Arithmetic Geometric Progression Past Year

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Contents


Preparation

  • Make sure you have have mastered all the materials in the previous parts.
  • If you are using the material here to learn this topic (as opposed to just a revision to supplement your school lessons), its' best you take a day (or more) off after learning the previous parts.
  • In fact, try take a week off, and then revise back all the materials before trying the following questions. This will be a good practice for how you are going to revise for the actual STPM.
  • Find a comfortable place & comfortable time.
  • DO NOT do this while you are half-awake, or directly after a long day in school. Else, you will frustrate yourself. Trust me.
  • Saving trees is a good thing, but DO NOT do this (in fact, any of the exercises) on rough paper/recycled paper.
  • "ROUGH PAPER = ROUGH WORK = CARELESS MISTAKES = LOSS OF MARKS"
  • Know that you will face questions that you have NEVER SEEN which will require you to adapt on the spot.
  • Keep a clock or watch handy and give yourself the suggested time to complete it. Check your answers too during that time limit.
  • After finishing, check the answers given. If you made mistakes or couldn't find the solution, you can refer to the answers/answers with guidance.

Questions

Estimated time :

1) a) The first three terms of an arithmetic series are -3\frac{1}{8},-1\frac{7}{8}, -\frac{5}{8} . Find the sum of the first 70\, terms of this series.

  • b) The first three terms of a geometric series are 2,-\frac{1}{2}, \frac{1}{8}. Find the sum to infinity of this series.

Find also the smallest value of n\, such that the difference between the sum of the first n\, terms and the sum to infinity is less than 10^{-5}\, .


2) A geometric series has the first term 1\, and common ratio r\,. Given that the sum of the first 3\, terms is seven times of the third term, find both the possible values of r\,.

Find the sum to infinity for each corresponding series.


3) In a geometric sequence with the common ratio r\, ( where r\, is real and r^{2}\neq 1), the sum of the first 16\, terms is three times the sum of the first 8\, terms. Find possible values of r\,.


4) Express the infinite decimal 0. 1 \dot 0 \dot 9 \left(=0.109090909\ldots\right) as a sum of a constant and an infinite geometric series. Hence, express 0. 1 \dot 0 \dot 9 as a fraction in the lowest term.


5) If S_{n}\, denotes the sum of the first n\, terms of a geometric series with the first term a\, and common ratio r\,, show that S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}.

If \left| r \right|<1 and S_{\infty} denotes the sum of the infinite series, find S_{\infty} in terms of a\, and r\,, and show that \frac{S_{\infty}-S_{n}}{S_{\infty}}=r^{n}.

For a geometric series with the third term 27\, and the sixth term 8\,,

  • i) find the value of S_{\infty}
  • ii) determine the smallest value of n\, such that \frac{S_{\infty}-S_{n}}{S_{\infty}}<0.001


6) a) Write the infinite decimal 0. 217 \dot 1 \dot 3 \left(=0.217131313\ldots\right) as a sum of a constant and two infinite geometric series. Hence, write 0. 217 \dot 1 \dot 3 as a fraction in the lowest terms.

  • b) The sum of n\, terms of a series is 3^{n}-1\,. Show that the terms of the series form a geometric sequence. State the first term and common ratio of the series.


7) The first, second and third terms of a geometric series are p,q\, and q^{2}\,, respectively where q<0\, . The first, second and third terms of an arithmetic series are p,q^{2}\, and q\, respectively. Find the values of p\, and q\,.


8) T_{n}\, and S_{n}\, respectively are the n^{th}\, term and the sum of the first n\, terms of a geometric series. The first term of the series is not zero, and the common ratio of the series is an integer greater than 1\,.

  • i) Show that \frac{T_{n+4}-T_{n+2}}{T_{n+1}} is an even number that is independent of n\,.
  • ii) Express \frac{S_{2n}}{S_{n}} in terms of r\, and n\,.


9) a) The sum of the first two terms of a geometric series is \frac{4}{3}, while the sum of the next two terms is 12\,.

  • i) Find the possible first term and common ratio of this geometric series.
  • ii) Find the sum from the 5^{th}\, term to the 8^{th}\, term of the series


b) S_{n}\, is the sum of the first n\, terms of a geometric series with the common ratio r\,, where 0<r<1\,

  • i) Show that \frac{S_{3n}-S_{2n}}{S_{n}}=r^{2n}
  • ii) If \sum_{n=1}^{\infty}\frac{S_{3n}-S_{2n}}{S_{n}}=\frac{1}{3}, find the value of r\, .


