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, $\begin{aligned} &\\ q-1 &=\dfrac{\log b-\log A}{\log R} \\\\ r-1 &=\dfrac{\log c-\log A}{\log R}\\\\ \end{aligned}$ $\begin{aligned} &(1)-(2) \Rightarrow p-q=\dfrac{\log a-\log b}{\log R} \\\\ &(2)-(3) \Rightarrow g-r=\dfrac{\log b-\log c}{\log R} \\\\ &(3)-(1) \Rightarrow r-p=\dfrac{\log c-\log a}{\log R}\\\\ \end{aligned}$ $\begin{aligned} \therefore\ & (p-q) \log c =\left(\dfrac{\log a-\log b}{\log R}\right) \log c \\\\ & (q-r) \log a =\left(\dfrac{\log b-\log c}{\log R}\right) \log a \\\\ & (r-p) \log b =\left(\dfrac{\log c-\log a}{\log R}\right) \log b \\\\ \end{aligned}$ $\begin{aligned} &\text{By } (1)+(2)+(3), \\\\ \therefore\ &(p-q) \log c+(q-r) \log a+(r-p) \log b \\\\ &=\left(\dfrac{\log a-\log b}{\log R}\right) \log c+\left(\dfrac{\log b-\log c}{\log R}\right) \log a+\left(\dfrac{\log c-\log a}{\log R}\right) \log b \\\\ &=\dfrac{\log a \cdot \log c-\log b \cdot \log c+\log a \cdot \log b-\log a \cdot \log c+\log b \cdot \log c-\log a \log b}{\log R} \\\\ &=0 \end{aligned}$

Geometric Progression : Problems and Solutions -Part (1)

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 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. Find
   (a) the common ratio,
   (b) the $10^{\text {th }}$ term and
   (c) the $n^{\text {th }}$ term of the following G.P.
   (i) $4,2,1, \dfrac{1}{2}, \ldots$
   (ii) $-2,4,-8,16, \ldots$
   (iii) $5,20,80,320, \ldots$
   (iv) $\dfrac{1}{3},-\dfrac{1}{9}, \dfrac{1}{27},-\dfrac{1}{81}, \ldots$
   (v) $\dfrac{8}{9}, \dfrac{4}{3}, 2,3, \ldots$
   (vi) $x^{5}, x^{4} y, x^{3} y^{2}, x^{2} y^{3}, \ldots$





=======Sol 1=======



2. If $3, x, y, z, w$ and 3072 are consecutive terms of a G.P., find the value of $x$, $y, z$ and $w$.





$3, x, y, z, w, 3072$ is a G.P. $\begin{aligned} & \\ \therefore\ &a=3 \text{ and }\\\\ &u_{6}=3072\\\\ &a r^{5}=3072\\\\ &3 r^{5}=3072\\\\ &r^{5}=1024\\\\ & r=(4)^{5} \\\\ & r= 4\\\\ &\therefore\ x=u_{2}=a r=-12 \\\\ & y=u_{3}=a r^{2}=48 \\\\ & z=u_{4}=a r^{3}=-192 \\\\ & w=u_{5}=a r^{4}=768\\\\ \end{aligned}$



3. The second term of a G.P. is 64 and the fifth term is 27 . Find the first 6 terms of the G.P.





In a G.P., $u_{2}=64$ $a r=64 \ldots(1)$ $u_{5}=27$ $a r^{4}=27 \ldots(2)$ $(2) \div(1)$ $\dfrac{a r^{4}}{a r}=\dfrac{27}{64}$ $r^{3}=\left(\dfrac{3}{4}\right)^{3}$ $r=\dfrac{3}{4}$ $\therefore\ a\left(\dfrac{3}{4}\right)=64$ $\therefore\ a=\dfrac{256}{3}$ $\begin{aligned} \therefore\ u_{1} &=a=\dfrac{256}{3} \\\\ u_{2} &=a r=64 \\\\ u_{3} &=a r^{2}=48 \\\\ u_{4} &=a r^{3}=36 \\\\ u_{5} &=a r^{4}=27 \\\\ u_{6} &=a r^{5}=\dfrac{81}{4} \end{aligned}$



4. Find the $10^{\text {th }}$ term of the G.P. $a^{5}, a^{4} b, a^{3} b^{2}, a^{2} b^{3}, \ldots$. Which term of the G.P. is $\dfrac{b^{20}}{a^{15}} ?$





