Sequence

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Introduction to sequence

In general, sequence is just simply a set of values that is written in such a way the values follows certain specific order or rule. A sequence can either be a finite set of numbers or an infinite set of numbers. A sequence is said to be infinite if there exist no end values and the numbers can continues infinitely. On the other hand, a finite is such that the set of numbers has and end value or stop at certain length. In this chapter, students will be introduced to two type of sequence; Arithmetic Sequence and Geometric Sequence. For both types, students will be exposed on the concepts of the \(n^{th}\ term\) of a sequence and the \(sum\ of\ the\ first\) \(n^{th}\ term\) of a sequence. Finally, students will be exposed on how to apply the concepts of arithmetic and geometric sequences to some common problems in daily life.

What is a sequence?

Sequence can be defined as a set of numbers which are written in specific order or following a specific rule. For example,
\[ \begin{aligned} &0,\ 2,\ 4,\ 6,\ 8,\ … \\ &1,\ 3,\ 9,\ 27,\ … \\ &17,\ 12,\ 7,\ 2,\ -3,\ -8,\ -13. \end{aligned} \] The three sets of numbers above are an example of sequence since between adjacent values, it follows certain order or rule.

Arithmetic sequence

Consider this sequence

\(5, 13, 21, 29,…\)

It starts with a particular first term, and to get the next term, add a fixed value to the previous term. In the above set of values, we add 8 to get the next term. The difference between consecutive terms is a constant.
i.e.,

\[ \begin{aligned} &5,\ 5 + 8,\ 5 + 8 + 8,\ 5 + 8 + 8 + 8,\ … \\ &5,\ 5 + 8,\ 5 + 2(8),\ 5 + 3(8),\ … \\ &5,\ 13,\ 21,\ 29,\ … \\ \end{aligned} \]

Definition
An arithmetic progression is a sequence where each new term after the first is obtained by adding a constant \(d\), called the common difference, to the preceding term. If the first term of the sequence is \(a\) then the arithmetic progression is

\(a, a + d, a + 2d, a + 3d,...\)

where the nth term is given by the formula

\[ T_n = a + (n − 1)d \]

Finding the nth term

Example 1
Write down the first five terms of the Arithmetic Sequence with first term 8 and common difference 7. So, here we have \(a = 8\) and \(d = 7\).
Using the general formula of \(n^{th}\) term,
\[ \begin{aligned} T_n &= a + (n - 1)d \\ T_1 &= a = 8 \\ T_2 &= a + d = 8 + 7 = 15 \\ T_3 &= a + 2d = 8 + 2(7) = 22 \\ T_4 &= a + 3d = 8 + 3(7) = 29 \\ T_5 &= a + 4d = 8 + 4(7) = 36 \\ \end{aligned} \]

The sequence is \(8, 15, 22, 29, 36,…\)

Example 2
Write down the first five terms of the Arithmetic Sequence with first term 2 and common difference −5. So, here we have \(a = 2\) and \(d = -5\).
Using the general formula of \(n^{th}\) term,

\[ \begin{aligned} T_n &= a + (n - 1)d \\ T_1 &= a = 2 \\ T_2 &= a + d = 2 + (-5) = -3 \\ T_3 &= a + 2d = 2 + 2(-5) = -8 \\ T_4 &= a + 3d = 2 + 3(-5) = -13 \\ T_5 &= a + 4d = 2 + 4(-5) = -18 \\ \end{aligned} \]

The sequence is \(2, -3, -8, -13, -18,…\)

Finding the sum of the first n terms

The sum of the terms of an arithmetic progression gives an arithmetic series. If the starting value is \(a\) and the common difference is \(d\) then the sum of the first \(n\) terms is given by,

\[ S_n = \frac{n}{2}[2a + (n - 1)d] \]

Example 3

Find the sum of the first 50 terms of the sequence

\(1, 3, 5, 7, 9,...\)

Here, we have \(a = 1\), \(d = 2\), and \(n = 50\). Use the formula \(S_n = \frac{n}{2}[2a + (n - 1)d]\), we get

\[ \begin{aligned} S_50 &= \frac{50}{2}[2(1) + (50-1)(2)] \\ &= 25[2+49(2)] \\ &= 25(100) \\ &= 2,500 \end{aligned} \]

Example 4
An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference.

Here, we are given \(a = 3\) and \(S_8 = 2S5\).
Thus,

\[ \begin{aligned} S_8 &= 2S_5 \\ \frac{8}{2}[2(3)+(8-1)d] &= 2\bigg{[}\frac{5}{2}[2(3)+(5-1)d]\bigg{]} \\ 4[6+7d] &= 5[6+4d] \\ 24+28d &= 30+20d \\ 8d &= 6 \\ d &= \frac{3}{4} \end{aligned} \]

Geometric sequence

A geometric progression, or GP, is a sequence where each new term after the first is obtained by multiplying the preceding term by a constant \(r\), called the common ratio.