10) For the geometric series 6+3+\frac{3}{2}+\ldots, obtain the smallest value of n\, if the difference between the sum of the first n+4\, terms and the sum of the first n\, terms is less than \frac{45}{64}.


11) If S_{n}\, denotes the sum of the first n\, terms of an arithmetic sequence with the first term a\, , last term l\,, and common difference d\,, show that S_{n}=\frac{n}{2}\left(a+l\right)=\frac{n}{2}\left[2a+\left(n-1\right)d\right]


  • i) Find the sum of all the positive integers less than 300\, but are multiples of 3\, or 5\, or both.
  • ii) The sum of the first 13\, terms of an arithmetic sequence is 312\, and the sum of the next 13\, terms is 819\,. Find the first term and common difference of the arithmetic sequence.


12) At the beginning of this year, Mr. Liu and Miss Dora deposited RM 10 000 and RM 2000 respectively in a bank. They receive an interest of 4% per annum. Mr. Liu does not make any additional deposit or withdrawal, whereas, Miss Dora continues to deposit RM 2000 at the beginning of each subsequent years without any withdrawal.

  • a) Calculate the total savings of Mr. Liu at the end of the n^{th}\, year.
  • b) Calculate the total savings of Miss Dora at the end of the n^{th}\, year.
  • c) Determine in which year the total savings of Miss Dora exceeds the total savings of Mr. Liu.


13) An investor intends to invest RM W\, in a finance company at the start of each year. The dividend of d%\, per year that he receives is reinvested in the company. Write the total of his investment that accumulates at the end of the first year, second year, and third year. Deduce that the total investment, in RM, that accumulates at the end of year n\, (including dividend for the n^{th}\, year) is W\left(1+\frac{100}{d}\right)\left[\left(1+\frac{d}{100}\right)^{n}-1\right]

If the amount invested each year is RM5000 and the dividend is 10% per year, find the total of his savings that he accumulates at the end of the tenth year.

If the investor stops depositing RM5000 every year after the tenth year, find the total of his investment at the end of the 12th year.

If the investor then withdraws RM30000 at the end of the 12th year and another RM30000 at the end of the 13th year, determine his total savings in the company at the end of the 13th year.

Answers

1) a) The first three terms of an arithmetic series are -3\frac{1}{8},-1\frac{7}{8}, -\frac{5}{8} . Find the sum of the first 70\, terms of this series.

  • 2800\,
    • \begin{align} & a= -\frac{25}{8}, d= -\frac{15}{8}-\left(-\frac{25}{8}\right)=\frac{5}{4}\\ & S_{70}=\frac{70}{2}\left[2\left(-\frac{25}{8}\right)+69\left(\frac{5}{4}\right)\right]\\ & \quad \; =2800 \end{align}
  • b) The first three terms of a geometric series are 2,-\frac{1}{2}, \frac{1}{8}. Find the sum to infinity of this series.
  • \frac{8}{5}
    • \begin{align} & a= 2, r= -\frac{1}{4}\\ & \left|r\right|<1\\ & \therefore S_{\infty}=\frac{2}{1-\left(-\frac{1}{4}\right)}\\ & \quad \quad =\frac{2}{\left(\frac{5}{4}\right)}=\frac{8}{5}\\ \end{align}

Find also the smallest value of n\, such that the difference between the sum of the first n\, terms and the sum to infinity is less than 10^{-5}\,.

  • 9\,
    • \begin{align} \left|S_{\infty}-S_{n}\right|&<10^{-5}\\ \left|\frac{a}{1-r}-\frac{a\left(1-r^{n}\right)}{1-r}\right|&<10^{-5}\\ \left|\frac{ar^{n}}{1-r}\right|&<10^{-5}\\ \left|\frac{2\left(-\frac{1}{4}\right)^{n}}{1-\left(-\frac{1}{4}\right)}\right|&<10^{-5}\\ \left|\left(-\frac{1}{4}\right)^{n}\right|&<\frac{5}{8}\times 10^{-5}\\ \left(\frac{1}{4}\right)^{n}&<\frac{5}{8}\times 10^{-5}\\ n&>\frac{\lg{\left(\frac{5}{8}\times 10^{-5}\right)}}{\lg{\frac{1}{4}}}\\ n&>8.6439 \\ \therefore \mbox{The smallest value of } n \mbox{ is } 9\\ \end{align}


2) A geometric series has the first term 1\, and common ratio r\,. Given that the sum of the first 3\, terms is seven times of the third term, find both the possible values of r\,.