$a^{5}$, $a^{4} b$, $a^{3} b^{2}$, $a^{2} b^{3}$, $\ldots$ is a GP. Let the first term be $A$ and the common ratio be $r$. Then, $\begin{aligned} &A=a^{5} \\\\ &r=\dfrac{a^{4} b}{a^{5}}=\dfrac{b}{a} \end{aligned}$ Let $\dfrac{b^{20}}{a^{15}}$ be the $n^{\text {th }}$ term. $\begin{aligned} \therefore u_{n} &=\dfrac{b^{20}}{a^{15}} \\\\ A r^{n-1} &=\dfrac{b^{20}}{a^{15}} \\\\ a^{5}\left(\dfrac{b}{a}\right)^{n-1} &=\dfrac{b^{20}}{a^{15}} \\\\ \left(\dfrac{b}{a}\right)^{n-1} &=\left(\dfrac{b}{a}\right)^{20}\\\\ \therefore\ n-1 &=20 \\\\ n &=21 \\\\ \therefore\ u_{21}&=\frac{b^{20}}{a^{15}} \end{aligned}$



5. The $4^{\text {th }}$ term of the G.P is 3 and the sixth term is 147 . Find the first three terms of the two possible geometric progressions.





In a G.P., $\begin{aligned} u_{4}&=3\\\\ a r^{3}&=3 \ldots(1)\\\\ u_{6}&=147\\\\ a r^{5}&=147 \ldots(2)\\\\ (2) \div &(1)\\\\ \dfrac{a r^{5}}{a r^{3}}&=\dfrac{147}{3}\\\\ r^{2} &=49 \\\\ r &=\pm 7 \\\\ \text { When } r &=-7 \\\\ a(-7)^{3} &=3 \\\\ a &=-\dfrac{3}{343} \\\\ \therefore\ u_{1}&=a=-\dfrac{3}{343} \\\\ u_{2}&=a r=\dfrac{3}{49} \\\\ u_{3}&=a r^{2}=-\dfrac{3}{7}\\\\ \text { When } r &=7 \\\\ a(7)^{3} &=3 \\\\ a &=\dfrac{3}{343} \\\\ \therefore \quad u_{1} &=a=\dfrac{3}{343} \\\\ u_{2} &=a r=\dfrac{3}{49} \\\\ u_{2} &=a r^{2}=\dfrac{3}{7} \end{aligned}$



6. The product of the first three terms of the G.P. is 1 and the product of the third, fourth and the fifth term is $11 \dfrac{25}{64}$. Find the fifth term of the G.P.





Let the given G.P. be $a, a r, a r^{2}, a r^{3}, a r^{4}, \ldots$ By the problem, $\begin{aligned} a \cdot a r \cdot a r^{2} &=1 \\\\ (a r)^{3} &=1 \\\\ a r &=1 \\\\ a &=\dfrac{1}{r}\\\\ a r^{2} \cdot a r^{3} \cdot a r^{4} &=11 \dfrac{25}{64} \\\\ a^{3} r^{9} &=\dfrac{729}{64} \\\\ \dfrac{1}{r^{3}} \cdot r^{9} &=\dfrac{729}{64} \\\\ r^{6} &=\dfrac{729}{64} \\\\ r^{6} &=\left(\pm \dfrac{3}{2}\right)^{6}\\\\ \therefore r &=\pm \dfrac{3}{2} \\\\ a &=\dfrac{1}{r} \\\\ \therefore a &=\pm \dfrac{2}{3} \\\\ u_{5} &=a r^{4} \\\\ &=\pm \dfrac{2}{3}\left(\pm \dfrac{3}{2}\right)^{4} \\\\ &=\pm \dfrac{27}{8} \end{aligned}$



7. Find two different values of $x$, so that $-\dfrac{3}{2}, x,-\dfrac{8}{27}$ will be a G.P.





$\begin{aligned} \text { If }-\dfrac{3}{2}, x,-\dfrac{8}{27} &\text { is a G.P., then } \\\\ \dfrac{x}{\left(-\dfrac{3}{2}\right)}&=\dfrac{\left(-\dfrac{8}{27}\right)}{x} \\\\ \therefore\ x^{2}&=\dfrac{4}{9} \\ x&=\pm \dfrac{2}{3} \end{aligned}$



