Arithmetic Geometric Progression Part2

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1 Geometric Progression
2 u_n for Geometric Progression
- 2.1 Examples
3 Definition of GP
- 3.1 Examples
4 Geometric Mean
5 Comparison
6 S_n for Geometric Progression
- 6.1 Prove
- 6.2 Examples
7 Sum to infinity for Geometric Progression
8 Rational Numbers as Fractions
9 Inequalities
- 9.1 Examples
10 Difference Between S_n and Sum to infinity
- 10.1 Examples
11 Calculating Interests
12 Exercise 2

Geometric Progression

Examples

$48,\,$ $96,\ldots$
- $3 \xrightarrow{{\color{blue}\times 2}} 6 \xrightarrow{{\color{blue}\times 2}} 12\xrightarrow{{\color{blue}\times 2}} 24\xrightarrow{{\color{blue}\times 2}} {\color{Red}48} \xrightarrow{{\color{blue}\times 2}} {\color{Red}96}$
$81,\,$ $-243,\ldots$
- $1 \xrightarrow{{\color{blue}\times -3}} -3 \xrightarrow{{\color{blue}\times -3}} 9\xrightarrow{{\color{blue}\times -3}} -27\xrightarrow{{\color{blue}\times -3}} {\color{Red}81} \xrightarrow{{\color{blue}\times -3}} {\color{Red}-243}$
$ar^{7},\,$ $ar^{9},\ldots$
- $ar \xrightarrow{{\color{blue}\times r^{2}}} ar^{3} \xrightarrow{{\color{blue}\times r^{2}}} ar^{5}\xrightarrow{{\color{blue}\times r^{2}}} {\color{Red}ar^{7}} \xrightarrow{{\color{blue}\times ar^{2}}} {\color{Red}ar^{9}}$

u_n for Geometric Progression

Geometric progression (GP from here onwards) takes the following form

$a, ar, ar^{2},\,$ $ar^{3}, ar^{4},\ldots$
where
- $a=\,$ first term
- $r=\,$
$\underbrace{a}_{1^{st} term},\underbrace{ar}_{2^{nd} term},\underbrace{ar^{2}}_{3^{rd} term},\underbrace{ar^{3}}_{4^{th} term},\ldots$ Once again, note the numbers $\underbrace{ar^{{\color{Blue}0}}}_{{\color{Blue}1}^{st} term},\underbrace{ar^{{\color{Blue}1}}}_{{\color{Blue}2}^{nd} term},\underbrace{ar^{{\color{Blue}2}}}_{{\color{Blue}3}^{rd} term},\underbrace{ar^{{\color{Blue}3}}}_{{\color{Blue}4}^{th} term},\ldots$
$u_{10}=\,$ $ar^{9}\,$

And thus, the general formula for a geometric progression,

$ar^{{\color{Red}??}}$
- $u_{n}=ar^{n-1}\,$

Examples

$u_{n+1}=\,$ $ar^{n}\,$
$u_{n-1}=\,$ $ar^{n-2}\,$
$u_{2n}=\,$ $ar^{2n-1}\,$

Definition of GP

Sequence where
$\therefore$ $\frac{u_{n+1}}{u_{n}}=$ $\mbox{constant}\,$
Use this definition when asked to prove a sequence is an GP.

Examples

Given $S_{n}=3^{n}-1\,$ . Prove that this is a geometric series and find term and common ratio.
- Analyze :
- $u_{n}=S_{n}-S_{n-1}\,$
- $=\left(3^{n}-1\right)-\left(3^{n-1}-1\right)$
- $=3^{n-1}\left(3-1\right)$ Note that $3^{n-1}\cdot3=3^{n}\,$
- $\frac{u_{n+1}}{u_{n}}=\frac{2\left(3^{n}\right)}{2\left(3^{n-1}\right)}$
- $=3\mbox{(constant)}\,$ Just use index law $\frac{3^{a}}{3^{b}}=3^{a-b}, \therefore \frac{3^{n}}{3^{n-1}}=3^{n-\left(n-1\right)}=3$
- $\therefore \mbox{It is a geometric series}$

Given $x+1,3x,x+10\,$ forms a GP with positive terms. Find value of $x\,$ .
- Analyze :
- $\frac{x+10}{3x}=\frac{3x}{x+1}$
- $x^{2}+11x+10=9x^{2}\,$
- $8x^{2}-11x-10=0\,$
- $\left(x-2\right)\left(8x-5\right)=0\,$
- $x=2, x=-\frac{5}{8}\,$ Done?
- $x=2, x=-\frac{5}{8}{\color{Red}(rejected)}\,$
- $\therefore x=2\,$ Done?

Geometric Mean

If $a,b,c\,$ forms an GP, $b\,$ is said to be the geometric mean of $a\,$ & $c\,$

$\mbox{Since}\;a, b, c\;\mbox{forms an GP}$
$\therefore$ $\frac{c}{b}=\frac{b}{a}\;$
$\therefore$ $b^{2}=ac\,$ $\therefore b=\sqrt{ac}$

Note : Geometric mean of ANY two values, $p\,$ & $q\,$ is $\sqrt{pq}$

Comparison

Let's take a breather and compare a few different things

To prove an AP, we use $u_{n+1}-u_{n}=\,$ $\mbox{constant}\,$
To prove an GP, we use $\frac{u_{n+1}}{u_{n}}=\,$ $\mbox{constant}\,$
To find $u_{n}\,$ from $S_{n}\,$ , we use $u_{n}=S_{n}-S_{n-1}\,$

