\\ A R^{p-q-1} &=b \ldots(2)\\\\ \text { Miltiply equation (1) } & \text { by equation (2), }\\\\ A R^{p+q-1} \cdot A R^{p-q-1} &=a b \\\\ A^{2} R^{2 p-2} &=a b \\\\ \left(A R^{p-1}\right)^{2} &=a b \\\\ A R^{p-1} &=\sqrt{a b} \\\\ \therefore\ u_p &=\sqrt{a b} \end{aligned}$



5. The third term of a G.P. is $6 \dfrac{1}{4}$ and the $7^{\text {th }}$ term is the reciprocal of the third. Which term of this GP is unity?





$\begin{aligned} \text { In a G.P., }& \\\\ u_{3}&=6 \dfrac{1}{4} \\\\ a r^{2}&=\dfrac{25}{4}-0 \\\\ u_{7}&=\dfrac{4}{25} \\\\ a r^{6}&=\dfrac{4}{25}-2\\\\ \therefore \dfrac{a r^{6}}{a r^{2}} &=\dfrac{\frac{4}{25}}{\frac{25}{4}} \\\\ r^{4} &=\dfrac{16}{625} \\\\ r &=\dfrac{2}{5} \\\\ \therefore\ a\left(\dfrac{2}{5}\right)^{2} &=\dfrac{25}{4} \\\\ \dfrac{4 a}{25} &=\dfrac{25}{4}\\\\ \therefore\ a &=\dfrac{625}{16} \\\\ \therefore\ a r^{4} &=\dfrac{625}{16} \times \dfrac{16}{625} \\\\ \therefore\ u_{5} &=1 \end{aligned}$



6. If $\dfrac{1}{x+y}, \dfrac{1}{2 y}, \dfrac{1}{y+z}$ are the three consecutive terms of an A.P., prove that $x, y, z$ are the three consecutive terms of a G.P.





$\dfrac{1}{x+y},\ \dfrac{1}{2 y},\ \dfrac{1}{y+z}$ are in A.P. $\begin{aligned} &\\ \dfrac{1}{2 y}-\dfrac{1}{x+y}&=\dfrac{1}{y+z}-\dfrac{1}{2 y} \\\\ \dfrac{x+y-2 y}{2 y(x+y)}&=\dfrac{2 y-y-z}{2 y(y+z)} \\\\ \dfrac{x-y}{2 y(x+y)}&=\dfrac{y-z}{2 y(y+z)} \\\\ \dfrac{x-y}{x+y}&=\dfrac{y-z}{y+z} \\\\ (x-y)(y+z)&=(x+y)(y-z) \\\\ x y-y^{2}+x z-y z&=x y+y^{2}-x z-y z \\\\ x z-y z+x z+y z&=2 y^{2} \\\\ 2 x z&=2 y^{2} \\\\ y^{2}&=x z\\\\ \therefore\ \dfrac{y}{x}&=\dfrac{z}{y}\\\\ \end{aligned}$ Hence, $x,\ y,\ z$ are in G.P.



7. If $x$ is the A.M. between $a$ and $b$ and $y$ is the A.M. between $b$ and $c$ while $a, b, c$ are in G.P., prove that $\dfrac{a}{x}+\dfrac{c}{y}=2$.





A.M between $a$ and $b=x\\\\$ $\dfrac{a+b}{2}=x\\\\$ A.M between $b$ and $c=y\\\\$ $\dfrac{b+c}{2}=y\\\\$ $a, b, c$ are in a G.P. $\\ $ $\dfrac{b}{a}=\dfrac{c}{b} \Rightarrow b^{2}=a c$ $\begin{aligned} &\\ & \dfrac{a}{x}+\dfrac{c}{y} \\\\ =& \dfrac{a}{\dfrac{a+b}{2}}+\dfrac{c}{\dfrac{b+c}{2}} \\\\ =& \dfrac{2 a}{a+b}+\dfrac{2 c}{b+c} \\\\ =& \dfrac{2(a b+a c+a c+b c)}{a b+a c+b^{2}+b c} \\\\ =& \dfrac{2\left(a b+2 b^{2}+b c\right)}{a b+2 b^{2}+b c}\left(\because b^{2}=a c\right) \\\\ =& 2 \end{aligned}$