8. Find which term of the G.P. $\dfrac{8}{9}$, $\dfrac{4}{3} \sqrt{\dfrac{2}{3}}$, $\dfrac{4}{3}, \ldots$ is $\sqrt{6}$.





$\dfrac{8}{9}$, $\dfrac{4}{3} \sqrt{\dfrac{2}{3}}$, $\dfrac{4}{3}$, $\ldots$, $\sqrt{6}$ is a G.P. $\begin{aligned} \therefore\ a &=\dfrac{8}{9} \\\\ r &=\dfrac{\dfrac{4}{3}}{\dfrac{4}{3} \sqrt{\dfrac{2}{3}}}=\sqrt{\dfrac{3}{2}}\\\\ \text{Let } u_{n}&=6, \text{then }\\\\ \therefore\ a^{n-1}&=\sqrt{6}\\\\ \dfrac{8}{9}\left(\sqrt{\dfrac{3}{2}}\right)^{n-1} &=\sqrt{6} \\\\ \left(\sqrt{\dfrac{3}{2}}\right)^{n-1} &=\dfrac{9 \sqrt{6}}{8} \\\\ &=\sqrt{\dfrac{9 \times 9 \times 6}{8 \times 8}} \\\\ &=\sqrt{\dfrac{3^{5}}{2^{5}}} \\\\ &=\left(\sqrt{\dfrac{3}{2}}\right)^{5}\\\\ \therefore\ n-1 &=5 \\\\ n &=6 \\\\ \therefore\ u_{6} &=\sqrt{6} \end{aligned}$



9. If $a, b, c, d$ is a G.P., show that $a^{2}-b^{2}$, $b^{2}-c^{2}$, $c^{2}-d^{2}$ is also a G.P.





$a, b, c, d$ is a G.P. Let the common ratio be $r$, then $\begin{aligned} a&=a \\\\ b&=a r \\\\ c&=a r^{2} \\\\ d&=a r^{3}\\\\ \dfrac{b^{2}-c^{2}}{a^{2}-b^{2}} &=\dfrac{a^{2} r^{2}-a^{2} r^{4}}{a^{2}-a^{2} r^{2}} \\\\ &=\dfrac{r^{2}\left(a^{2}-a^{2} r^{2}\right)}{a^{2}-a^{2} r^{2}} \\\\ &=r^{2} \\\\ \dfrac{c^{2}-d^{2}}{b^{2}-c^{2}} &=\dfrac{a^{2} r^{4}-a^{2} r^{6}}{a^{2} r^{2}-a^{2} r^{4}} \\\\ &=\dfrac{r^{2}\left(a^{2} r^{2}-a^{2} r^{4}\right)}{a^{2} r^{2}-a^{2} r^{4}} \\\\ &=r^{2}\\\\ \therefore\ \dfrac{b^{2}-c^{2}}{a^{2}-b^{2}}&=\dfrac{c^{2}-d^{2}}{b^{2}-c^{2}} \end{aligned}$ $\therefore\ a^{2}-b^{2}, b^{2}-c^{2}, c^{2}-d^{2}$ is a G.P.



10. If $a, b, c, d$ is a G.P., show that
    (i) $\dfrac{b+c}{c+d}=\dfrac{a+c}{b+d}$.
    (ii) $(a+d)(b+c)-(a+c)(b+d)=(b-c)^{2}$.





$a, b, c, d$ is $a G . P .$ Let the common ratio be $r$, then $\begin{aligned} a&=a \\\\ b&=a r \\\\ c&=a r^{2} \\\\ d&=a r^{3}\\\\ \text { (i) }\quad \dfrac{b+c}{c+d} &=\dfrac{a r+a r^{2}}{a r^{2}+a r^{3}} \\\\ &=\dfrac{a r+a r^{2}}{r\left(a r+a r^{2}\right)} \\\\ &=\dfrac{1}{r} \\ \dfrac{a+c}{b+d} &=\dfrac{a+a r^{2}}{a r+a r^{3}} \\\\ &=\dfrac{a+a r^{2}}{r\left(a+a r^{2}\right)} \\\\ &=\dfrac{1}{r}\\\\ \therefore\ \dfrac{b+c}{c+d}&=\dfrac{a+c}{b+d} \\\\ \text { (ii) }\quad\quad\quad &(a+d)(b+c)-(a+c)(b+d) \\\\ =\ & a b+a c+b d+c d-a b-a d-b c-c d \\\\ =\ & a c+b d-a d-b c \\\\ =\ & a\left(a r^{2}\right)+(a r)\left(a r^{3}\right)-a\left(a r^{3}\right)-(a r)\left(a r^{2}\right) \\\\ =\ &(a r)^{2}+\left(a r^{2}\right)^{2}-2 a^{2} r^{3} \\\\ =\ &(a r)^{2}-2(a r)\left(a r^{2}\right)+\left(a r^{2}\right)^{2} \\\\ =\ & b^{2}-2 b c+c^{2} \\\\ =\ &(b-c)^{2} \end{aligned}$