  • r=-\frac{1}{3},r=\frac{1}{2}
    • \begin{align} & S_{3}=7u_{3}, a=1\\ & 1+r+r^{2}=7r^{2}\\ & 6r^{2}-r-1=0\\ & \left(3r+1\right)\left(2r-1\right)=0\\ & r=-\frac{1}{3}, r=\frac{1}{2} \\ & \therefore r=-\frac{1}{3}, r=\frac{1}{2} \\ \end{align}
    • OR
    • \begin{align} & S_{3}=7u_{3}, a=1\\ & \frac{r^{3}-1}{r-1}=7r^{2}\\ & r^{3}-1=7r^{3}-7r^{2}\\ & 6r^{3}-7r^{2}+1=0\\ & \left(r-1\right)\left(6r^{2}-r-1\right)=0\\ & \left(r-1\right)\left(3r+1\right)\left(2r-1\right)=0\\ & r=1\mbox{(rejected)}, r=-\frac{1}{3}, r=\frac{1}{2} \\ & \therefore r=-\frac{1}{3}, r=\frac{1}{2} \\ \end{align}

Find the sum to infinity for each corresponding series.

  • S_{\infty}=\frac{3}{4},S_{\infty}=2
    • \begin{align} & \mbox{When } r=-\frac{1}{3},\left|r\right|<1, S_{\infty}=\frac{1}{1-\left(-\frac{1}{3}\right)}=\frac{3}{4}\\ & \mbox{When } r=\frac{1}{2}, \left|r\right|<1, S_{\infty}=\frac{1}{1-\left(\frac{1}{2}\right)}=2\\ \end{align}


3) In a geometric sequence with the common ratio r\, ( where r\, is real and r^{2}\neq 1), the sum of the first 16\, terms is three times the sum of the first 8\, terms. Find possible values of r\,.

  • r= 2^{\frac{1}{8}}\mbox{ or } r= -2^{\frac{1}{8}}
    • \begin{align} & S_{16}=3S_{8}\\ & \frac{a\left(r^{16}-1\right)}{r-1}=\frac{a\left(r^{8}-1\right)}{r-1}\\ & \frac{r^{16}-1}{r^{8}-1}=3 \\ & \frac{\left(r^{8}-1\right)\left(r^{8}+1\right)}{r^{8}-1}=3 \\ & r^{8}+1 =3\\ & r^{8}=2 \\ & r=\pm 2^{\frac{1}{8}}\\ & r= 2^{\frac{1}{8}}\mbox{ or } r= -2^{\frac{1}{8}}\\ \end{align}


4) Express the infinite decimal 0. 1 \dot 0 \dot 9 \left(=0.109090909\ldots\right) as a sum of a constant and an infinite geometric series. Hence, express 0. 1 \dot 0 \dot 9 as a fraction in the lowest term.

  • \frac{6}{55}
    • \begin{align} 0. 1 \dot 0 \dot 9 & = 0.109090909\ldots \\ & = 0.1 + \underbrace{0.009 +0.00009+0.0000009+\ldots...}_{\mbox{GP with }a=0.09, r=0.01}\\ & =0.1 +\frac{0.009}{1-0.01}\\ & =0.1 +\frac{0.009}{0.99}\\ & =\frac{1}{10}+\frac{9}{990}\\ & =\frac{6}{55} \end{align}


5) If S_{n}\, denotes the sum of the first n\, terms of a geometric series with the first term a\, and common ratio r\,, show that S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}.

    • \begin{align} & S_{n}= a+ar+ar^{2} + \ldots +ar^{n-2} + ar^{n-1} \frac{\qquad}{}(1)\\ & rS_{n}= ar+ar^{2}+ar^{3} + \ldots +ar^{n-1} + ar^{n} \frac{\qquad}{}(2)\\ & (1)-(2): \\ & S_{n}-rS_{n} = a-ar^{n} \\ & S_{n}\left(1-r\right)=a\left(1-r^{n}\right) \\ & S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}\quad \frac{\qquad}{}\quad \mbox{shown} \end{align}

If \left| r \right|<1 and S_{\infty} denotes the sum of the infinite series, find S_{\infty} in terms of a\, and r\,, and show that \frac{S_{\infty}-S_{n}}{S_{\infty}}=r^{n}.

