Arithmetic Geometric Progression Part2

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Geometric Progression

Examples

  • 3,6,12,24,\, 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}
  • 1,-3,9,-27,\, 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,ar^{3},ar^{5},\, 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}}

un 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=\, common ratio We get back the value of r\, when we take the ratio of a term to the term before it
  • \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,

  • u_{n}\,= 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 ratio between successive terms is constant.
  • \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 : We need u_{n}\, formula to prove that it is GP, but we are only given S_{n}\,, thus first we find u_{n}\, from S_{n}\,
    • u_{n}=S_{n}-S_{n-1}\,
    • =\left(3^{n}-1\right)-\left(3^{n-1}-1\right)
    • =3^{n}-3^{n-1}\, How do we simplify these two exponents? Compare with 3^{9}-3^{8}=\, 3^{8}\left(3-1\right) Where we factored the smaller index out. Which is the smaller index in this case? 3^{n-1}\,
    • =3^{n-1}\left(3-1\right) Note that 3^{n-1}\cdot3=3^{n}\,
    • =2\left(3^{n-1}\right) Ok, we now have u_{n}\,. To prove that it is GP, we need to prove \frac{u_{n+1}}{u_{n}} is constant.
    • \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 : We know they form a GP, thus, we can equate the ratios. Note that we shouldn't in this case go and try use GP formula such as ar^{2}=x+10\,
    • \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?
    • Always remember the two things to do : 1) Check that we have answered the question. 2) Check that our answers is correct
    • The condition was positive terms. Now, it doesn't mean we will directly reject x\, negative (or even accept any x\, positive), we must check that every term will be positive, and reject it if it is not.
    • x=2, x=-\frac{5}{8}{\color{Red}(rejected)}\,
    • \therefore x=2\, Done?
    • Check : 3, 6, 12\, is indeed a GP with positive terms

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}\,

Sn 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}
    • Note the equal terms between them
      • \begin{array}{ccccccccccc} S & = & 1 & + & {\color{Red}2} & + & {\color{Blue}2^{2}} & + & \ldots & + & {\color{Purple}2^{9}} \\ 2S & = & {\color{Red}2} & + & {\color{Blue}2^{2}} & + & \ldots & + & {\color{Purple}2^{9}} & + & 2^{10} \\ \end{array}
      • Since the "middle" terms are all the same, we can get rid of them by subtracting one sum from the other. In fact, only two terms will be left.
      • It will be easier to take the bottom minus the top (to avoid negative)
    • \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}
    • (2)-(1):\, The left hand side will be 2S-S\,, while the right hand side will only have 2^{10}\, and 1\, left, and it is 2^{10}{\color{Red}-}1
    • 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}\,
    • What should we multiply to get the same effect as the example above? r\,
      • Left hand side will be rS_{n}\,. Note that S_{n}\, is just a value and can be manipulated as any unknowns.
    • \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}
    • (2)-(1):\, Note that we are assuming r>1\,, to avoid negative
    • 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 S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}
    • Alternatively, if we have taken (1)-(2):\, we will get S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}