11. If $a, b, c$ is an A.P. and $x, y, z$ is a G.P., show that $x^{b-c} y^{c-a} z^{a-b}=1$.





$a, b, c$ is an $A P$. Let $d$ be the common difference. $\begin{aligned} \therefore a =\ &a \\\\ b &= a+d \\\\ c &= &a+2 d \end{aligned}$ $x, y, z$ is a G.P Let $r$ be the common ratio. $\begin{aligned} \therefore x =\ &x \\\\ y =\ &x r \\\\ z =\ &x r^{2} \\\\ & x^{b-c} y^{c-a} z^{a-b} \\\\\ =\ &x^{a+d-a-2 d}(x r)^{a+2 d-a}\left(x r^{2}\right)^{a-a-d} \\\\ =\ &x^{-d} \cdot x^{2 d} \cdot r^{2 d} \cdot x^{-d} \cdot r^{-2 d} \\\\ =\ &x^{0} r^{0} \\\\ =\ &1 \end{aligned}$



12. In a G.P. the product of any three consecutive terms is $512$ . When 8 is added to the first term and $6$ to the second, then the terms form an A.P. Find the terms of the G.P.





Let $a$, ar, ar be ginen G.P., then $\begin{aligned} &a \cdot a r \cdot a r^{2}=512 \\\\ &(a r)^{3}=8^{3} \\\\ &a r=8 \\\\ &a=\dfrac{8}{r} \end{aligned}$ $\therefore$ The terms of given G.P. are $\dfrac{8}{r}, 8,8 r .$ By the problem, $\begin{aligned} \dfrac{8}{r}+8,(8+6), 8 r & \text { is an AP. } \\\\ \therefore 14-\dfrac{8}{r}-8 &=8 r-14 \\\\ 6-\dfrac{8}{r}&=8 r-14 \\\\ 6 r-8&=8 r^{2}-14 r \\\\ 8 r^{2}-20 r+8&=0 \\\\ 2 r^{2}-5 r+2&=0 \\\\ (2 r-1)(r-2)&=0\\\\ \end{aligned}$ $\therefore r=\dfrac{1}{2}$ or $r=2$. When $r=\dfrac{1}{2}$ the terms of G.P ane $16,8,4$. When $r=2$, the terms of G.P. are $4,8,16$.



13. In a G.P. the product of any three consecutive terms is $216$ . When 1 is added to the first term and $2$ to the second, then the terms form an A.P. Find the terms of the G.P.





Let $a$, ar, ar be ginen G.P., then $\begin{aligned} &a \cdot a r \cdot a r^{2}=216 \\\\ &(a r)^{3}=6^{3} \\\\ &a r=8 \\\\ &a=\dfrac{6}{r} \end{aligned}$ $\therefore$ The terms of given G.P. are $\dfrac{6}{r}, 6,6 r .$ By the problem, $\begin{aligned} \dfrac{6}{r}+1,(6+2), 6 r & \text { is an AP. } \\\\ \therefore 8-\dfrac{6}{r}-1 &=6 r-8 \\\\ 7-\dfrac{6}{r}&=6 r-8 \\\\ 7 r-6&=6 r^{2}-8 r \\\\ 6 r^{2}-15 r+6&=0 \\\\ 2 r^{2}-5 r+2&=0 \\\\ (2 r-1)(r-2)&=0\\\\ \end{aligned}$ $\therefore r=\dfrac{1}{2}$ or $r=2$. When $r=\dfrac{1}{2}$ the terme of G.P ane $12,6,3$. When $r=2$, the terms of G.P. are $3,6,12$.



14. $8, x, y$ are three consecutive terms of an A.P. while $x, y, 36$ are three consecutive terms of a G.P., find the possible values of $x$ and $y$.





$8, x, y$ is an A.P. $\begin{aligned} \therefore\ x -8 & =y-x \\\\ y &=2 x-8 \\\\ &=2(x-4)--(1) \\\\ x, y &, 36 \text { is a G.P. } \\\\ \dfrac{y}{x} &=\dfrac{36}{y} \\\\ \therefore\ y^{2} &=36 x-(2) \end{aligned}$ Substituting $y=2 x-8$ in equation (2), $\begin{aligned} 4(x-4)^{2} &=36 x \\\\ x^{2}-8 x+16 &=9 x \\\\ x^{2}-17 x+16 &=0 \\\\ \therefore\ (x-1)(x-16) &=0 \\\\ x=1 \text { or } x &=16 \end{aligned}$ When $x=1, y=2(1-4)=-6$ When $x=16, y=2(16-4)=24$



15. In a G.P., whose first term is positive, the sum of the first and the third term is $\dfrac{13}{9}$ and the product of the second and fourth term is $\dfrac{16}{81}$. Find the common ratio and the $6^{\text {th }}$ term.