S_n for Geometric Progression

Example

$1+2+2^{2}+\ldots+2^{9}$
- First, note that the AP method doesn't work, as $1+2^{9}\neq 2+2^{8}$ Instead, we find that if we multiply the sum with another $2\,$ , something interesting happens
- $\mbox{Let }S=1+2+2^{2}+\ldots+2^{9}$
- $2S=2+2^{2}+\ldots+2^{10}$
- $\begin{array}{ccccccccccc} S & = & 1 & + & \cancel{{\color{Red}2}} & + & \cancel{{\color{Red}2^{2}}} & + & \ldots & + & \cancel{{\color{Red}2^{9}}}\;\frac{\qquad}{}(1) \\ 2S & = & \cancel{{\color{Red}2}} & + & \cancel{{\color{Red}2^{2}}} & + & \ldots & + & \cancel{{\color{Red}2^{9}}} & + & 2^{10}\;\frac{\qquad}{}(2) \\ \end{array}$
- $2S-S=2^{10}-1\,$
- $S=2^{10}-1\,$

Prove

The above is a actually a geometric series $(a=1, r=2)\,$ , thus we can use the same method to derive a general formula for $S_{n}\,$ for any geometric series.

- $S_{n}=\,$ $a+ar+ar^{2}+\ldots+$ $ar^{n-1}\,$ Note that it is NOT $ar^{n}\,$
- $\begin{array}{ccccccccccc} S_{n} & = & a & + & ar & + & ar^{2} & + & \ldots & + & ar^{n-1}\;\frac{\qquad}{}(1) \\ rS_{n} & = & ar & + & ar^{2} & + & \ldots & + & ar^{n-1} & + & ar^{n}\;\frac{\qquad}{}(2) \\ \end{array}$
- $rS_{n}-S_{n} = ar^{n}-a\,$ Factorize and rearrange
- $\left(r-1\right)S_{n}=a\left(r^{n}-1\right)$
- $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
Thus, we have

Note:

The two formulas are actually the same.
- $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{-a\left(1-r^{n}\right)}{r-1}=\frac{-a\left(1-r^{n}\right)}{-\left(1-r\right)}=\frac{a\left(1-r^{n}\right)}{1-r}$
- When we say
  - $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \mbox{ if } r>1$
  - $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r} \mbox{ if } r<1$
The value of
cannot be , that is , since denominator can't be zero.
- A value of $r=1\,$ will result in a sequence of
Don't mix up : $u_{n}=ar^{{\color{Blue}n}{\color{Red}-1}}, S_{n}=\frac{a\left(r^{{\color{Blue}n}}-1\right)}{r-1}$

Useful Formulas:

$r^{2}-1=\,$ $\left(r-1\right)\left(r+1\right)$
- $r^{4}-1=\left({\color{Blue}r^{2}}\right)^{2}-1=\left({\color{Blue}r^{2}}-1\right)\left({\color{Blue}r^{2}}+1\right)$
- $r^{6}-1=\left({\color{Blue}r^{3}}\right)^{2}-1=\left({\color{Blue}r^{3}}-1\right)\left({\color{Blue}r^{3}}+1\right)$

Examples

Find the sum of the following series

$1-2+4-\ldots$ until the $20^{th}\,$ term
- Analyze :
- $\underbrace{1-2+4-\ldots}_{\mbox{GP with}\;a=1,\; r=-2,\; n=20}$
- $S_{20}=\frac{\left[1-\left(-2\right)^{20}\right]}{1-\left(-2\right)}$
- $=-349525\,$

$2+4+8+\ldots+4096$
- $\underbrace{2+4+8+\ldots+4096}_{\mbox{GP with}\;a=2,\; r=2,\; u_{n}=4096}$
- $u_{n}=4096\,$
- $2\left(2^{n-1}\right)=4096\,$
- $n=\frac{\lg 4096}{\lg 2}=12$
- $S_{12}=\frac{2\left(2^{12}-1\right)}{2-1}$
- $=8190\,$

$1+3+3^{2}+3^{3}+\ldots+3^{n-1}$
- Analyze :
- $\underbrace{1+3+3^{2}+3^{3}+\ldots+3^{n-1}}_{\mbox{GP with}\;a=1,\; r=3,\; n \mbox{ terms}}$
- $S_{n}=\frac{1\left(3^{n}-1\right)}{3-1}=\frac{1}{2}\left(3^{n}-1\right)$

Given sum of the first $3\,$ terms in a GP is $7\,$ and sum of the first $6\,$ terms is $63\,$ . Find its common ratio and first term.
- Analyze
- $S_{3}=7 \qquad \qquad S_{6}=63$
- $\begin{align} & \frac{a\left(r^{3}-1\right)}{r-1}=7\frac{\quad}{}(1)\quad \frac{a\left(r^{6}-1\right)}{r-1}=63\frac{\quad}{}(2)\\ & \frac{(2)}{(1)}: \frac{\frac{a\left(r^{6}-1\right)}{r-1}}{\frac{a\left(r^{3}-1\right)}{r-1}}=\frac{63}{7} \\ &\frac{r^{6}-1}{r^{3}-1}=9\\ \end{align}$
  Note that $r^{6}-1\,$ can be factorized to ${\color{Blue}\left(r^{3}-1\right)}\left(r^{3}+1\right)$
- $\begin{align} & \frac{\left(r^{3}-1\right)\left(r^{3}+1\right)}{r^{3}-1}=9\\ & r^{3}+1 =9\\ & r^{3} = 8\\ & r = 2\\ & \therefore \frac{a\left(8-1\right)}{2-1}=7\\ & a=1 \\ & \therefore a=1, r=2 \\ \end{align}$
- Check $\underbrace{\overbrace{1,2,4}^{7},8,16,32}_{63}$

Given sum of first three terms in a GP is $13\,$ and its first term is $1\,$ . Find its common ratio.
- $S_{3}=13 \quad a=1$
- $\frac{r^{3}-1}{r-1}=1$
- Hmmm....
- $\begin{align} & r^{2}+r -12=0 \\ & \left(r-3\right)\left(r+4\right)=0 \\ & r=3, r=-4 \\ \end{align}$
- Check : $1,3,9 \mbox{ or } 1,-4,9\,$