8. If the arithmetic mean and geometric mean between the two numbers are $30$ and $18$ respectively, find the two numbers





Let the two numbers be $a$ and $b\\\\ $. A.M between $a$ and $b=30\\\\ $ $\begin{aligned} \dfrac{a+b}{2} &=30 \\\\ a+b &=60 \\\\ b &=60-a\\\\ \end{aligned}$ G.M. between $a$ and $b=18\\\\ $ $\begin{aligned} \sqrt{a b}&=18\\\\ a b &=324 \\\\ a(60-a) &=324 \\\\ 60 a-a^{2} &=324 \\\\ a^{2}-60 a+32 y &=0 \\\\ (a-6)(a-54) &=0 \\\\ a=6 \text { or } a &=54\\\\ \end{aligned}$ When $a=6, b=60-6=54\\\\ $. When $a=54, b=60-54=6\\\\ $. $\therefore$ The two numbers are $6$ and $54$.



9. Let $a, b, c$ are three numbers between $2$ and $18$ , such that their sum is $25$. If $2, a, b$ are in A.P. and $b, c, 18$ are in G.P., then find $c$.





$2, a, b$ are in A.P. $\begin{aligned} &\\ \therefore a &=\dfrac{2+b}{2} \\\\ &=1+\dfrac{b}{2}\\\\ \end{aligned}$ $b, c, 18$ are in G.P. $\begin{aligned} &\\ \dfrac{c}{b} &=\dfrac{18}{c} \\\\ \therefore b &=\dfrac{c^{2}}{18} \\\\ \therefore a &=1+\dfrac{c^{2}}{36} \cdot \\\\ a+b+c &=25 \text { (given) }\\\\ 1+\dfrac{c^{2}}{36}+\dfrac{c^{2}}{18}+c &=24 \\\\ \dfrac{3 c^{2}}{36}+c &=24 \\\\ \dfrac{c^{2}}{12}+c &=24 \\\\ c^{2}+12 c-288 &=0 \\\\ (c+24)(c-12) &=0 \\\\ c=-24\ \text { or }\ c &=12\\\\ \end{aligned}$ Since $2 < c < 18, c=-24$ is impossible. $\begin{aligned} &\\ &\therefore\ c=12 \end{aligned}$



10. An A.P. has first term $a$ and common difference $d, d \neq 0$. If the $3^{\text {rd }}, 4^{\text {th }}$ and $7^{\text {th }}$ terms of this A.P. are the first three terms of a G.P., show that $a=-\dfrac{3}{2} d$. Hence show that the $4^{\text {th }}$ term of the G.P. is the $16^{\text {th }}$ term of the A.P.





In an A.P., $\begin{aligned} &\\ a&=\text { first term } \\\\ d&=\text { common difference } \\\\ u_{3}&=a+2 d \\\\ u_{4}&=a+3 d \\\\ u_{7}&=a+6 d \\\\ \end{aligned}$ By the problem, $a+2 d, a+3 d, a+6 d$ are in a G.P. $\begin{aligned} &\\ \therefore \dfrac{a+3 d}{a+2 d} &=\dfrac{a+6 d}{a+3 d} \\\\ \therefore a^{2}+6 a d+9 d^{2} &=a^{2}+8 a d+12 d^{2} \\\\ \therefore \quad-2 a d &=3 d^{2} \\\\ a &=-\dfrac{3}{2} d\\\\ \end{aligned}$ Let the common ratio of G.P. be $r$. $\begin{aligned} &\\ \text { then, } r &=\dfrac{a+3 d}{a+2 d} \\\\ &=\dfrac{-\dfrac{3}{2} d+3 d}{-\dfrac{3}{2} d+2 d} \\\\ &=3 \\\\ \therefore\ 4^{\text {th }} \text { term of G.P.} &=3(a+6 d) \\\\ &=3\left(-\dfrac{3}{2} d+6 d\right) \\\\ &=\dfrac{27}{2} d\\\\ 16^{\text {th }} \text { term of A.P. } &=a+15 d \\\\ &=-\dfrac{3}{2} d+15 d \\\\ &=\dfrac{27}{2} d\\\\ \end{aligned}$ Hence, proved.