Let $a$, ar, $a r^{2}$ be given GI.P., where $a>0$. By the problem, $\begin{aligned} a+a r^{2}&=\dfrac{13}{9} \ldots (1)\\\\ ar \cdot ar^{2}&=\dfrac{16}{81}\\\\ \left(a r^{2}\right)^{2}&=\dfrac{16}{81}\\\\ \therefore\ a r^{2}&=\dfrac{4}{9} \left(\because a>0 \text { and } r^{2}>0\right)\ldots(2) \end{aligned}$ Substituting $a r^{2}=\dfrac{4}{9}$ in equation $(1)$, $\begin{aligned} a+\dfrac{4}{9} &=\dfrac{13}{9} \\\\ a &=1\\\\ \therefore 1\left(r^{2}\right) &=\dfrac{4}{9} \\\\ r &=\pm \dfrac{2}{3} \end{aligned}$ When $a=1$, and $r=-\dfrac{2}{3}$, $\begin{aligned} u_{6} &=a r^{5} \\\\ &=1\left(-\dfrac{2}{3}\right)^{5} \\\\ &=-\dfrac{32}{243} \end{aligned}$ When $a=1$, and $r=\dfrac{2}{3}$ $\begin{aligned} u_{6} &=a r^{5} \\\\ &=1\left(\dfrac{2}{3}\right)^{5} \\\\ &=\dfrac{32}{243} \end{aligned}$



16. If $\log _{3} 2$, $\log _{4} x$, $\log _{2} 81$ is a G.P. then find the possible values of $x$.





$\log _{3} 2, \log _{4} x, \log _{2} 81$ is a G.P. $\begin{aligned} \therefore\ \dfrac{\log _{4} x}{\log _{3} 2} &=\dfrac{\log _{2} x}{\log _{4} x} \\\\\ \therefore\ \left(\log _{4} x\right)^{2} &=\log _{2} 81 \times \log _{3} 2 \\\\\ &=4 \log _{2} 3 \times \dfrac{1}{\log _{2} 3}\\\\\ \left(\log _{4} x\right)^{2}&=4 \\\\\ \therefore\ \log _{4} x&=\pm 2 \\\\\ \therefore\ x &=4^{-2} \text { or } x=4^{2} \\\\\ x &=\dfrac{1}{16} \text { or } x=16 \end{aligned}$



17. If $\dfrac{1}{b+a}, \dfrac{1}{2 b}$ and $\dfrac{1}{b+c}$ are in A.P., prove that $a, b$ and $c$ are in G.P.





$\begin{aligned} \dfrac{1}{b+a}, \dfrac{1}{2 b}, \dfrac{1}{b+c}\ & \text {is an A.P } \\\\\ \therefore \dfrac{1}{2 b}-\dfrac{1}{b+a} &=\dfrac{1}{b+c}-\dfrac{1}{2 b} \\\\\ \dfrac{b+a-2 b}{2 b(b+a)} &=\dfrac{2 b-b-c}{2 b(b+c)} \\\\\ \dfrac{a-b}{a+b} &=\dfrac{b-c}{b+c}\\\\\ \text {By Componendo} &- \text {Dividendo}\\\\ \dfrac{(a-b)+(a+b)}{(a-b)-(a+b)} &=\dfrac{(b-c)+(b+c)}{(b-c)-(b+c)} \\\\\ -\dfrac{2 a}{2 b} &=-\dfrac{2 b}{2 c} \\\\\ \dfrac{a}{b} &=\dfrac{b}{c} \\\\\ \therefore \dfrac{b}{a} &=\dfrac{c}{b}\\\\\ \therefore a, b, c & \text { are in G.P.} \end{aligned}$



18. In a G.P., the first term exceeds the third term by $72$ and the sum of the second and third term is $36$ . Find the first term.