    • \begin{align} & \mbox{If} \left| r \right|<1, \mbox{as}\quad n \to \infty,r^{n} \to 0 \\ & S_{\infty}= \lim_{n \to \infty} S_{n} =\lim_{n \to \infty} \frac{a\left(1-r^{n}\right)}{1-r}= \frac{a}{1-r} \\ & \frac{S_{\infty}-S_{n}}{S_{\infty}} = \frac{\frac{a}{1-r}-\frac{a\left(1-r^{n}\right)}{1-r}}{\frac{a}{1-r}}\\ & \qquad \qquad = \frac{\frac{a}{1-r}\left[1-\left(1-r^{n}\right)\right]}{\frac{a}{1-r}}\\ & \qquad \qquad = r^{n}\\ \end{align}

For a geometric series with the third term 27\, and the sixth term 8\,,

  • i) find the value of S_{\infty}
  • \frac{729}{4}
    • \begin{array}{ll} u_{3}=27 & u_{6}= 8 \\ ar^{2} =27\frac{\quad}{}(1) & ar^{5} =8\frac{\quad}{}(1) \\ \end{array}
    • \begin{align} & \frac{(2)}{(1)}:\frac{ar^{5}}{ar^{2}}=\frac{8}{27}\\ & \qquad r^{3} = \frac{8}{27} \\ & \qquad r = \frac{2}{3}, \therefore a=\frac{243}{4}\\ & S_{\infty} = \frac{a}{1-r} = \frac{\frac{243}{4}}{1-\frac{2}{3}}=\frac{729}{4} \\ \end{align}
  • ii) determine the smallest value of n\, such that \frac{S_{\infty}-S_{n}}{S_{\infty}}<0.001
  • 18\,
    • \begin{align} \frac{S_{\infty}-S_{n}}{S_{\infty}}& <0.001 \\ r^{n} & < 0.001 \\ \left(\frac{2}{3}\right)^{n} & < 0.001 \\ n & > 17.037 \\\end{align}
    • \therefore \mbox{Smallest value of } n \mbox{ is } 18


6) a) Write the infinite decimal 0. 217 \dot 1 \dot 3 \left(=0.217131313\ldots\right) as a sum of a constant and two infinite geometric series. Hence, write 0. 217 \dot 1 \dot 3 as a fraction in the lowest terms.

  • \frac{2687}{12375}
    • \begin{align} 0. 217 \dot 1 \dot 3 & =0.217131313\ldots \\ & =0.217 + 0.0001 +0.00003+ 0.000001 +0.0000003 +0.00000001 +0.000000003 +\ldots \\ & =0.217 + \underbrace{\left(0.0001 + 0.000001 + 0.00000001 + \ldots \right)}_{\mbox{GP with } a=0.0001, r=0.01} + \underbrace{\left(0.00003 + 0.0000003 + 0.000000003 +\ldots \right)}_{\mbox{GP with }a=0.00003, r=0.01} \\ & =0.217 +\frac{0.0001}{1-0.01}+\frac{0.00003}{1-0.01} \\ & =0.217 + \frac{1}{9900} +\frac{3}{99000} \\ & =\frac{2687}{12375} \end{align}
  • b) The sum of n\, terms of a series is 3^{n}-1\,. Show that the terms of the series form a geometric sequence. State the first term and common ratio of the series.
  • a=2, r=3\,
    • \begin{align} u_{n} &= S_{n}-S_{n-1}\\ & =\left(3^{n}-1\right)-\left(3^{n-1}-1\right) \\ & = 3^{n} - 3^{n-1}\\ & = 3^{n-1}\left(3-1\right) \\ & = 2\left(3^{n-1}\right)\\ \frac{u_{n+1}}{u_{n}}& =\frac{2\left(3^{n}\right)}{2\left(3^{n-1}\right)} = 3^{n-\left(n-1\right)}\\ & = 3 \mbox{(constant)}\\ & \therefore \mbox{the terms form a geometric sequence} \end{align}
    • a=u_{1}=2, r=3\,


7) The first, second and third terms of a geometric series are p,q\, and q^{2}\,, respectively where q<0\, . The first, second and third terms of an arithmetic series are p,q^{2}\, and q\, respectively. Find the values of p\, and q\,.