11. In a set of four numbers, the first three are in G.P. and the last three are in A.P. with a common difference 6 . If the first number is the same as the $4^{\text {th }}$, find the four numbers.





Let $A=\{w, x, y, z\}\\\\$ By the problem, $w, x, y$ are in GP. $\\ \dfrac{x}{w}=\dfrac{y}{x}\\\\$ $x, y, z$ are in A.P. Let the common difference be $d.\\\\$ $\begin{aligned} \therefore\ x &=x, y=x+6, z=x+12 \\\\ w &=z(\text { given }) \\\\ \therefore\ w &=x+12 \\\\ \dfrac{x}{x+12} &=\dfrac{x+6}{x} \\\\ x^{2} &=x^{2}+18 x+72 \\\\ x &=-4 \\\\ \therefore\ y &=2, z=w=8 \end{aligned}$



12. If the A.M. between two positive numbers $a$ and $b$, where $a>b$, is twice the G.M. between them, prove that $a: b=\left(2+\sqrt{3}\right): \left(2-\sqrt{3}\right)$





A.M. between $a$ and $b=\dfrac{a+b}{2}\\\\ $ G.M. between $a$ and $b =\sqrt{a b} \\\\ $ $\begin{aligned} \text{By the problem} &\\\\ \text{A.M.} &= 2\cdot\text {G.M.}\\\\ \dfrac{a+b}{2}&=2 \sqrt{a b} \\\\ a+b&=4 \sqrt{a b} \\\\ (a+b)^{2}&=16 a b \dots(1)\\\\ a^{2}+2 a b+b^{2}&=16 a b \\\\ a^{2}-14 a b+b^{2}&=0\\\\ a^{2}-2 a b+b^{2}&=12 a b \\\\ (a-b)^{2}&=12 a b\ldots(2) \\\\ \text{By } (1)\div (2), &\\\\ \dfrac{(a+b)^{2}}{(a-b)^{2}}&=\dfrac{16 a b}{12 a b} \\\\ \left(\dfrac{a+b}{a-b}\right)^{2}&=\dfrac{4}{3}\\\\ \dfrac{a+b}{a-b}&=\dfrac{2}{\sqrt{3}} \\\\ 2 a-2 b&=\sqrt{3} a+\sqrt{3} b \\\\ 2 a-\sqrt{3} a&=2 b+\sqrt{3} b \\\\ a(2-\sqrt{3})&=b(2+\sqrt{3}) \\\\ \therefore \dfrac{a}{b}&=\dfrac{2+\sqrt{3}}{2-\sqrt{3}} \end{aligned}$



13. If $A_{1}, A_{2}$ are the two A.M.s between two numbers $a$ and $b$ and $G_{1}, G_{2}$ are two G.M.s between the same two numbers, then prove that $\dfrac{A_{1}+A_{2}}{G_{1} G_{2}}=\dfrac{a+b}{a b}$.





$A_{1}, A_{2}$ are the two A.M.s between two numbers $a$ and $b\\\\ $ $\therefore a, A_{1}, A_{2}, b$ is an A.P. $\begin{aligned} &\\ \therefore \quad A_{1}-a &=b_{-} A_{2} \\\\ A_{1}+A_{2} &=a+b \\\\ \end{aligned}$ $G_{1}, G_{2}$ are the two G.M.s between $a$ and $b.\\\\ $ $a, G_{1}, G_{2}, b$ is a G.P. $\begin{aligned} \dfrac{G_{1}}{a} &=\dfrac{b}{G_{2}} \\\\ G_{1} G_{2} &=a b . \\\\ \therefore \dfrac{A_{1}+A_{2}}{G_{1} G_{2}} &=\dfrac{a+b}{a b} \end{aligned}$