Let $a$, $ar$, $ar^{2}$ be given G.P. By the problem, $\begin{aligned} a-a r^{2}&=72 \\\\ a\left(1-r^{2}\right)&=72\ldots(1) \\\\ a r+a r^{2}&=36 \\\\ a r(1+r)&=36 \ldots(2)\\\\ (1) \div(2) & \\\\ \dfrac{a\left(1-r^{2}\right)}{\operatorname{ar}(1+r)} &=\dfrac{72}{3 b} \\\\ \dfrac{(1-r)(1+r)}{r(1+r)} &=2 \\\\ \therefore \quad 1-r &=2 r \\\\ 3 r &=1 \\\\ r &=\dfrac{1}{3}\\\\ \text{Substituting } r=\dfrac{1}{3}& \text{ in equation} (1),\\\\ a\left(1-\dfrac{1}{9}\right)=72 \\\\ \dfrac{8 a}{9}=72 \\\\ a=81 \end{aligned}$



19. The fourth term of a G.P. is $9$ and the ninth term is $2187$. Find the first 4 terms of the G.P.





$\begin{aligned} \text{In a G.P, } & \\\\ u_{4}&=9 \\\\ a r^{3}&=9\ldots(1) \\\\ u_{9}&=2187 \\\\ a{r}^{8}&=2187\ldots(2) \\\\ (2) \div(1),& \\\\ \dfrac{a r^{8}}{a r^{3}}&=\dfrac{2187}{9}\\\\ r^{5}&=243 \\\\ r^{5}&=3^{5} \\\\ r&=3 \end{aligned}$ Substituting $r=3$ in equation (1), $\begin{aligned} a\left(3^{3}\right)&=9 \\\\ a&=\dfrac{1}{3}\\\\ \therefore\ u_{1} &=a=\dfrac{1}{3} \\\\ u_{2} &=a r=1 \\\\ u_{3} &=a r^{2}=3 \\\\ u_{4} &=a^{3}=9 \end{aligned}$



20. The fourth term of a G.P. exceeds the third by $\dfrac{3}{44}$ and the third term exceeds the second term by $\dfrac{1}{22}$. Find the first term and the sixth term of the G.P.





In a G.P, $\begin{aligned} u_{4}-u_{3}&=\dfrac{3}{44} \\\\ a r^{3}-a r^{2}&=\dfrac{3}{44} \\\\ a r^{2}(r-1)&=\dfrac{3}{44}\ldots(1)\\\\ u_{3}-u_{2} &=\dfrac{1}{22} \\\\ {ar}^{2}-a r &=\dfrac{1}{22} \\\\ {ar}(r-1) &=\dfrac{1}{22}\ldots(2) \\\\ (1) \div(2), & \\\\ {ar}^{2}(r-1) & \\\\ {ar}(r-1) &=\dfrac{1}{\dfrac{3}{22}}\\\\ r &=\dfrac{3}{44} \times 22 \\\\ &=\dfrac{3}{2}\\\\ \end{aligned}$ Substituting $r=\dfrac{3}{2}$ in equation (2), $\begin{aligned} \dfrac{3 a}{2}\left(\dfrac{3}{2}-1\right) &=\dfrac{1}{22} \\\\ \dfrac{3 a}{4} &=\dfrac{1}{22} \\\\ a &=\dfrac{2}{33} \\\\ \therefore \quad u_{6} &=a r^{5} \\\\ &=\left(\dfrac{2}{33}\right)\left(\dfrac{3}{2}\right)^{5} \\\\ &=\dfrac{81}{176} \end{aligned}$



21. Three consecutive terms of a G.P. are $3^{2 x-1}, 9^{x}$ and $243$ . Find the value of $x$. If $243$ is the fifth term of the G.P., find the seventh term.





$3^{2 x-1}, 9^{x}, 243$ are in a G.P. $\begin{aligned} \therefore\ \dfrac{9^{x}}{3^{2 x-1}} &=\dfrac{243}{9^{x}} \\\\ \dfrac{3^{2 x}}{3^{2 x-1}} &=\dfrac{3^{5}}{3^{2 x}} \\\\ 3 &=3^{5-2 x} \\\\ \therefore \quad 5-2 x &=1 \\\\ x &=2\\\\ \therefore\ u_{3} &=3^{4-1}=27 \\\\ u_{4} &=9^{2}=81 \\\\ u_{5} &=243 \\\\ \therefore\ r &=\dfrac{81}{27} \\\\ &=3 \\\\ \therefore u_{7} &=a r^{6} \\\\ &=\left(a r^{4}\right) r^{2} \\\\ &=u_{5} \times r^{2}=243 \times 9=2187 \end{aligned}$