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
      • We DO NOT mean there are two different formulas for two situation, but simply two different forms so that we can choose one that is more convenient (avoid negative on denominator). Either form will still give us the same value, regardless of the value of r\,
  • The value of r\, cannot be 1\,, that is \left(r\neq1\right), since denominator can't be zero.
    • A value of r=1\, will result in a sequence of a,a,a,a,\ldots, which doesn't need to be considered as geometric progression.
  • 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(r^{2}-1\right)\left(r^{2}+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(r^{3}-1\right)\left(r^{3}+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 :
      • First identify the separate terms which might be more difficult to see here 1{\color{Blue}-2}{\color{Red}+4}-\ldots 1\xrightarrow{\times -2}{\color{Blue}-2}\xrightarrow{\times -2}{\color{Red}+4}-\ldots
      • We can now see that it is a GP. Since r\, <1\,, it is more convenient to use the second form of the formula.
    • \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\,
    • 2^{n}=4096\, If you can see that 4096=2^{12}\,, then we can write it directly. Otherwise, we will just use logarithms (just take note that n\, is a positive integer)
    • 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 : We have to be careful when determining the number of terms. Is it n-1\, terms or n\, terms?
      • Note that from 3^{{\color{Blue}0}}+3^{{\color{Blue}1}}+3^{{\color{Blue}2}}+3^{{\color{Blue}3}}+\ldots+3^{{\color{Blue}n-1}}, there are n\, terms
      • alternatively Since a=1, r=3\,, u_{n}\, =1\left(3^{n-1}\right). The last term given is 3^{n-1}\,, thus we know that there are n\, terms.
    • \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 Not too difficult but involves an algebraic trick
    • S_{3}=7 \qquad \qquad S_{6}=63
    • \frac{a\left(r^{3}-1\right)}{r-1}=7 \quad \frac{a\left(r^{6}-1\right)}{r-1}=63
      • How should we solve this simultaneous equation?
      • Note the common terms, \frac{{\color{Blue}a}\left(r^{3}-1\right)}{{\color{Red}r-1}}=7 \quad \frac{{\color{Blue}a}\left(r^{6}-1\right)}{{\color{Red}r-1}}=63
      • Thus, the best way is to divide them
    • \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.... we will end up having to solve a cubic equation.
      • But since S_{3}\, involves only the first 3 terms, it might be easier just to write out the terms. Let's try
    • \begin{align} & S_{3}=13 \qquad a=1 \\ & a + ar + ar^{2}=13 \\ & \therefore 1 + r + r^{2}=13 \\ \end{align}
      • Note : We can see that we will now only need to solve a quadratic equation. Thus, depending on situation, it is sometimes easier to use the direct sum instead of the formula.
    • \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 \left|r\right|<1 \left(-1<r<1\right),
    • S_{n}=\, \frac{a\left(1-r^{n}\right)}{1-r}
  • as n \to \infty, r^{n}\to 0\,
    • \therefore S_{\infty}
      • =\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}
    • ar=2\qquad \frac{a}{1-r}=-\frac{9}{2}\frac{\qquad}{}(2) substitute a\, into second equation (dividing will also work but might be more confusing)
    • \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}
    • Can we accept both answers? Check that it satisfies the condition. In this case, there is a hidden condition, which is \left|r\right|<1
    • \mbox{Since }\left|r\right|<1, \therefore r=-\frac{1}{3}
    • Check : a=-6, S_{\infty}=-\frac{9}{2}
      • Note that r=\frac{4}{3}, a = \frac{3}{2} will also give us the stated value for \frac{a}{1-r}\,, which is why we have to be extra careful of condition needed for sum to infinity to exist


  • State the set of values of x\, if the series x+2x^{2}+4x^{3}+8x^{4}+\ldots is converging
    • Analyze converging means limit to infinity exists
      • The series is a geometric series x \xrightarrow{\times 2x} 2x^{2} \xrightarrow{\times 2x} 4x^{3} \xrightarrow{\times 2x} 8x^{4}
      • Thus S_{\infty} exist, \therefore \left|r\right|<1
    • \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 : Note that only 63\, is repeated. Again, keep in mind that checking at the end detect any mistakes
    • 0.5\overline{63}= 0.5636363636\ldots write the ones that do not repeat separately as a term, be careful not to miss out any 0\,'s
    • =0.5 + 0.063+0.00063+0.0000063+\ldots The first term is not part of the GP, so just leave it as it is
    • =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}
    • =0.5 + \frac{0.063}{0.99} actually, with calculators nowadays that can handle fractions, not too much manual work is needed....
    • =\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}\, alternates between positive and negative values
  • 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 : Nothing new here(yet). Just be careful and remember to check answers
    • \begin{align} u_{n} & >500\\ 2^{n-1} & >500 \\ \end{align} As always, be careful when dividing values over inequality
    • \begin{align} n-1 & > \frac{\lg 500}{\lg 2}\\ n & > 9.966 \end{align}
    • \mbox{ least value of } n \mbox{ is } 10\,
    • Check : Check that u_{10}\, is larger than 500\,, and u_{9}\, isn't
      • u_{10}=512, u_{9}=256\,
  • 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 : This is more interesting. Keep in mind first, that unless we know something about n\,, we can't simplify \left(-2\right)^{n-1}. Also, when the situation is tricky like this, it pays even more to check that our final answer really does satisfy the conditions given
    • \begin{align} u_{n} & >500 \\ \left(-2\right)^{n-1} & >500 \\ \end{align}
    • Analyze : What do we know about \left(-2\right)^{n-1}?
      • It is sometimes positive, sometimes negative, depending on whether n-1\, is odd or even
      • Since we need it to be larger than 500\,, it means that we need it to be positive
      • For that to happen n-1\, must be even, meaning n\, must be odd
      • And when that happens, we can simplify \left(-2\right)^{n-1} to 2^{n-1}\, For example : \left(-2\right)^{2}=2^{2}, \left(-2\right)^{4}=2^{4}, \left(-2\right)^{8}=2^{8}
    • \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 2^{n-1} >500 We continue to solve, keeping in mind that n\, is odd
    • \begin{align} n-1 & > \frac{\lg 500}{\lg 2}\\ n & > 9.966 \end{align} The smallest integer is 10\,, but remember we need an odd number
    • \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 Sn and Sum to infinity