Sum to infinity for Geometric Progression

if ,
- $S_{n}=\,$ $\frac{a\left(1-r^{n}\right)}{1-r}$
as $0\,$
- - $=\lim_{n\to\infty}S_{n}$
  - $=\lim_{n\to\infty}\frac{a\left(1-\cancelto{0}{r^{n}}\right)}{1-r}$
  - $=\frac{a}{1-r}$
The series is said to be converging and it converges and to the value of $S_{\infty}$

Note

It is important to note the condition where $S_{\infty}\,$ exists, which is $\left|r\right|<1$ .
Don't mix up $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}=\frac{a\left(r^{n}-1\right)}{r-1}$ but $S_{\infty}=\frac{a}{1-r}{\color{Red} \neq\frac{a}{r-1}}\,$

Examples

In a GP, its second term is $2\,$ and the sum to infinity is $-\frac{9}{2}$ . Find its common ratio
- Analyze : Since there is a sum to infinity, $\left|r\right|<1$
- $u_{2}=2\qquad S_{\infty}=-\frac{9}{2}$
- $\begin{align} & a=\frac{2}{r}\frac{\qquad}{}(1)\\ & (1)\to(2): \\ & \frac{2}{r\left(1-r\right)}=-\frac{9}{2} \end{align}$
- $\begin{align} & 4=-9r+9r^{2} \\ & 9r^{2}-9r-4=0 \\ &\left(3r-4\right)\left(3r+1\right)=0\\ & r=\frac{4}{3} \mbox{ or } r=-\frac{1}{3} \end{align}$
- $\mbox{Since }\left|r\right|<1, \therefore r=-\frac{1}{3}$
- Check :

State the set of values of $x\,$ if the series $x+2x^{2}+4x^{3}+8x^{4}+\ldots$ is converging
- Analyze
- $\underbrace{x+2x^{2}+4x^{3}+8x^{4}+\ldots}_{\mbox{GP with }a=x,\;r=2x}$
- $\mbox{Series is converging if } \left|r\right|<1$
- $\therefore \left|2x\right|<1$
- $\left|x\right|<\frac{1}{2}$ Final answer in set form
- $\therefore \left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}$

Rational Numbers as Fractions

Notation

$0. \dot 3$ or $0.\overline{3}=$ $0.333333333\ldots$
$0. \dot 1 \dot 2$ or $0.\overline{12}=$ $0.12121212\ldots$
$1.13 \dot 4 \dot 5 \dot 7$ or $1.13\overline{457}=$ $1.13457457457\ldots$

Example

Express the following as fraction in its simplest form
$0.\dot 7$
- $0.\dot 7=$ $0.777777\ldots$ Lets write it out as a series
- $=0.7+0.07+0.007+0.0007+\ldots$ What kind of series is this? GP, $a=0.7, r=\,$ $\frac{0.07}{0.7}=0.1$
- $=\underbrace{0.7+0.07+0.007+0.0007+\ldots}_{\mbox{GP } a =0.7\; r=0.1}$ it is sum to infinity
- $=\frac{0.7}{1-0.1}$
- $=\frac{0.7}{0.9}$
- $=\frac{7}{9}$
- Check : Simply use calculator to check that $\frac{7}{9}=0.77777777\ldots$

$0.5\overline{63}$
- Analyze :
- $0.5\overline{63}=$
- $=0.5 + \underbrace{0.063+0.00063+0.0000063+\ldots}_{\mbox{GP } a =0.063\; r=0.01}$
- $=0.5 + \frac{0.063}{1-0.01}$
- $=\frac{1}{2} + \frac{63}{990}$
- $=\frac{31}{55}$
- Check : $\frac{31}{55}=0.56363636\ldots$

Inequalities

Negative $r\,$

If $r<0\,$

$r^{n}>0\,$ when $n\,$ is even. Ex : $\left(-2\right)^{2}, \left(-2\right)^{4}, \left(-2\right)^{6}$
$r^{n}<0\,$ when $n\,$ is odd. Ex : $\left(-2\right)^{3}, \left(-2\right)^{5}, \left(-2\right)^{7}$

Thus, when we have a GP with $r<0\,$

$u_{n}\,$
$S_{n}\,$ increases and decrease alternately
Example : $8,-4,2,-1,\ldots\,$

Examples

In a GP, find the least value of $n\,$ such that

$u_{n}>500 \left(a=1, r=2\right)$
- Analyze :
- $\begin{align} n-1 & > \frac{\lg 500}{\lg 2}\\ n & > 9.966 \end{align}$
- $\mbox{ least value of } n \mbox{ is } 10\,$
- Check :

$u_{n}<0.001 \left(a=1, r=0.5\right)$
- $\begin{align} u_{n} & < 0.001 \\ \left(0.5\right)^{n-1} & <0.001 \\ \end{align}$ Careful!
- $\begin{align} n-1 & > \frac{\lg 0.001}{\lg 0.5}\\ n & > 10.966 \end{align}$
- $\therefore \mbox{ least value of } n \mbox{ is } 11\,$
- Check : $u_{11}=0.000977, u_{10}=0.00195\,$

$u_{n}>500 \left(a=1, r=-2\right)$
- Analyze :
- $\begin{align} u_{n} & >500 \\ \left(-2\right)^{n-1} & >500 \\ \end{align}$
- Analyze :
- $\mbox{If } n-1 \mbox{ is even }\left(n \mbox{ is odd }\right)$
- $\left(-2\right)^{n-1}>0, \left(-2\right)^{n-1}=2^{n-1}$
- $\therefore \mbox{ least value of } n \mbox{ is } 11\,$
- Check : We will see why we can't accept $n=10\,$ , $u_{11}=1024, u_{10}=-512\,$