22. Find three numbers in a G.P. such that their sum is $42$ and their product is $512$.





Let the three numbers in a G.P. be $a$, $ar$, $ar^2$. By the problem, $\begin{aligned} a+a r+a r^{2}&=42 \\\\ a\left(1+r+r^{2}\right)&=42 \ldots(1)\\\\ a\cdot ar \cdot ar^{2}&=512\\\\ (a r)^{3}&=512\\\\ a r&=8 \\\\ a&=\dfrac{8}{r} \ldots(2) \end{aligned}$ Substituting $a=\dfrac{8}{r}$ in equation (1). $\begin{aligned} \dfrac{8}{r}\left(1+r+r^{2}\right)&=42 \\\\ 4+4 r+4 r^{2}&=21 r \\\\ 4 r^{2}-17 r+4&=0 \\\\ (4 r-1)(r-4)&=0\\\\ \therefore\ r=\frac{1}{4} \text { or } r&=4 \end{aligned}$ When $r=\dfrac{1}{4}$, $a=\dfrac{8}{1/4}=32$ Then three numbers are $32,8,2$. When $r=4$, $a=\dfrac{8}{4}=2$ Then three numbers are $2,8,32$.



23. Insert two geometric means between $2$ and $128$ .





Let the required geometric means be $x$ and $y$, $\therefore 2, x, y, 128$ is a G.P. Let the common ratio be $r$. $\begin{aligned} \therefore\ 2 &=2 \\\\ x &=2 r \\\\ y &=2 r^{2} \\\\ 128 &=2 r^{2}\\\\ \therefore\ r^{3} &=64 \\\\ r &=4 \\\\ \therefore\ x &=8 \\\\ y &=32 \end{aligned}$



24. The ratios of two numbers is $9: 1$. The sum of the arithmetic mean and geometric mean between the two numbers is $96$ , find the two numbers.





Let the two numbers be $9 x$ and $x$. $\begin{aligned} \text { A.M between } 9 x \text { and } x&=\frac{9 x+x}{2}\\\\ &=5 x \\\\ \text { G.M between } 9 x \text { and } x&=\sqrt{9 x^{2}}\\\\ &=3 x \\\\ \text { A.M } + \text { G.M } &=96 \\\\ 5 x+3 x &=96 \\\\ 8 x &=96 \\\\ x &=12 \\\\ 1^{\text {st} } \text {number } &=108 \\\\ 2^{\text {n d} } \text { number } &=12 \end{aligned}$



25. If the arithmetic mean between $x$ and $y$ is $15$ and the geometric mean is $9$ , find $x$ and $y$.





$\begin{aligned} \text{A.M between } x \text{ and } y &=15\\\\ \therefore \ \dfrac{x+y}{2} &=15 \\\\ x+y &=30\ldots(1)\\\\ \text{G.M between } x \text{ and } y = 9\\\\ \therefore \ \sqrt{x y} &=9 \\\\ x y &=81 \\\\ y &=\dfrac{81}{x}\ldots(2)\\\\ \text{Substituting } y=\dfrac{81}{x} &\text{ in equation } (1),\\\\ x+\dfrac{81}{x}&=30 \\\\ x^{2}+81&=30 x \\\\ x^{2}-30 x+81&=0 \\\\ (x-3)(x-27)&=0 \\\\ x=3 \text { or } x&=27 \\\\ \text { When } x=3, y&=\dfrac{81}{3}=27 \\\\ \text { When } x=27, y&=\dfrac{81}{27}=3 \end{aligned}$



26. Show that the products of the corresponding terms of the sequences $a, a r, a r^{2}, \ldots, a r^{n-1}$ and $A, A R, A R^{2}, \ldots, A R^{n-1}$ form a G.P, and find the common ratio.





The first sequence is $a, a r, a r^{2}, \ldots, a r^{n-1}$. The second sequence is $A, A R, A R^{2}, \ldots, A R^{n-1}$. The sequence formed by the products of the corresponding terms of above sequences is $a A, a r A R, a r^{2} A R^{2}, \ldots, a r^{n-1} A R^{n-1}$. $\begin{aligned} &\frac{u_{2}}{u_{1}}=\frac{a r A R}{a A}=r R \\\\ &\frac{u_{3}}{u_{2}}=\frac{a r^{2} A R^{2}}{a r A R}=r R \end{aligned}$ $\therefore$ The ratio of any two successive terms is the same. $\therefore$ The products of the corresponding terms of the sequences $a, a r, a r^{2}, \ldots, a r^{n-1}$ and $A, A R, A R^{2}, \ldots, A R^{n-1}$ form a G.P.