  • p=1, q=-\frac{1}{2}
    • \begin{align} & \mbox{GP}: p,q, q^{2}\\ & \therefore \frac{q^{\cancel{2}}}{\cancel{q}}=\frac{q}{p}\\ & p\cancel{q}=\cancel{q} \left(\because q \neq 0 \right)\\ & p=1 \\ & \mbox{AP}: p,q^{2}, q\\ & \therefore q-q^{2}=q^{2}-p \\ & q=2q^{2}-1 \\ & 2q^{2}-q-1=0 \\ & \left(2q+1\right)\left(q-1\right)=0 \\ & q=-\frac{1}{2}, q=1\mbox{(rejected)}\\ & \therefore p=1, q=-\frac{1}{2} \\ \end{align}


8) T_{n}\, and S_{n}\, respectively are the n^{th}\, term and the sum of the first n\, terms of a geometric series. The first term of the series is not zero, and the common ratio of the series is an integer greater than 1\,.

  • i) Show that \frac{T_{n+4}-T_{n+2}}{T_{n+1}} is an even number that is independent of n\,.
    • \begin{align} \frac{T_{n+4}-T_{n+2}}{T_{n+1}} & = \frac{ar^{n+3}-ar^{n+1}}{ar^{n}}\\ & = \frac{ar^{n}\left(r^{3}-r\right)}{ar^{n}}\\ & = r^{3}-r \\ & = r\left(r^{2}-1\right)\\ & = r\left(r-1\right)\left(r+1\right)\\ \end{align}
    • \mbox{When } r \mbox{ is odd}, r-1 \mbox{ and } r+1 \mbox{ is even}, \therefore r\left(r-1\right)\left(r+1\right) \mbox{ is even}
    • \mbox{When } r \mbox{ is even}, r-1 \mbox{ and } r+1 \mbox{ is odd}, \therefore r\left(r-1\right)\left(r+1\right) \mbox{ is even}
    • \therefore \frac{T_{n+4}-T_{n+2}}{T_{n+1}} \mbox{ is an even number that is independent of }n
  • ii) Express \frac{S_{2n}}{S_{n}} in terms of r\, and n\,.
    • \begin{align} \frac{S_{2n}}{S_{n}} & = \frac{\frac{a\left(r^{2n}-1\right)}{r-1}}{\frac{a\left(r^{n}-1\right)}{r-1}}\\ & = \frac{r^{2n}-1}{r^{n}-1} \\ & = \frac{\left(r^{n}-1\right)\left(r^{n}+1\right)}{r^{n}-1} \\ & = r^{n}+1 \end{align}


9) a) The sum of the first two terms of a geometric series is \frac{4}{3}, while the sum of the next two terms is 12\,.

  • i) Find the possible first term and common ratio of this geometric series.
    • \begin{array}{ll} u_{1}+u_{2} =\frac{4}{3} & u_{3}+u_{4} =12 \\ a+ar =\frac{4}{3} & ar^{2}+ar^{3}=12 \\ a\left(1+r\right)=\frac{4}{3}\frac{\quad}{}(1) & ar^{2}\left(1+r\right)\frac{\quad}{}(2)\\ \end{array}
    • \begin{align} & \frac{(2)}{(1)}: \frac{ar^{2}\left(1+r\right)}{a\left(1+r\right)}=\frac{12}{\left(\frac{4}{3}\right)}\\ & r^{2}=9 \\ & r=\pm 3 \\ & \therefore r=3, a=\frac{1}{3}; r=-3, a = -\frac{2}{3}\\ \end{align}
  • ii) Find the sum from the 5^{th}\, term to the 8^{th}\, term of the series
  • 1080\,
    • \mbox{When } r=3, a=\frac{1}{3}
    • \begin{align} S_{8}-S_{4} & = \frac{\frac{1}{3}\left(3^{8}-1\right)}{3-1}-\frac{\frac{1}{3}\left(3^{4}-1\right)}{3-1} \\ & = \frac{3280}{3}-\frac{40}{3} \\ & = 1080 \\ \end{align}
    • \mbox{When } r=-3, a=-\frac{2}{3}
    • \begin{align} S_{8}-S_{4} & = \frac{-\frac{2}{3}\left[1-\left(-3\right)^{8}\right]}{1-\left(-3\right)}-\frac{-\frac{2}{3}\left[1-\left(-3\right)^{4}\right]}{1-\left(-3\right)} \\ & = \frac{3280}{3}-\frac{40}{3} \\ & = 1080 \\ \end{align}

b) S_{n}\, is the sum of the first n\, terms of a geometric series with the common ratio r\,, where 0<r<1\,