14. If $A$ and $G$ be A.M and G.M respectively between two positive numbers. Prove that the numbers are $A \pm \sqrt{A^{2}-G^{2}}$





Let the two positive numbers be $a$ and $b$ where $a>b\\\\ $ $\begin{aligned} \therefore\ \dfrac{a+b}{2} &=A \\\\ a+b &=2 A \\\\ \sqrt{a b} &=G \\\\ a b &=G^{2} \\\\ \left(a+b^{2}\right) &=4 A^{2} \\\\ a^{2}+2 a b+b^{2} &=4 A^{2}\\\\ a^{2}+b^{2} &=4 A^{2}-2 a b \\\\ a^{2}+b^{2}-2 a b &=4 A^{2}-4 a b \\\\ (a-b)^{2} &=4 A^{2}-4 G^{2} \\\\ a-b &=2 \sqrt{A^{2}-G^{2}} \\\\ \text{By } (1)+(2), & \\\\ 2 a &=2 A+\sqrt{A^{2}-G^{2}} \\\\ \therefore\ a &=A+\sqrt{A^{2}-G^{2}} \\\\ \text{By } (1)-(2),\\\\ 2b &=2 A-2 \sqrt{A^{2}-G^{2}} \\\\ \therefore\ b &=A-\sqrt{A^{2}-G^{2}} \end{aligned}$



15. If the ratio of A.M. and G.M. between two positive numbers is $m: n$, then prove that the numbers are in the ratio $\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$.





Let the two numbers be $a$ and $b\\\\ $. A.M. between $a$ and $b=\dfrac{a+b}{2}\\\\ $ G.M. between $a$ and $b=\sqrt{a b}\\\\ $ By the problem, $\begin{aligned} &\\ \dfrac{\text{A.M}}{\text{G.M}}&=\dfrac{m}{n} \\\\ \dfrac{a+b}{2 \sqrt{a b}}&=\dfrac{m}{n}\\\\ \text{Let } a+b=k m, 2 \sqrt{a b}&=k n\\\\ \therefore\ (a+b)^{2} &=k^{2} m^{2} \\\\ 4 a b &=k^{2} n^{2} \\\\ a^{2}+2 a b+b^{2} &=k^{2} m^{2} \\\\ a^{2}+2 a b+b^{2}-4 a b &= k^{2} n^{2}-4 a b \\\\ a^{2}-2 a b+b^{2} &=k^{2} m^{2}-k^{2} n^{2} \\\\ (a-b)^{2} &=k^{2}\left(n^{2}-n^{2}\right) \\\\ a-b &=k \sqrt{m^{2}-n^{2}}\\\\ \dfrac{a+b}{a-b} &=\dfrac{k m}{k \sqrt{m^{2}-n^{2}}} \\\\ &=\dfrac{m}{\sqrt{m^{2}-n^{2}}}\\\\ \end{aligned}$ By Componendo and Dividendo, $\begin{aligned} &\\ \dfrac{(a+b)+(a-b)}{(a+b)-(a-b)} &=\dfrac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}} \\\\ \dfrac{2 a}{2 b} &=\dfrac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}} \\\\ \dfrac{a}{b} &=\dfrac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}} \end{aligned}$



16. Given that the system of linear equations
    $\begin{aligned} x+y+z &=a+4 \\\\ 2 x-y+2 z &=2 a+2 \\\\ 3 x+2 y-3 z &=1-2 a\\\\ \end{aligned}$
    where $x, y, z$ in this order are in G.P. and $a$ is a positive real number. Find the value of $a$.