27. Given that $\dfrac{a+b x}{a-b x}=\dfrac{b+c x}{b-c x}=\dfrac{c+d x}{c-d x}$ where $x \neq 0$, prove that $a, b, c, d$ are in G.P.





$\begin{aligned} \text { Let } \dfrac{a+b x}{a-b x}&=\dfrac{b+c x}{b-c x}=\dfrac{c+d x}{c-d x}=k. \\\\ \therefore\ a+b x &=k(a-b x) \\\\ a+b x &=k a-k b x \\\\ k b x+b x &=k a-a \\\\ b x(k+1) &=a(k-1) \\\\ \dfrac{b}{a} &=\dfrac{k-1}{(k+1) x} \\\\ \text { Similarly, } & \\\\ b+c x &=k(b-c x) \\\\ b+c x &=k b-k c x \\\\ k c x+c x &=k b-b \\\\ c x(k+1) &=b(k-1)\\\\ \dfrac{c}{b} & =\dfrac{k-1}{(k+1) x} \\\\ \text { Again, } & \\\\ c+d x & =k(c-d x) \\\\ c+d x & =k c-k d x \\\\ k d x+d x & =k c-c \\\\ d x(k+1) & =c(k-1) \\\\ \dfrac{d}{c} & =\dfrac{k-1}{(k+1) x} \\\\ \therefore\ \dfrac{b}{a}=\dfrac{c}{b} & =\dfrac{d}{c} \\\\ \therefore\ a, b, c, d & \text { is in a G.P.} \end{aligned}$



28. If $a, b, c, d$ are in G.P., prove that $a+b, b+c$ and $c+d$ are also in G.P.





$a, b, c, d$ are in G.P. Let the common ratio be $r$. $\begin{aligned} \therefore a &=a \\\\ b &=a r \\\\ c &=a r^{2} \\\\ d &=a r^{3} \\\\ \dfrac{b+c}{a+b} &=\dfrac{a r+a r^{2}}{a+a r} \\\\ &=\dfrac{r(a+a r)}{a+a r} \\\\ &=r\\\\ \dfrac{c+d}{b+c} &=\dfrac{a r^{2}+a r^{3}}{a r+a r^{2}} \\\\ &=\dfrac{r\left(a r+a r^{2}\right)}{a r+a r^{2}} \\\\ &=r \\\\ \therefore\ \dfrac{b+c}{a+b} &=\dfrac{c+d}{b+c}\\\\ \therefore\ a+b,\ & b+c,\ c+d \text{ are in G.P.} \end{aligned}$



29. If $a, b, c$ are in A.P., show that $2^{a x+1}$, $2^{b x+1}$, $2^{c x+1}$ are in G.P.





$a, b, c$ are in A.P. Let the common difference be $d$. $\begin{aligned} \therefore a &=a \\\\ b &=a+d \\\\ c &=a+2 d \\\\ \dfrac{2^{b x+1}}{2^{a x+1}} &=2^{(b-a) x}=2^{dx}\\\\ \dfrac{2^{c x+1}}{2^{b x+1}} &=2^{(c-b) x}=2^{dx} \\\\ \therefore\ \dfrac{2^{b x+1}}{2^{a x+1}}&=\dfrac{2^{c x+1}}{2^{b x+1}} \\\\ \therefore\ 2^{a x+1},\ & 2^{b x+1},\ 2^{c x+1} \text { are in G.P. } \end{aligned}$



30. The first three terms of an infinite geometric progression are $m-1,6, m+4$, where $m$ is an integer. Show that $m$ satisfies the equation $m^{2}+3 m-40=0$. Hence find the possible values of $m$ and the common ratio of the progression.





\begin{aligned} m-1,\ 6,\ m+4,& \cdots \text { is a GP. } \\\\ \therefore\ \dfrac{6}{m-1} &=\dfrac{m+4}{6} \\\\ m^{2}+3 m-4 &=36 \\\\ \therefore m^{2}+3 m-40 &=0 \\\\ (m+8)(m-5) &=0 \\\\ \therefore\ m=-8 \text { or } m &=5\\\\ \text { When } m &=-8, \\\\ \text { common ratio } &=\dfrac{6}{-8-1} \\\\ &=-\dfrac{2}{3} \\\\ \text { When } m &=5, \\\\ \text { common ratio } &=\dfrac{6}{5-1} \\\\ &=\dfrac{3}{2} \end{aligned}