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

  • \therefore S_{\infty} >\,S_{n}\,
    • We are comparing adding positive numbers till a certain n\, with adding indefinitely
  • \therefore Difference = S_{\infty}-S_{n}
    • Difference between two numbers = larger number - smaller number

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

Notes

  • \left(-3\right)^{n} \neq-3^{n}\,
    • \left(-3\right)^{n} 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 )


  • -\left(-3\right)^{n} \neq3^{n}\,
    • We cannot "bring out" the negative, neither can be "bring in" another negative to cancel it


  • \left|3^{n}\right|= 3^{n}\,
    • 3^{n}\, is already always positive
  • \left|-2\left(3^{n}\right)\right|= 2\left(3^{n}\right)\,


  • \left|\left(-3\right)^{n}\right|= 3^{n}\,
    • if n\, is even, \left(-3\right)^{n}= 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 : The difference between S_{n}\, and S_{\infty}\, is NOT S_{n}-S_{\infty}. We need to take the larger minus the smaller. Since both a\, and r\, are positive, all terms are positive, thus S_{\infty} > S_{n}. Thus the difference is S_{\infty}-S_{n}. Alternatively, we can write \left|S_{n}-S_{\infty}\right|.
    • S_{\infty}-S_{n}<0.01
    • \frac{a}{1-r}-\frac{a\left(1-r^{n}\right)}{1-r}<0.01 Normally, we will substitute the values before continuing, but in this case, its actually easier to simplify first
    • \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 : Since r<0\,, we no longer know S_{\infty} > S_{n} (in fact, S_{n}\, will be alternately larger and smaller than S_{\infty}), thus we write the difference as \left|S_{\infty} - S_{n}\right|.
      • Note : we can actually always use \left|S_{\infty} - S_{n}\right|, even in the question above.
    • \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 DO NOT multiply the 2\, into the power, such as 2\left(-\frac{1}{3}\right)^{n}=\left(-\frac{2}{3}\right)^{n}, in order to do that, we need a^{{\color{Blue}n}}b^{{\color{Blue}n}}=\left(ab\right)^{n}
        • Similarly, if you have chosen to put in the values first, \left|\frac{2}{1-\left(-\frac{1}{3}\right)}-\frac{2\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1-\left(-\frac{1}{3}\right)}\right|, DO NOT change 1-\left(-\frac{1}{3}\right)^{n} to 1+\left(\frac{1}{3}\right)^{n}
    • \left|\frac{3}{2}\left(-\frac{1}{3}\right)^{n}\right|<0.01
    • \left|\left(-\frac{1}{3}\right)^{n}\right|<\frac{1}{150} As we have seen earlier \left|\left(-\frac{1}{3}\right)^{n}\right| will become \left(\frac{1}{3}\right)^{n}
    • \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 : We cannot see whether it is an AP or GP yet, so let's study the situation first
      • Interest was calculated at the end of the year, while the savings was put at the beginning
      • At the beginning of the first year, 200\, is saved. At the end of it, 3%\, interest is given
      • What is the fastest way to count a value after increasing by 3%\, ? By multiplying with 103%\,, which is 1.03\,
      • Thus, we will have 200\left(1.03\right)=206 at the end of the first year
      • At the beginning of the 2nd year, we put another 200\,, meaning, we now have 206+200=406\,, and at the end of the year, another 3%\, will be given, and this 3%\, will be calculated based on total savings, which was now at 406\,, thus 406\left(1.03\right)
      • The process is repeated every year. We can see that there is a pattern, but not just a simple AP or GP. Thus, we will try to deduce the pattern out. Note that in order to see the pattern of how the value depends on 200\, and 1.03\,, we will not calculate to actual values yet (such as 200\left(1.03\right)=206)
      • Firstly, we must find a way to "label" the values from year to year. The value that is needed is the total savings at the end of the year
    • \mbox{Let } T_{n} = \mbox{total savings at the end of the } n^{th} \mbox{ year}\, Let's start with the first year
    • T_{1}=200\left(1.03\right) At the beginning of the next year, we will have T_{1}\, plus another 200\, and then total of this two, multiply with another 1.03\,
    • T_{2}=\left(T_{1}+200\right)\left(1.03\right) Note that its not T_{1}+200\left(1.03\right). Let's substitute T_{1}\, into it
    • =\left[200\left(1.03\right)+200\right]\left(1.03\right) We simplify, keeping in mind that we need to see the 200\, and 1.03\, "un-calculated" out
    • =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(1.03\right)^{3}+200\left(1.03\right)^{2}+200\left(1.03\right)
      • Can see the pattern already? Let's look at the final form of each year
      • \begin{align} T_{1} =& 200\left(1.03\right) \\ T_{2} =& 200\left(1.03\right)^{2}+200\left(1.03\right) \\ T_{3} =& 200\left(1.03\right)^{3}+200\left(1.03\right)^{2}+200\left(1.03\right) \\ \end{align} \begin{align} T_{{\color{Red}1}} =& 200\left(1.03\right)^{{\color{Red}1}} \\ T_{{\color{Blue}2}} =& 200\left(1.03\right)^{{\color{Blue}2}}+200\left(1.03\right) \\ T_{{\color{Purple}3}} =& 200\left(1.03\right)^{{\color{Purple}3}}+200\left(1.03\right)^{2}+200\left(1.03\right) \\ \end{align}
      • Thus, we should be able to deduce the value that we want, which is T_{10}\,
    • \therefore T_{10}=200\left(1.03\right)^{10}+\ldots+200\left(1.03\right)^{2}+200\left(1.03\right) Can you see the pattern of T_{10}\,? It is a GP (but we need to rearrange it for easier calculations). We might as well also factorize the 200\,
    • =200\left[\left(1.03\right)+\left(1.03\right)^{2}+\ldots+\left(1.03\right)^{10}\right] Be careful, especially with determining the number of terms.
    • =200\underbrace{\left[\left(1.03\right)+\left(1.03\right)^{2}+\ldots+\left(1.03\right)^{10}\right]}_{{\color{Red}\mbox{GP},\;a=1.03,\;r=1.03,\;n=10}} Note that the 200\, is not part of the GP, so just leave it as it is
    • =200\left\{\frac{1.03\left[\left(1.03\right)^{10}-1\right]}{1.03-1}\right\}
    • =2361.56\, Final answer to statement questions should be a statement
    • \therefore \mbox{the total savings at end of 10th year is RM }2361.56
    • Check : There is no easy way to check the answer (actually calculating it 10 times will be too wasting time and also error prone by itself). Just look through the working and maybe repeat the last calculation step. That said, we can also check that our answer makes sense. After 10 years, total savings without interest is 20000, so the rest is from interest.