$u_{n}>500 \left(a=-1, r=-2\right)$
- Try it on your own first! Answer $10\,$
- $\begin{align} & u_{n} >5000 \\ & -1\left(-2\right)^{n-1} >500 \\ & \left(-2\right)^{n-1} < -500 \\ & \mbox{If } n-1 \mbox{ is odd }\left(n \mbox{ is even }\right),\left(-2\right)^{n-1} <0, \left(-2\right)^{n-1}= -\left(2^{n-1}\right) \\ & \therefore -\left(2^{n-1}\right) < -500 \\ & 2^{n-1} > 500 \\ & n-1 > \frac{\lg 500}{\lg 2}\\ & n > 9.966 \\ \therefore & \mbox{ least }n \mbox{ is } 10 \\ \end{align}$
- Check : $u_{10}=512, u_{9}=-256\,$

Difference Between S_n and Sum to infinity

If $a>0\,$ and every $r>0\,$ , then every term is positive. Assuming $\left|r\right|<1$ ,

$>\,$
Difference =

Otherwise, if $\left(r<0\right)$ and/or $\left(a<0\right)$ , we are not sure which will be larger, thus we put a modulus sign, $\left|S_{\infty}-S_{n}\right|$

Interpreting Information

Difference between and is less than
- $\left|S_{\infty}-S_{n}\right|<0.01$
is more than of
- $S_{n}>\frac{99}{100}S_{\infty}$
Difference between and is less than of
- $\left|S_{\infty}-S_{n}\right|<\frac{1}{100}S_{\infty}$

Notes

$\neq$
- means $\left(-3\right)\left(-3\right)\left(-3\right)\left(-3\right)\ldots\left(-3\right)$
  - $-3^{n}\,$ means $-\left(3\times 3\times 3\times 3\times \ldots 3\right)$
  - DO NOT change one to the other (unless we have more information about whether $n\,$ is odd or even )

$\neq$

$3^{n}\,$
- $3^{n}\,$ is already always positive

$\left|-2\left(3^{n}\right)\right|=$ $2\left(3^{n}\right)\,$

$3^{n}\,$
- if is even, $3^{n}\,$
  - if $n\,$ is odd, $\left(-3\right)^{n}=$ $-3^{n}\,$
  - In both cases, $\left|\left(-3\right)^{n}\right|=$ $3^{n}\,$

Examples

Find the least value of $n\,$ such that the difference between $S_{n}\,$ and $S_{\infty}\,$ is a GP is less than $0.01\,$ if

$a=1, r=\frac{1}{2}$
- Analyze :
- $S_{\infty}-S_{n}<0.01$
- $\frac{a-a+ar^{n}}{1-r}<0.01$
- $\frac{ar^{n}}{1-r}<0.01$
- $\frac{\left(\frac{1}{2}\right)^{n}}{1-\frac{1}{2}}<0.01$
- $\left(\frac{1}{2}\right)^{n}<0.005$
- $\begin{align} & n > \frac{\lg 0.005}{\lg \frac{1}{2}} \\ & n> 7.644 \end{align}$
- $\therefore \mbox{The least value of } n \mbox{ is } 8$
- Check : $S_{\infty}-S_{8}=0.00781$

$a=2, r=-\frac{1}{3}$
- Analyze :
- $\left|S_{\infty} - S_{n}\right|<0.01$
- $\left|\frac{a}{1-r}-\frac{a\left(1-r^{n}\right)}{1-r}\right|<0.01$
- $\left|\frac{ar^{n}}{1-r}\right|<0.01$
- $\left|\frac{2\left(-\frac{1}{3}\right)^{n}}{1-\left(-\frac{1}{3}\right)}\right|<0.01$ Simplify everything except $\left|\frac{2{\color{Red}\left(-\frac{1}{3}\right)^{n}}}{1-\left(-\frac{1}{3}\right)}\right|<0.01$
  - Note
- $\left|\frac{3}{2}\left(-\frac{1}{3}\right)^{n}\right|<0.01$
- $\therefore \left(\frac{1}{3}\right)^{n} < \frac{1}{150}$
- $\begin{align} & n > \frac{\lg \frac{1}{150}}{\lg\frac{1}{3}} \\ & n > 4.561 \\ & \therefore \mbox{The least value of } n \mbox{ is } 5\\ \end{align}$
- Check : $\left|S_{\infty} - S_{5}\right|=\left|-0.00617\right|=0.00617$

Calculating Interests

A bank gives 3% interest for its saving account, calculated based on the total savings at the end of the year. If you put in savings of RM 200 in the beginning of every year, what is the total savings at the end of the 10th year? At the end of which year will the total savings be more than RM 10000 for the first time?
- Analyze :
- $\mbox{Let } T_{n} = \mbox{total savings at the end of the } n^{th} \mbox{ year}\,$ Let's start with the first year
- $=200\left(1.03\right)^{2}+200\left(1.03\right)$ Let's continue with the 3rd year
- $T_{3}=\left(T_{2}+200\right)\left(1.03\right)$
- $=\left[200\left(1.03\right)^{2}+200\left(1.03\right)+200\right]\left(1.03\right)$
- $=200\left\{\frac{1.03\left[\left(1.03\right)^{10}-1\right]}{1.03-1}\right\}$
- $\therefore \mbox{the total savings at end of 10th year is RM }2361.56$

Let's do the second part
- Analyze :
- $T_{n} = 200\underbrace{\left[\left(1.03\right)+\left(1.03\right)^{2}+\ldots+\left(1.03\right)^{n}\right]}_{\mbox{GP},\;a=1.03,\;r=1.03,\;n \mbox{ terms}}$
- $\therefore T_{n} = 200\left\{\frac{1.03\left[\left(1.03\right)^{n}-1\right]}{1.03-1}\right\}$ Let's put it into the condition
- $T_{n}>10000\,$
- $200\left\{\frac{1.03\left[\left(1.03\right)^{n}-1\right]}{1.03-1}\right\}>10000$
- $\left(1.03\right)^{n} > 2.456$
- $\begin{align} & n > \frac{\lg 2.456}{\lg 1.03} \\ & n > 30.40 \end{align}$
- $\therefore \mbox{at end of 31st year}$

