Arithmetic Geometric Progression Past Year

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1997-S2-10

ii) An education award fund is started in a school with an initial amount of RM 20 000.00, to provide annual awards of RM 1 800.00 to a best student. The fund's money is kept in a bank with an annual interest of 6%. If the first award is given exactly one year after the money is kept at the bank, find the number of years the awards can be given consecutively. [ 8 Marks ]

$\mbox{The award can be given consecutively for 18 years.}\,$
- $\mbox{Let } T_{n} = \mbox{Remaining amount in the fund at the end of the } n^{\mbox{th}} \mbox{ year the award is given}\,$
- $\begin{align} T_{0} & = 20000\left(1.06\right)\\ T_{1} & = \left[20000\left(1.06\right)-1800\right]\left(1.06\right) \\ & = 20000\left(1.06\right)^{2}-1800\left(1.06\right)\\ T_{2} & = \left[20000\left(1.06\right)^{2}-1800\left(1.06\right)-1800\right]\left(1.06\right)\\ & = 20000\left(1.06\right)^{3}-1800\left(1.06\right)^{2}-1800\left(1.06\right)\\ T_{3} &= \left[20000\left(1.06\right)^{3}-1800\left(1.06\right)^{2}-1800\left(1.06\right)-1800\right]\left(1.06\right)\\ &= 20000\left(1.06\right)^{4}-1800\left(1.06\right)^{3}-1800\left(1.06\right)^{2}-1800\left(1.06\right)\\ \therefore T_{n} & = 20000\left(1.06\right)^{n+1}-1800\left(1.06\right)^{n}-1800\left(1.06\right)^{n-1}-\ldots-1800\left(1.06\right)^{2}-1800\left(1.06\right)\\ & = 20000\left(1.06\right)^{n+1}-1800\underbrace{\left[1.06+\left(1.06\right)+\left(1.06\right)^{2}+\ldots+\left(1.06\right)^{n-1}+\left(1.06\right)^{n}\right]}_{\mbox{GP }a=1.06,\;r=1.06,\;\mbox{number of term } = n}\\ & = 20000\left(1.06\right)^{n+1}-\frac{1800\left(1.06\right)\left[\left(1.06\right)^{n}-1\right]}{1.06-1}\\ & = 21200\left(1.06\right)^{n}- 318000\left[\left(1.06\right)^{n}-1\right]\\ & = 31800-10600\left(1.06\right)^{n} \\ \end{align}$
- $\mbox{At the the end of the last year the award is given, }\,$
- $\begin{align} T_{n} & <1800 \\ 31800-10600\left(1.06\right)^{n} & < 1800 \\ 30000 & <10600\left(1.06\right)^{n} \\ \left(1.06\right)^{n} & > \frac{150}{53} \\ n \lg 1.06 & > \lg \frac{150}{53} \\ n & > \frac{\lg \frac{150}{53}}{\lg 1.06}\\ n & > 17.854 \\ \therefore n =18 \\ \end{align}$
- $\therefore \mbox{The award can be given consecutively for 18 years.}\,$
OR
- $\mbox{Let } T_{n} = \mbox{Remaining amount in the fund at the beginning of the } n^{\mbox{th}} \mbox{ year the award is given}\,$
- $\begin{align} T_{0} & = 20000\\ T_{1} & = 20000\left(1.06\right)-1800\\ T_{2} & = \left[20000\left(1.06\right)-1800\right]\left(1.06\right)-1800 \\ & = 20000\left(1.06\right)^{2}-1800\left(1.06\right)-1800\\ T_{3} &= \left[20000\left(1.06\right)^{2}-1800\left(1.06\right)-1800\right]\left(1.06\right)-1800\\ &= 20000\left(1.06\right)^{3}-1800\left(1.06\right)^{2}-1800\left(1.06\right)-1800\\ \therefore T_{n} & = 20000\left(1.06\right)^{n}-1800\left(1.06\right)^{n-1}-1800\left(1.06\right)^{n-2}-\ldots-1800\left(1.06\right)-1800\\ & = 20000\left(1.06\right)^{n}-1800\underbrace{\left[1+1.06+\left(1.06\right)+\ldots+\left(1.06\right)^{n-2}+\left(1.06\right)^{n-1}\right]}_{\mbox{GP }a=1,\;r=1.06,\;\mbox{number of term } = n}\\ & = 20000\left(1.06\right)^{n}-\frac{1800\left[\left(1.06\right)^{n}-1\right]}{1.06-1}\\ & = 20000\left(1.06\right)^{n}- 300000\left[\left(1.06\right)^{n}-1\right]\\ & = 30000-10000\left(1.06\right)^{n} \\ \end{align}$
- $\mbox{For award to be given, }\,$
- $\begin{align} T_{n} & >0 \\ 30000-10000\left(1.06\right)^{n} & >0 \\ 30000 & >10000\left(1.06\right)^{n} \\ \left(1.06\right)^{n} & < 3 \\ n \lg 1.06 & < \lg 3 \\ n & < \frac{\lg 3}{\lg 1.06}\\ n & < 18.854 \\ \therefore n =18 \\ \end{align}$
- $\therefore \mbox{The award can be given consecutively for 18 years.}\,$
- Analyze
- The first step of "letting" is actually the hardest

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