Finding the nth term

If the first term of the sequence is \(a\) then the geometric progression is

\(a,\ ar,\ ar_2,\ ar_3,\ ...\)

where the n-th term is given by
\[T_n\ =\ ar^{n-1}\]

Finding the sum of the first n terms

The sum of the terms of a geometric progression gives a geometric series. If the starting value is \(a\) and the common ratio is \(r\) then the sum of the first \(n\) terms is
\[ \begin{aligned} S_n\ &=\ a\bigg(\ \frac{r^n\ -\ 1}{r\ -\ 1}\ \bigg)\ \ \ where\ r\ >\ 1 \\ \\ S_n\ &=\ a\bigg(\ \frac{1\ -\ r^n}{1\ -\ r}\ \bigg)\ \ \ where\ r\ <\ 1 \\ \end{aligned} \]

Exercises

1. Find the sum of the first 23 terms of the AP \(4, −3, −10,...\)
   

Solution

$$ \[\begin{aligned} a\ &=\ 4 \\ d\ &=\ -3\ -\ 4\ =\ -7 \\ S_{23}\ &=\ \frac{n}{2}(2a\ +\ (n-1)d) \\ &=\ \frac{23}{2}(2(4)\ +\ 22(-7)) \\ &=\ -1,679 \end{aligned}\]

$$

2. An arithmetic series has first term 4 and common difference \(\frac{1}{2}\). Find,

   1. the sum of the first 20 terms,

   2. the sum of the first 100 terms.

Solution

\[ \begin{aligned} a&=4 \\ d&=\frac{1}{2} \\ \\ S_{20}&=\frac{20}{2}[2a+(20-1)d] \\ &=10[2(4)+19(\frac{1}{2})] \\ &=10[\frac{35}{2}] \\ &=175 \\ \\ S_{100}&=\frac{100}{2}[2a+(100-1)d] \\ &=50[2(4)+99(\frac{1}{2})] \\ &=50[\frac{115}{2}] \\ &=2,875 \\ \end{aligned} \]

3. Find the sum of the arithmetic series with first term 1, common difference 3, and last term 100.

Solution

\[ \begin{aligned} a&=1 \\ d&=3 \\ T_{n}&=100 \\ \\ Firstly,\ find\ the\ value\ of\ n\ using\ the\ last\ term, \\ \\ a+(n-1)d&=100 \\ 1+(n-1)(3)&=100 \\ n-1&=\frac{100-1}{3} \\ n&=34 \\ \\ S_{34}&=\frac{34}{2}[a+l] \\ &=17(1+100) \\ &=1,717 \end{aligned} \]

4. The sum of the first 20 terms of an arithmetic series is identical to the sum of the first 22 terms. If the common difference is −2, find the first term.

Solution

\[ \begin{aligned} S_{20}&=S_{22} \\ d&=-2 \\ \\ 10(2a+19(-2))&=11(2a+21(-2)) \\ 20a-380&=22a-462 \\ 2a&=82 \\ a&=41 \end{aligned} \]

5. Write down the first five terms of the geometric progression which has first term 1 and common ratio 12.

Solution

\[ \begin{aligned} a&=1 \\ r&=12 \\ \\ T_1,...,T_5&=a,ar,ar^2,ar^3,ar^4, ... \\ &=1,1(12),1(12)^2,1(12)^3,1(12)^4, ... \\ &=1,12,144,1728,20736, ... \end{aligned} \]

6. Find the 10th and 20th terms of the GP with first term 3 and common ratio 2.

Solution

\[ \begin{aligned} a&=3 \\ r&=2 \\ \\ T_{10}&=3(2)^9=1,536 \\ T_{20}&=3(2)^{19}=1,572,864 \end{aligned} \]

7. Find the 7th term of the GP 2, −6, 18, …,

Solution

\[ \begin{aligned} a&=2 \\ r&=\frac{-6}{2}=\frac{18}{-6}=-3 \\ \\ T_7&=ar^6=2(-3)^6=1,458 \end{aligned} \]

8. Find the sum of the first five terms of the GP with first term 3 and common ratio 2.

Solution

\[ \begin{aligned} a&=3 \\ r&=2 \\ \\ S_5&=\frac{a(r^5-1)}{r-1} \\ &=\frac{3(2^5-1)}{2-1} \\ &=93 \end{aligned} \]

9. Find the sum of the first 20 terms of the GP with first term 3 and common ratio 1.5.

Solution

\[ \begin{aligned} a&=3 \\ r&=1.5 \\ \\ S_{20}&=\frac{a(r^{20}-1)}{r-1} \\ &=\frac{3(1.5^{20}-1)}{1.5-1} \\ &=19,945.5403804779\ @\ \frac{10457207475}{524288} \end{aligned} \]

10. The sum of the first 3 terms of a geometric series is \(\frac{37}{8}\). The sum of the first six terms is \(\frac{3367}{512}\). Find the first term and common ratio.