A bank gives out loan with 5% interest, calculated based on the remaining loan at the beginning of the year. For a loan of RM20000 what is the monthly installment (to the nearest ringgit) if it is to be settled in 10 years?
- Analyze :
- $=20000\left(1.05\right)^{2}-\left(1.05\right)x-x$
- $T_{3}=\left[20000\left(1.05\right)^{2}-\left(1.05\right)x-x\right]\left(1.05\right)-x$
- $=20000\left(1.05\right)^{3}-\left(1.05\right)^{2}x-\left(1.05\right)x-x$
  See the pattern $\begin{align} T_{{\color{Red}1}}& =20000^{{\color{Red}1}}\left(1.05\right)-x\\ T_{{\color{Blue}2}}& =20000\left(1.05\right)^{{\color{Blue}2}}-\left(1.05\right)^{\color{Blue}1}x-x\\ T_{{\color{Purple}3}}& =20000\left(1.05\right)^{{\color{Purple}3}}-\left(1.05\right)^{{\color{Purple}2}}x-\left(1.05\right)x-x\\ \end{align}$
- $T_{10}=20000\left(1.05\right)^{10}-x\underbrace{\left[1+\left(1.05\right)+\left(1.05\right)^{2}+\ldots+\left(1.05\right)^{9}\right]}_{{\color{Red}\mbox{GP }a=1,\;r=1.05,\;n=10}}$
- $=20000\left(1.05\right)^{10} - \frac{x\left[\left(1.05\right)^{10}-1\right]}{1.05-1}$
- $T_{10}=0\,$
- $\therefore \frac{x\left[\left(1.05\right)^{10}-1\right]}{1.05-1}=20000\left(1.05\right)^{10}$
- $\therefore x = 2590.09$
- $\frac{x}{12} = 215.84$
- $\therefore \mbox{monthly payment = RM } 216$

Exercise 2

Notes

As usual, go through all the notes and examples and make sure you understand all of it BEFORE attempting this, as well as memorize all the needed formulas.
As always with questions with lots of calculation, count carefully and deliberately, but don't overwrite too many calculation steps.
Questions that can be checked directly quickly should be checked. Questions that can't, try to check solution to simultaneous equation, etc, or just go through the workings once again to spot careless mistakes.

1) Find sum of

i) $-3+6-12+\ldots$ until the $20^{th}\,$ term
$1048575\,$
- $\begin{align} & \underbrace{-3+6-12+\ldots}_{\mbox{GP with}\;a=-3,\; r=-2,\; n=20} \\ & S_{20} = \frac{-3\left[1-\left(-2\right)^{20}\right]}{1-\left(-2\right)} \\ & \qquad = 1048575 \end{align}$

ii) $50+25+\frac{25}{2}+\ldots+\frac{25}{256}$
$\frac{25575}{256}$
- $\begin{align} & \underbrace{50+25+\frac{25}{2}+\ldots+\frac{25}{256}}_{\mbox{GP with}\;a=50,\; r=2,\; u_{n}=\frac{25}{256}} \\ & u_{n}=\frac{25}{256} \\ & 50\left(\frac{1}{2}\right)^{n-1}=\frac{25}{256}\\ & \left(\frac{1}{2}\right)^{n-1}=\frac{1}{512}\\ & \left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{9}\\ & n-1=9\\ & n=10 \\ & S_{10} = \frac{50\left[1-\left(\frac{1}{2}\right)^{10}\right]}{1-\left(\frac{1}{2}\right)} \\ & \qquad = \frac{25575}{256} \end{align}$

iii) $1+5+5^{2}+5^{3}+\ldots+5^{9}$
$2441406\,$
- $\begin{align} & \underbrace{1+5+5^{2}+5^{3}+\ldots+5^{9}}_{\mbox{GP with}\;a=1,\; r=5,\; n=10} \\ & S_{10} = \frac{1\left(5^{10}-1\right)}{5-1} \\ & \qquad = 2441406 \end{align}$

iv) $7+7^{2}+7^{3}+\ldots+7^{9}$
$47079207\,$
- $\begin{align} & \underbrace{7+7^{2}+7^{3}+\ldots+7^{9}}_{\mbox{GP with}\;a=7,\; r=7,\; n=9} \\ & S_{9} = \frac{7\left(7^{9}-1\right)}{7-1} \\ & \qquad = 47079207 \end{align}$

2) Find the first term and common ratio in the following GP if

i) Its $3^{rd}\,$ term is $20\,$ and its $5^{th}\,$ term is $80\,$ . Find also the possible sums of the first $10\,$ terms.
$r=2, a=5 ; r=-2, a=5; S_{10}=5115 ; S_{10}=-1705\,$
- $\begin{array}{ll} u_{3}=20 & u_{5}=80 \\ ar^{2}=20\frac{\quad}{}(1) & ar^{4}=80\frac{\quad}{}(2) \\ \end{array}$
- $\begin{align} &\frac{(2)}{(1)}: r^{2}=4\\ & r=\pm 2 \\ & \mbox{When } r=2, a=5 \\ & \mbox{When } r=-2, a=5 \\ & \therefore a=5, r=2 ; a=5, r=-2 \\ & \mbox{When } r=2, S_{10}=\frac{5\left[\left(2\right)^{10}-1\right]}{2-1}=5115 \\ & \mbox{When } r=-2, S_{10}=\frac{5\left[1-\left(-2\right)^{10}\right]}{1-\left(-2\right)}=-1705 \\ \end{align}$