  • i) Show that \frac{S_{3n}-S_{2n}}{S_{n}}=r^{2n}
    • \begin{align} \frac{S_{3n}-S_{2n}}{S_{n}} & = \frac{\frac{a\left(1-r^{3n}\right)}{1-r}-\frac{a\left(1-r^{2n}\right)}{1-r}}{\frac{a\left(1-r^{n}\right)}{1-r}}\\ & = \frac{-r^{3n}+r^{2n}}{1-r^{n}}\\ & = \frac{r^{2n}\left(1-r^{n}\right)}{1-r^{n}}\\ & = r^{2n} \frac{\quad}{}\mbox{(Shown)} \end{align}
  • ii) If \sum_{n=1}^{\infty}\frac{S_{3n}-S_{2n}}{S_{n}}=\frac{1}{3}, find the value of r\, .
  • r=\frac{1}{2}
    • \begin{align} \sum_{n=1}^{\infty}\frac{S_{3n}-S_{2n}}{S_{n}} & = \sum_{n=1}^{\infty}r^{2n} \\ & = \underbrace{r^{2} + r^{4} +r^{6}+\ldots}_{\mbox{GP } a= r^{2} , \mbox{ common ratio } = r^{2} } \\ & = \frac{r^{2}}{1-r^{2}}\\ \end{align}
    • \begin{align} \therefore \frac{r^{2}}{1-r^{2}} =\frac{1}{3}\\ 3r^{2}=1-r^{2}\\ 4r^{2}=1\\ r^{2}=\frac{1}{4} \\ r=\pm\frac{1}{2}\\ \therefore r=\frac{1}{2}\\ \end{align}


10) For the geometric series 6+3+\frac{3}{2}+\ldots, obtain the smallest value of n\, if the difference between the sum of the first n+4\, terms and the sum of the first n\, terms is less than \frac{45}{64}.

  • 5\,
    • \begin{align} & a=6, r=\frac{1}{2} \\ & S_{n+4}-S_{n}<\frac{45}{64}\\ & \frac{6\left[1-\left(\frac{1}{2}\right)^{n+4}\right]}{1-\frac{1}{2}}-\frac{6\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}<\frac{45}{64}\\ & -\left(\frac{1}{2}\right)^{n+4}+\left(\frac{1}{2}\right)^{n}<\frac{45}{64} \times \frac{1}{12}\\ & \left(\frac{1}{2}\right)^{n}\left[1-\left(\frac{1}{2}\right)^{4}\right]<\frac{15}{256}\\ & \left(\frac{1}{2}\right)^{n} < \frac{1}{16}\\ & \left(\frac{1}{2}\right)^{n} < \left(\frac{1}{2}\right)^{4}\\ & n>4 \\ & \therefore \mbox{smallest value of }n \mbox{ is } 5 \end{align}


11) If S_{n}\, denotes the sum of the first n\, terms of an arithmetic sequence with the first term a\, , last term l\,, and common difference d\,, show that S_{n}=\frac{n}{2}\left(a+l\right)=\frac{n}{2}\left[2a+\left(n-1\right)d\right]

