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.

Solution $x, y, z$ 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 $n$.

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



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

SOLUTION 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 $\sqrt{a b}$.



Solution
Let the first term and the common ratio of given G.P.
be $A$ and $R$ respectively.


By the problem,


$\begin{aligned}
&\\
u_{p+q}&=a \\\\
A R^{p+q-1}&=a\ldots(1) \\\\
u_{p-q}&=b\\\\
A R^{p-q-1} &=b \ldots(2)\\\\
\text { Miltiply equation (1) } & \text { by equation (2), }\\\\
A R^{p+q-1} \cdot A R^{p-q-1} &=a b \\\\
A^{2} R^{2 p-2} &=a b \\\\
\left(A R^{p-1}\right)^{2} &=a b \\\\
A R^{p-1} &=\sqrt{a b} \\\\
\therefore\ u_p &=\sqrt{a b}
\end{aligned}$