ii) The sum of the first and $2^{nd}\,$ term is $108\,$ and the sum of the $3^{rd}\,$ and $4^{th}\,$ term is $12\,$ . Find also the possible sum to infinity.
$a=81, r=\frac{1}{3} ; a=162, r=-\frac{1}{3} ; S_{\infty}=\frac{243}{2} ; S_{\infty}=\frac{243}{2}$
- $\begin{array}{ll} u_{1}+u_{2}=108 & u_{3}+u_{4}=12 \\ a+ar=108 & ar^{2}+ar^{3}=12 \\ a\left(1+r\right)=108\frac{\quad}{}(1) & ar^{2}\left(1+r\right)=12\frac{\quad}{}(2) \\ \end{array}$
- $\begin{align} &\frac{(2)}{(1)}: \frac{ar^{2}\left(1+r\right)}{a\left(1+r\right)}=\frac{12}{108}\\ & r^{2} =\frac{1}{9} \\ & r=\pm \frac{1}{3} \\ & \mbox{When } r=\frac{1}{3}, a=81 \\ & \mbox{When } r=-\frac{1}{3}, a=162 \\ & \therefore a=81, r=\frac{1}{3} ; a=162, r=-\frac{1}{3} \\ & \mbox{When } r=\frac{1}{3}, \left|r\right|<1 ,\therefore S_{\infty}=\frac{81}{1-\left(\frac{1}{3}\right)}=\frac{243}{2} \\ & \mbox{When } r=-\frac{1}{3}, \left|r\right|<1 ,\therefore S_{\infty}=\frac{162}{1-\left(-\frac{1}{3}\right)}=\frac{243}{2} \\ \end{align}$
- Check: $\overbrace{81,27}^{108},\overbrace{9,3}^{12}; \overbrace{162,-54}^{108},\overbrace{18,-6}^{12}$

iii) The sum of the first $3\,$ terms in a GP is $26\,$ and sum of the first $6\,$ terms is $728\,$ .
$a=2 , r=3\,$
- $\begin{array}{ll} S_{3}=26 & S_{6}=728 \\ \frac{a\left(r^{3}-1\right)}{r-1}=26\frac{\quad}{}(1) & \frac{a\left(r^{6}-1\right)}{r-1}=728\frac{\quad}{}(2) \\ \end{array}$
- $\begin{align} &\frac{(2)}{(1)}: \frac{\frac{a\left(r^{6}-1\right)}{r-1}}{\frac{a\left(r^{3}-1\right)}{r-1}}=\frac{728}{126}\\ &\frac{r^{6}-1}{r^{3}-1}=\frac{52}{9} \\ &\frac{\left(r^{3}-1\right)\left(r^{3}+1\right)}{r^{3}-1}=28 \\ & r^{3}+1=28 \\ & r^{3}=27 \\ & r = 3, \therefore a = 2 \\ \end{align}$
- Check: $\underbrace{\overbrace{2,6,18}^{26}, 54,162,486}_{728}$

iv) Its second term is $2.5\,$ and the sum to infinity is $10\,$ .
$a=5 , r=\frac{1}{2}\,$
- $\begin{array}{ll} u_{2}=2.5 & S_{\infty}=10, \left|r\right|<1 \\ ar=\frac{5}{2} & \frac{a}{1-r}=10\frac{\quad}{}(2) \\ a =\frac{5}{2r}\frac{\quad}{}(1) \end{array}$
- $\begin{align} &(1)\to(2):\\ & \frac{5}{2r\left(1-r\right)}=10\\ & 1 = 4r\left(1-r \right) \\ & 1 = 4r -4r^{2}\\ & 4r^{2}-4r+1=0\\ & \left(2r-1\right)^{2}=0 \\ & r =\frac{1}{2}, \therefore a= 5\\ \\ \end{align}$

v) Sum of the first $3\,$ terms is $19\,$ and the sum to infinity is $27\,$ .
$a=9 , r=\frac{2}{3}\,$
- $\begin{array}{ll} S_{3}=19 & S_{\infty}=27, \left|r\right|<1 \\ \frac{a\left(1-r^{3}\right)}{1-r}=19\frac{\quad}{}(1) & \frac{a}{1-r}=27\frac{\quad}{}(2) \\ \end{array}$
- $\begin{align} &\frac{(2)}{(1)}:1-r^{3}=\frac{19}{27}\\ & r^{3} =\frac{8}{27} \\ & r=\frac{2}{3}\\ & \therefore \frac{a}{1-\frac{2}{3}}=27 \\ & a = 9 \\ & \therefore a=9 , r=\frac{2}{3}\\ \end{align}$
- Check : $\overbrace{9, 6, 3}^{19}$

3) Given $S_{n}=\frac{2}{3}\left(4^{n}-1\right)$ , Prove that it is a geometric series and find its first term and common ratio.

$a=2, r=4\,$
- $\begin{align} u_{n}& = S_{n}-S_{-1}\\ & =\frac{2}{3}\left(4^{n}-1\right)-\frac{2}{3}\left(4^{n-1}-1\right) \\ & =\frac{2}{3}\left(4^{n}- 4^{n-1}\right)\\ & =\frac{2}{3}\left[4^{n-1}\left(4-1\right)\right]\\ & = 2\left(4^{n-1}\right)\\ \end{align}$
- $\begin{align} \frac{u_{n+1}}{u_{n}}& = \frac{2\left(4^{n}\right)}{2\left(4^{n-1}\right)} \\ & =4 \mbox{(constant)} \end{align}$
- $\therefore S_{n} \mbox{ is a geometric series}$
- $a=u_{1}=2, r=4\,$
- Check : $S_{n}=\frac{2\left(4{n}-1\right)}{4-1}=\frac{2}{3}\left(4^{n}-1\right)$