$\begin{aligned} x+y+z=a+4 \ldots(1)&\\\\ 2 x-y+2 z=2 a+2 \ldots(2)&\\\\ 3 x+2 y-3 z=1-2 a\ldots(3)& \\\\ \text{By } (1)+(2),& \\\\ 3 x+3 z =3 a+6 &\\\\ \therefore\ x+z =a+2& \\\\ \text { Substituting } x+z=a+2 &\text { in (1) } \\\\ a+2+y&=a+4 \\\\ \therefore\ y&=2\\\\ (2) \times 2 \Rightarrow\ 4 x-2 y+4 z&=4 a+4 \\\\ (3) \times 1 \Rightarrow\ 3 x+2 y-3 z&=1-2 a \\\\ \text { By addition, } 7 x+z&=2 a+5 \\\\ \text { By }(5)-(4), 6 x&=a+3 \\\\ \text { Substituting } x&=\dfrac{a+3}{6} \\\\ \dfrac{a+3}{6}+z&=a+2 \\\\ z=\dfrac{6 a+12-a-3}{6} &\\\\ z=\dfrac{5 a+9}{6}\hspace{2cm}&\\\\ \end{aligned}$ Since $x, y, z$ are in G.P., $\begin{aligned} &\\ \dfrac{y}{x}&=\dfrac{z}{y} \\\\ x z&=y^{2} \\\\ \dfrac{a+3}{6} \times \dfrac{5 a+9}{6}&=2^{2} \\\\ 5 a^{2}+2 y a+27&=144 \\\\ 5 a^{2}+2 y a-117&=0 \\\\ (5 a+3 q)(a-3)&=0\\\\ \therefore\ a &=-\dfrac{39}{5} \text { (or) } \\\\ a &=3\\\\ \end{aligned}$ Since $a>0, a=-\dfrac{39}{5}$ is rejected. $\begin{aligned} &\\ &\therefore\ a=3 \end{aligned}$



17. If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a GP be $a, b, c(a, b, c>0)$, then prove that $(q-r) \log a+(r-p) \log b+(p-q) \log c=0$.





Let the first term be $A$ and the cominon ratio be $R$ of the given G.P. $\begin{aligned} &\\ \therefore u_p &=a \\\\ A R^{\ p-1} &=a \\\\ R^{\ p-1} &=\dfrac{a}{A} \\\\ \log \left(R^{\ p-1}\right) &=\log \left(\dfrac{a}{A}\right) \\\\ (p-1) \log R &=\log a-\log A\\\\ \end{aligned}$ $p-1=\dfrac{\log a-\log A}{\log R} \\\\ $ Similarly, $\begin{aligned} &\\ q-1 &=\dfrac{\log b-\log A}{\log R} \\\\ r-1 &=\dfrac{\log c-\log A}{\log R}\\\\ \end{aligned}$ $\begin{aligned} &(1)-(2) \Rightarrow p-q=\dfrac{\log a-\log b}{\log R} \\\\ &(2)-(3) \Rightarrow g-r=\dfrac{\log b-\log c}{\log R} \\\\ &(3)-(1) \Rightarrow r-p=\dfrac{\log c-\log a}{\log R}\\\\ \end{aligned}$ $\begin{aligned} \therefore\ & (p-q) \log c =\left(\dfrac{\log a-\log b}{\log R}\right) \log c \\\\ & (q-r) \log a =\left(\dfrac{\log b-\log c}{\log R}\right) \log a \\\\ & (r-p) \log b =\left(\dfrac{\log c-\log a}{\log R}\right) \log b \\\\ \end{aligned}$ $\begin{aligned} &\text{By } (1)+(2)+(3), \\\\ \therefore\ &(p-q) \log c+(q-r) \log a+(r-p) \log b \\\\ &=\left(\dfrac{\log a-\log b}{\log R}\right) \log c+\left(\dfrac{\log b-\log c}{\log R}\right) \log a+\left(\dfrac{\log c-\log a}{\log R}\right) \log b \\\\ &=\dfrac{\log a \cdot \log c-\log b \cdot \log c+\log a \cdot \log b-\log a \cdot \log c+\log b \cdot \log c-\log a \log b}{\log R} \\\\ &=0 \end{aligned}$