    • \begin{array}{ccccccccccccccc} S_{n}= & a & + & \left(a+d \right) & + & \left(a+2d \right) & + & \ldots & + & \left(l-2d \right) & + & \left(l-d \right) & + & l & \frac{\quad}{}(1) \\ S_{n}= & l & + & \left(l-d \right) & + & \left(l-2d \right) & + & \ldots & + & \left(a+2d \right) & + & \left(a+d \right) & + & a & \frac{\quad}{}(2)\\ \end{array}
    • \begin{align} &(1)+(2): \\ 2S_{n}& =\underbrace{\left(a+l\right)+\left(a+l\right)+\left(a+l\right)+\ldots+\left(a+l\right)+\left(a+l\right)+\left(a+l\right)}_{n\mbox{ terms}} \\ & = n \left(a+l\right) \\ S_{n}& =\frac{n}{2}\left(a+l\right) \\ \end{align}
    • \begin{align} & \mbox{Since } l=a+\left(n-1\right)d \\ & S_{n} =\frac{n}{2}\left[a+a+\left(n-1\right)d\right] =\frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ & \therefore S_{n}=\frac{n}{2}\left(a+l\right)=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ \end{align}
  • i) Find the sum of all the positive integers less than 300\, but are multiples of 3\, or 5\, or both.
  • 20850\,
    • \begin{align} \mbox{Let } S_{a}&=\underbrace{3+6+9+\ldots+297}_{\mbox{AP with}\;a=3,\; d=3,\; n=99} \\ S_{b} & = \underbrace{5+10+15+\ldots+295}_{\mbox{AP with}\;a=5,\; d=5,\; n=59}\\ S_{c} & = \underbrace{15+45+60+\ldots+285}_{\mbox{AP with}\;a=15,\; d=15,\; n=19}\\ & \mbox{Sum} = S_{a}+S_{b}-S_{c}\\ & \qquad =\frac{99}{2}\left(3+297\right)+\frac{59}{2}\left(5+295\right)-\frac{19}{2}\left(15+285\right)\\ & \qquad =14850+8850-2850=20850 \end{align}
  • ii) The sum of the first 13\, terms of an arithmetic sequence is 312\, and the sum of the next 13\, terms is 819\,. Find the first term and common difference of the arithmetic sequence.
  • a=6,d=3\,
    • \begin{array}{ll} S_{13} = 312 \qquad& S_{26}-S_{13}= 819 \\ & S_{26}-312= 819 \\ & S_{26}=1131 \\ \frac{13}{2}\left(2a+12d\right)=312 & \frac{26}{2}\left(2a+25d\right)=1131\\ 2a+12d=48\frac{\quad}{}(1) & 2a+25d =87\frac{\quad}{}(2) \\ \end{array}
    • \begin{align} & (2)-(1): \\ & 13d = 39 \\ & d=3 , a=6 \end{align}


12) At the beginning of this year, Mr. Liu and Miss Dora deposited RM 10 000 and RM 2000 respectively in a bank. They receive an interest of 4% per annum. Mr. Liu does not make any additional deposit or withdrawal, whereas, Miss Dora continues to deposit RM 2000 at the beginning of each subsequent years without any withdrawal.

  • a) Calculate the total savings of Mr. Liu at the end of the n^{th}\, year.
  • 10000\left(1.04\right)^{n}
    • \begin{align} u_{1}& =10000\left(1.04\right) \\ u_{2}& =\left[10000\left(1.04\right)\right]\left(1.04\right) \\ & = 10000\left(1.04\right)^{2} \\ u_{3}& =10000\left(1.04\right)^{2}\left(1.04\right) \\ & = 10000\left(1.04\right)^{3} \\ \therefore u_{n} & = 10000\left(1.04\right)^{n} \\ \end{align}
  • b) Calculate the total savings of Miss Dora at the end of the n^{th}\, year.
  • 52000\left[\left(1.04\right)^{n}-1\right]
    • \begin{align} T_{1}& =2000\left(1.04\right) \\ T_{2}& =\left[2000\left(1.04\right)+2000\right]\left(1.04\right) \\ & = 2000\left(1.04\right)^{2} + 2000\left(1.04\right) \\ T_{3}& =\left[2000\left(1.04\right)^{2} + 2000\left(1.04\right)+2000\right]\left(1.04\right) \\ & = 2000\left(1.04\right)^{3}+ 2000\left(1.04\right)^{2} + 2000\left(1.04\right) \\ \therefore T_{n} & = 2000\left(1.04\right)^{n}+2000\left(1.04\right)^{n-1}+\ldots + 2000\left(1.04\right)^{2} + 2000\left(1.04\right) \\ & = 2000\underbrace{\left[\left(1.04\right)+\left(1.04\right)^{2}+\left(1.04\right)^{3}+\ldots +\left(1.04\right)^{n}\right]}_{\mbox{GP} a=1.04, r=1.04 } \\ & = \frac{2000\left(1.04\right)\left[\left(1.04\right)^{n}-1\right]}{1.04-1} \\ & = 52000\left[\left(1.04\right)^{n}-1\right] \\ \end{align}
  • c) Determine in which year the total savings of Miss Dora exceeds the total savings of Mr. Liu.
  • \mbox{at the end of 6th year}\,
    • \begin{align} 52000\left[\left(1.04\right)^{n}-1\right] & > 10000\left(1.04\right)^{n} \\ 52000\left(1.04\right)^{n} - 52000 & > 10000\left(1.04\right)^{n} \\ 42000\left(1.04\right)^{n} & > 52000 \\ \left(1.04\right)^{n} & > \frac{26}{21}\\ n & >\frac{\lg \frac{26}{21}}{\lg 1.04 } \\ n & > 5.4454 \\ \therefore \mbox{at the end of 6th year}& \end{align}