4) Express the following as fractions in the lowest terms

i) $0.\dot 5$
$\frac{5}{9}$
- $\begin{align} 0. \dot 5 & = 0.5555555\ldots \\ & = \underbrace{0.5 +0.05+0.005+\ldots...}_{\mbox{GP with }a=0.5, r=0.1}\\ & =\frac{0.5}{1-0.1}\\ & =\frac{0.5}{0.9}\\ & =\frac{5}{9}\\ \end{align}$
- Check : $\frac{5}{9}=0.555555\ldots$

ii) $0.0\overline{34}$
$\frac{17}{495}$
- $\begin{align} 0.0\overline{34} & = 0.0343434\ldots \\ & = \underbrace{0.034 +0.00034+0.0000034+\ldots...}_{\mbox{GP with }a=0.034, r=00.1}\\ & =\frac{0.034}{1-0.01}\\ & =\frac{0.034}{0.99}\\ & =\frac{34}{990}\\ & =\frac{17}{495}\\ \end{align}$
- Check : $\frac{17}{495}=0.0343434\ldots$

iii) $1.3\dot 4\dot 7$
$\frac{667}{495}$
- $\begin{align} 1.3\dot 4\dot 7 & = 1.347474747\ldots \\ & = 1.3 + \underbrace{0.047+0.00047+0.0000047\ldots...}_{\mbox{GP with }a=0.047, r=00.1}\\ & =1.3 + \frac{0.047}{1-0.01}\\ & =1.3+ \frac{0.047}{0.99}\\ & =\frac{13}{10} + \frac{47}{990}\\ & =\frac{667}{495}\\ \end{align}$
- Check : $\frac{667}{495}=1.347474747\ldots$

5) In a GP, find least $n\,$ such that

i) $u_{n}>1000 \left(a=2,r=3\right)$
$n=7\,$
- $\begin{align} u_{n} & >1000 \\ 2\left(3^{n-1}\right) & >1000 \\ 3^{n-1} & > 500 \\ n-1 & > \frac{\lg 500}{\lg 3}\\ n & > 6.657\\ \therefore & \mbox{ least }n \mbox{ is } 7 \\ \end{align}$
- Check : $u_{7}=1458;u_{6}=486\,$

ii) $u_{n}>-0.001 \left(a=-6,r=0.2\right)$
$n=7\,$
- $\begin{align} u_{n} & >-0.001 \\ -6\left(0.2\right)^{n-1} & >-0.001 \\ \left(0.2\right)^{n-1} & < \frac{1}{6000} \\ n-1 & > \frac{\lg \frac{1}{6000}}{\lg 0.2}\\ n & > 6.405\\ \therefore & \mbox{ least }n \mbox{ is } 7 \\ \end{align}$
- Check : $u_{7}=-0.000384;u_{6}=-0.00192\,$

iii) $S_{n}>5000 \left(a=2,r=3\right)$
$n=8\,$
- $\begin{align} S_{n} & >5000 \\ \frac{2\left(3^{n}-1\right)}{3-1} & >5000 \\ 3^{n}-1 & < 5000 \\ 3^{n} & < 5001 \\ n & > \frac{\lg 5001}{\lg 3}\\ n & > 7.753\\ \therefore & \mbox{ least }n \mbox{ is } 8 \\ \end{align}$
- Check : $S_{8}=6560;S_{7}=2186\,$

iv) $u_{n}>2000 \left(a=2,r=-3\right)$
$n=9\,$
- $\begin{align} & u_{n} >2000 \\ & 2\left(-3\right)^{n-1} >2000 \\ & \left(-3\right)^{n-1} > 1000 \\ & \mbox{If } n-1 \mbox{ is even }\left(n \mbox{ is odd }\right),\left(-3\right)^{n-1} > 0, \therefore \left(-3\right)^{n-1}= 3^{n-1} \\ & \therefore 3^{n-1} > 1000 \\ & n-1 > \frac{\lg 1000}{\lg 3}\\ & n > 7.288 \\ \therefore & \mbox{ least }n \mbox{ is } 9 \\ \end{align}$
- Check : $u_{9}=13122;u_{8}=-4374;u_{7}=1458\,$

v) $S_{n}\,$ is more than $99%\,$ of $S_{\infty}\left(a=4,r=\frac{1}{3}\right)$
$n=5\,$
- $\begin{align} S_{n} & > \frac{99}{1000}S_{\infty} \\ \frac{4\left[1-\left(\frac{1}{3}\right)^{n}\right]}{1-\frac{1}{3}}&>\frac{99}{100}\left(\frac{4}{1-\frac{1}{3}}\right)\\ n &> \frac{\lg\frac{1}{100}}{\lg \frac{1}{3}} \\ 1- \left(\frac{1}{3}\right)^{n} & > \frac{99}{100}\\ \left(\frac{1}{3}\right)^{n} & < \frac{1}{100}\\ n & > \frac{\lg \frac{1}{100}}{\lg \frac{1}{3}}\\ n & > 4.192 \\ \therefore & \mbox{ least }n \mbox{ is } 5 \\ \end{align}$
- Check : $S_{5}=5.975;\frac{99}{100}S_{\infty}=5.94\,S_{4}=5.925$

vi) the difference between $S_{\infty}$ and $S_{n}\,$ is less than $0.01\,$ $\left(a=3,r=-\frac{1}{2}\right)$
$n=8\,$
- $\begin{align} \left|S_{\infty}-S_{n}\right| & < 0.01 \\ \left|\frac{a}{1-r}-\frac{a\left(1-r^{n}\right)}{1-r}\right|&<0.01 \\ \left|\frac{ar^{n}}{1-r}\right|& <0.01 \\ \left|\frac{3\left(-\frac{1}{2}\right)^{n}}{1-\left(-\frac{1}{2}\right)}\right|& <0.01 \\ \left|\left(-\frac{1}{2}\right)^{n}\right|&<0.005 \\ \left(\frac{1}{2}\right)^{n}&<0.005 \\ n & >\frac{\lg 0.005}{\lg\left(\frac{1}{2}\right)} \\ n & > 7.644 \\ \therefore & \mbox{ least }n \mbox{ is } 8 \\ \end{align}$
- Check : $\left|S_{\infty} - S_{8}\right|=0.00781\,$

6) A bank gives 3% interest for its saving account, calculated based on the total savings at the end of the year. If you put in savings of RM 100 in the beginning of every month, what is the total savings at the end of the 5th year? At the end of which year will the total savings be more than RM 30000 for the first time?