Solution

\[ \begin{aligned} S_3&=\frac{37}{8} \\ S_6&=\frac{3367}{512} \\ \\ Take\ the\ ratio\ of\ S_6\ over\ S_3, \\ \\ \frac{\frac{a(1-r^6)}{1-r}}{\frac{a(1-r^3)}{1-r}}&=\frac{\frac{3367}{512}}{\frac{37}{8}} \\ \frac{1-r^6}{1-r^3}&=\frac{91}{64} \\ \frac{(1-r^3)(1+r^3)}{1-r^3}&=\frac{91}{64} \\ r^3&=\frac{27}{64} \\ r^3&=(\frac{3}{4})^3 \\ r&=\frac{3}{4} \\ \\ when\ r&=\frac{3}{4} \\ \frac{a(1-(\frac{3}{4})^3)}{1-\frac{3}{4}}&=\frac{37}{8} \\ \frac{37}{16}a&=\frac{37}{8} \\ a&=2 \end{aligned} \]

11. How many terms in the GP 4, 3.6, 3.24, … are needed so that the sum exceeds 35?

Solution

\[ \begin{aligned} a&=4 \\ r&=\frac{3.6}{4}=\frac{3.24}{3.6}=0.9 \\ \\ S_n&\gt35 \\ \frac{a(1-r^n)}{1-r}&\gt35 \\ \frac{4(1-0.9^n)}{1-0.9}&\gt35 \\ 1-0.9^n&\gt\frac{3.5}{4} \\ -0.9^n&\gt0.875-1 \\ n&\gt\frac{log(0.125)}{log(0.9)} \\ n&\gt19.73644... \\ \therefore n&=20 \end{aligned} \]

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Past Semester Exam Questions

MAT112 - MAC 2015
The 4th term of a geometric sequence is -96 and the 9th term is -3. Find:
a. the common ratio and the first term
b. the sum of the first five terms.

Solution

\[ \begin{aligned} T_4&=-96 \\ T_9&=-3 \\ \\ \frac{T_9}{T_4}&=\frac{-3}{-96} \\ \frac{ar^8}{ar^3}&=\frac{-3}{-96} \\ r^5&=\frac{1}{32} \\ r^5&=(\frac{1}{2})^5 \\ r&=\frac{1}{2} \\ \\ when\ r&=\frac{1}{2} \\ a(\frac{1}{2})^3&=-96 \\ a&=-96(8)=-768 \\ \\ S_5&=\frac{a(1-r^5)}{1-r} \\ &=\frac{-768(1-(\frac{1}{2})^5)}{1-\frac{1}{2}} \\ &=\frac{-768(\frac{31}{32})}{\frac{1}{2}} \\ &=-1,488 \end{aligned} \]

SEPT 2014
a. Given a sequence: 10, 30, 90, 270, 810, 2430….., find the sum of the first 10 terms

2. The fourth term of an arithmetic sequence is less than the fifth term by 3. The seventh term is three times the fifth term. Find
   1. the common difference and the first term
      
   2. the sum of the first 12 terms

Solution

a.
\[ \begin{aligned} a&=10 \\ r&=3 \\ \\ S_{10}&=\frac{a(r^{10}-1)}{r-1} \\ &=\frac{10(3^{10}-1)}{3-1} \\ &=295,240 \end{aligned} \]

b.
\[ \begin{aligned} T_4&=T_5-3\ Eq.1 \\ T_7&=3T_5 \\ \\ d&=3\ from\ Eq.1 \\ a+6d&=3(a+4d) \\ a+6(3)&=3(a+4(3)) \\ a+18&=3a+36 \\ 2a&=-18 \\ a&=-9 \\ \\ S_{12}&=\frac{12}{2}(2a+11d) \\ &=6(2(-9)+11(3)) \\ &=90 \end{aligned} \]

MAC 2014
a. Determine the number of terms for the following sequence:

71, 65, 59, …, -13

2. The fifth term of a geometric sequence is 504 and the first term is 8064. Determine the common ratio and the sum of the first ten terms of the sequence. Assume that the common ratio is positive.

Solution

a.
\[ \begin{aligned} a&=71 \\ d&=-6 \\ T_n&=-13 \\ \\ a+(n-1)d&=-13 \\ 71+(n-1)(-6)&=-13 \\ n-1&=14 \\ n&=15 \end{aligned} \]
b.
\[ \begin{aligned} a&=8064 \\ T_5&=504 \\ \\ ar^4&=504 \\ 8064r^4&=504 \\ r^4&=\frac{504}{8064} \\ r^4&=(\frac{1}{2})^4 \\ r&=\frac{1}{2} \\ \\ S_{10}&=\frac{8064(1-(\frac{1}{2})^{10})}{1-\frac{1}{2}} \\ &=\frac{64449}{4}\ @\ 16112.25 \end{aligned} \]