13) An investor intends to invest RM W\, in a finance company at the start of each year. The dividend of d%\, per year that he receives is reinvested in the company. Write the total of his investment that accumulates at the end of the first year, second year, and third year. Deduce that the total investment, in RM, that accumulates at the end of year n\, (including dividend for the n^{th}\, year) is W\left(1+\frac{100}{d}\right)\left[\left(1+\frac{d}{100}\right)^{n}-1\right]

    • \mbox{Let } T_{n} = \mbox{Total investment at the end of the } n^{th} \mbox{ year}\,
    • \begin{align} T_{1} & = W\left(1+\frac{d}{100}\right) \\ T_{2} & = \left[W+W\left(1+\frac{d}{100}\right)\right]\left(1+\frac{d}{100}\right) \\ & = W\left(1+\frac{d}{100}\right)+W\left(1+\frac{d}{100}\right)^{2} \\ T_{3} & = \left[W+ W\left(1+\frac{d}{100}\right)+W\left(1+\frac{d}{100}\right)^{2}\right]\left(1+\frac{d}{100}\right)\\ & = W\left(1+\frac{d}{100}\right)+W\left(1+\frac{d}{100}\right)^{2}+W\left(1+\frac{d}{100}\right)^{3} \\ T_{n} & = W\left(1+\frac{d}{100}\right)+W\left(1+\frac{d}{100}\right)^{2}+W\left(1+\frac{d}{100}\right)^{3}+\ldots+ W\left(1+\frac{d}{100}\right)^{n} \\ & = W \underbrace{\left[\left(1+\frac{d}{100}\right)+\left(1+\frac{d}{100}\right)^{2}+\left(1+\frac{d}{100}\right)^{3}+\ldots+ \left(1+\frac{d}{100}\right)^{n}\right]}_{\mbox{GP }\;a=1+\frac{d}{100},\;r=1+\frac{d}{100},\; n \mbox{ terms }}\\ & = W\left(1+\frac{d}{100}\right)\frac{\left[\left(1+\frac{d}{100}\right)^{n}-1\right]}{\left(1+\frac{d}{100}\right)-1}\\ & = W \left(1+\frac{d}{100}\right)\left(\frac{100}{d}\right)\left[\left(1+\frac{d}{100}\right)^{n}-1\right]\\ & = W\left(1+\frac{100}{d}\right)\left[\left(1+\frac{d}{100}\right)^{n}-1\right]\frac{\qquad}{}\mbox{(deduced)} \end{align}

If the amount invested each year is RM5000 and the dividend is 10% per year, find the total of his savings that he accumulates at the end of the tenth year.

  • \mbox{RM }87655.84\,
    • \begin{align} T_{10} & = 5000\left(1+\frac{100}{10}\right)\left[\left(1+\frac{10}{100}\right)^{10}-1\right] \\ & = 87655.84 \end{align}
    • \mbox{total savings at end of tenth year is RM 87655.84}\,

If the investor stops depositing RM5000 every year after the tenth year, find the total of his investment at the end of the 12th year.

  • \mbox{RM }106063.56\,
    • \begin{align} T_{11} & = T_{10}\left(1.10\right) \\ T_{12} & = T_{10}\left(1.10\right)^{2} \\ & = 106063.56 \end{align}
    • \mbox{total savings at end of 12th year is RM 106063.56}\,

If the investor then withdraws RM30000 at the end of the 12th year and another RM30000 at the end of the 13th year, determine his total savings in the company at the end of the 13th year.

  • \mbox{RM }53669.92\,
    • \begin{align} T_{13} & = \left(T_{12}-30000\right)\left(1.10\right)-30000 \\ & = 53669.92 \end{align}
    • \mbox{total savings at end of 12th year is RM 53669.92}\,
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