  • Let's do the second part
    • Analyze : Now, we have to find the year, meaning we have to find n\,, thus we will need the formula for T_{n}\,. We can probably just directly deduce from what we did for T_{10}\,, though I guess one or two more lines wouldn't hurt that much... (in case the marking scheme needs it to be shown explicitly)
    • 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 : We can see from the previous example that we will need to
      • 1) Define T_{n}\,
      • 2) List out a few years until we see the pattern
      • But how should we define T_{n}\,?
        • Note that the interest is calculated on the remaining loan at the beginning of the year
        • We need to settle in 10 years, meaning remaining loan is zero at the end of the 10th year
        • Thus, we should define T_{n}\, to be remaining loan at the end of the year, so that we will be able to use the value T_{10}=\, 0\, (it is slightly more difficult, but actually still possible to use beginning of the year)
      • How should we take into account the payments? The total remaining loans decreases as payments are made, but we don't know how much is the payment, thus we represent it with an unknown first.
        • Payments are being made monthly, but interest calculated yearly (at the end of the year, after 12 payments have been made). Thus instead of subtracting 12 times, we find out yearly payment first.
    • \begin{align} \mbox{Let } T_{n} & = \mbox{ total remaining loan at the end of the year}\\ x & = \mbox{ yearly payment} \end{align} Interest calculated at beginning of first year, by the end of the year, x\, have been paid, thus it should be subtracted from the total
    • T_{1}=20000\left(1.05\right)-x This will carry forward to the early year, where interest is calculated again, and then another x\, is subtracted
    • T_{2}=\left[20000\left(1.05\right)-x\right]\left(1.05\right)-x We should expand without actual calculating
    • =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}
    • \therefore T_{10}=20000\left(1.05\right)^{10}-\left(1.05\right)^{9}x-\ldots-\left(1.05\right)^{2}x-\left(1.05\right)x-x We see that the first term doesn't fit the rest, which is GP. Let's rearrange the rest of the terms first
    • T_{10}=20000\left(1.05\right)^{10}-x\left[1+\left(1.05\right)+\left(1.05\right)^{2}+\ldots+\left(1.05\right)^{9}\right] The number of terms in the GP is 10\,, not 9\, \left(1.05\right)^{{\color{Blue}0}}+\left(1.05\right)^{{\color{Blue}1}}+\left(1.05\right)^{{\color{Blue}2}}+\ldots+\left(1.05\right)^{{\color{Blue}9}}
    • 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}
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