$\mbox{RM } 6562.09, \mbox{ end of 19th year}\,$
- $\begin{align} \mbox{Let } T_{n} & = \mbox{total savings at the end of } n^{th} \mbox{ year}\\ \mbox{Savings } & \mbox{ per year }= 1200 \\ T_{1} & = 1200\left(1.03\right) \\ T_{2} & = \left[1200\left(1.03\right)+1200\right]\left(1.03\right) \\ & = 1200\left(1.03\right)^{2} + 1200\left(1.03\right) \\ T_{3} & = \left\{\left[1200\left(1.03\right)^{2} + 1200\left(1.03\right)\right]+120\right\}\left(1.03\right)\\ & = 1200\left(1.03\right)^{3} + 1200\left(1.03\right)^{2} + 1200\left(1.03\right) \\ \therefore T_{5} & = 1200\left(1.03\right)^{5} +\ldots+ 1200\left(1.03\right)^{2} + 1200\left(1.03\right)\\ & = 1200\underbrace{\left[\left(1.03\right)+\left(1.03\right)^{2}+\ldots+\left(1.03\right)^{5}\right]}_{\mbox{GP with}\;a=1.03,\; r=1.03,\; n=5}\\ & = \frac{1200\left(1.03\right)\left[\left(1.03\right)^{5}-1\right]}{1.03-1}\\ & = 6562.09 \\ \end{align}$
- $\therefore \mbox{Total savings = RM } 6562.09$
- $\begin{align} T_{n}& =1200\left[\left(1.03\right)+\left(1.03\right)^{2}+\ldots+\left(1.03\right)^{n}\right]\\ & = \frac{1200\left(1.03\right)\left[\left(1.03\right)^{n}-1\right]}{1.03-1}\\ \end{align}$
- $\begin{align} T_{n} & > 30000 \\ \frac{1200\left(1.03\right)\left[\left(1.03\right)^{n}-1\right]}{1.03-1} & >30000\\ \left(1.03\right)^{n} & > \frac{178}{103}\\ n & > \frac{\lg \frac{178}{103} }{\lg 1.03 } \\ n & > 18.507 \end{align}$
- $\therefore \mbox{At the end of the 19th year}\,$

7) A bank gives out loan with 4% interest, calculated based on the remaining loan at the beginning of the year. For a loan of RM200,000 what is the monthly installment (to the nearest ringgit) if it is to be settled in 30 years?

$\mbox{RM } 6964\,$
- $\begin{align} \mbox{Let } T_{n} & = \mbox{remaining loan at the end of } n^{th} \mbox{ year}\\ \mbox{Let } x & = \mbox{ yearly payment }\\ T_{1} & = 200000\left(1.04\right)-x \\ T_{2} & = \left[200000\left(1.04\right)-x\right]\left(1.04\right)-x \\ & = 200000\left(1.04\right)^{2}-\left(1.04\right)x-x \\ T_{3} &= \left[200000\left(1.04\right)^{2}-\left(1.04\right)x-x\right]\left(1.04\right)-x \\ & = 200000\left(1.04\right)^{3}-\left(1.04\right)x^{2}-\left(1.04\right)x-x \\ \therefore T_{30} &= 200000\left(1.04\right)^{30}-\left(1.04\right)x^{29}-\ldots-\left(1.04\right)x-x \\ &= 200000\left(1.04\right)^{30}-x\underbrace{\left[1+\left(1.04\right)+\left(1.04\right)^{2}+\ldots +\left(1.04\right)^{29}\right]}_{\mbox{GP with}\;a=1,\; r=1.04,\; n=30}\\ &= 200000\left(1.04\right)^{30}-\frac{x\left[\left(1.04\right)^{30}-1\right]}{1.04-1} \end{align}$
- $\begin{align} T_{30}& =0 \\ \therefore \frac{x\left[\left(1.04\right)^{30}-1\right]}{1.04-1}&=200000\left(1.04\right)^{30}\\ x & = 11566.02\\ \mbox{monthly payment }&= \frac{x}{12} \\ & = \mbox{RM }964 \end{align}$

Arithmetic Geometric Progression Part2

Contents

Geometric Progression

u_n for Geometric Progression

Examples

Definition of GP

Examples

Geometric Mean

Comparison

S_n for Geometric Progression

Prove

Examples

Sum to infinity for Geometric Progression

Rational Numbers as Fractions

Inequalities

Examples

Difference Between S_n and Sum to infinity

Examples

Calculating Interests

Exercise 2

Personal tools

Namespaces

Variants

Views

Actions

Search

Navigation

Toolbox

Arithmetic Geometric Progression Part2

Contents

Geometric Progression

un for Geometric Progression

Examples

Definition of GP

Examples

Geometric Mean

Comparison

Sn for Geometric Progression

Prove

Examples

Sum to infinity for Geometric Progression

Rational Numbers as Fractions

Inequalities

Examples

Difference Between Sn and Sum to infinity

Examples

Calculating Interests

Exercise 2

Personal tools

Namespaces

Variants

Views

Actions

Search

Navigation

Toolbox

u_n for Geometric Progression

S_n for Geometric Progression

Difference Between S_n and Sum to infinity