Suppose that the nth term is the biggest term is n in {a}, thus
an ≤ an-1
an ≤ an+1
Namely
9n (n + 1) / 10n ≤ 9n-1 n / 10n-1
9n (n + 1) / 10n ≤ 9(n + 1)(n + 2) / 10n+1
So, 8 ≤ n ≤ 9 which means a8 and a9 are the biggest terms.
- From the first several terms we can summary a general term of the sequence. But it may not just have one probability.
Example 4
The first 4 terms are 1, 2, 4, and 8. Find a possible nth term.
Rewrite the terms as:
20, 21, 22, 23
That is: 21-1, 22-1, 23-1, 24-1
From this pattern, you can see that 2n-1 is a possible nth term for the sequence with first 4 terms 1, 2, 4, and 8.
Example 5
Given the first 4 terms of a sequence√ 2, √ 5, 2√ 2, √ 11…, find that which term has the corresponding value 2√ 5.
Rewrite the terms as:
√ 2, √ 5, √ 8, √ 11
Suppose the general term of the sequence is √ 3n -1.
an = 2√ 5 = √ 20 = √ 3 × 7 – 1
Thus, n = 7, 2√ 5 is the 7th term of the sequence.
Example 6
a) (-1) ⁿ
The terms for the question will be –1, 1, –1, 1 ……. this sequence is successively oscillating between -1 and 1. This type of sequence is called oscillate.
b) (1 / 4) ⁿ
The terms of the sequence are 1/2, 1/4, 1/8, 1/16, 1/32 …… each term is half of the immediately preceding term. As n increases, the succeeding term becomes closer to zero. The type sequence is called convergent whose nth term approaches a finite number as n approaches infinity. The convergent sequence is converging on or get closer to a number which is called the limit or limited value.
c) sin (30nº)
To calculate the first 9 terms of the sequence we can find they are1/2, –1/2, –1, –1/2, 1/2, 1, 1/2, –1/2, –1 ……... and this terms will repeat at 7th term to 12th term which the sequence repeated at 6 terms’ intervals. A sequence repeated in a set number of terms is called a periodic sequence.
- Representing a sequence graphically
It is a useful way to represent a sequence graphically. By plotting points of each term can help us understand the behavior of a sequence. These graphs shown below are the first few terms of the tree sequences discussed in example 6.
f (n) = (–1)n oscillating sequence
f (n) = (1/2)n convergent sequence
f (n) = sin (30nº) periodic sequence
A sequence is arithmetic if the difference between any two consecutive terms is the same. Thus, a sequence is arithmetic if there is a number d such that an+1-an = d for any n positive integer. The number d is the common difference of the arithmetic sequence.
a1 = a1
a2 = a1 + d
a3 = a2 + d = a1 + 2d
a4 = a3 + d = a2 + 2d = a1 + 3d
……
Therefore, the general term for any arithmetic series is: an = a1 + (n – 1) d
Example 7
An arithmetic series has first term 23 and common difference – 5. Find the second, third and fourth terms of the series and the nth term.
In this series, the first term is 23, the common difference – 5 are given. Using the formula an = a1 + (n – 1) × d
Second term = 23 − (2 – 2) × 5 = 18
Third term = 23 – (3 – 1) × 5 = 13
Forth term = 23 – (4 – 1) × 5 = 8
nth term = 23 – (n – 1) × 5
If you write down the sum of the first n terms of the sequence you obtain:
a1 + a2 + a3 +……+ an
This is a finite series of n terms. It is often written as:
The sign ∑ is a Greek capital letter S stands for sum. The sign shown above means the sum of the terms obtained by substituting 1, 2, 3 ……, n in turn for r in ar. The number below and above it show the lower and upper limits between which the variable is being summed.
If the terms of a sequence continue infinitely without stopping
a1 + a2 + a3 +……+ an + ……
This is an infinite series which is written as:
Example 8
Write down the first 4 terms and the nth term of the series
Taking r = 1, 2, 3, and 4 in 3r + 2 gives: (3 + 2), (6 + 2), (9 + 2), and (12 + 2) as the first four terms. That is 5, 8, 11, and 14 are the first four terms and the nth term of the series is 3r + 2 and each term are 3 to the previous term.
Example 9
Find the common difference and the 10th term of the arithmetic series given by
Taking r = 1, 2, and 3 in turn. The first three terms of the series are:
(7 – 13), (14 – 13) and (21 – 13)
That is: –6, 1, and 8
The common difference is:
2nd term – 1st term = 1 – (–6)
= 1 + 6
= 7
The 10th term is found by taking r = 10 and it is
a10 = 7 × 10 – 13 = 57
- The sum of an arithmetic series
To find the sum Sn of the first n terms of the general arithmetic series can use the method just shown follow:
Sn = a + (a + d) + (a + 2d) + …… + (L – d) + L
where L is the nth term (L for ‘last’).
Rewriting the series in the reverse order gives:
Sn = L + (L + d) + …… + (a + d) + a
Add these two together to get:
2Sn = (a + L) + (a + L) + ……+ (a + L)
That is,
2Sn = n(a + L) and Sn = n(a + L) / 2
Now L = a + (n + 1)d because L is the nth term of the series.
Substituting for L in the expression for Sn gives:
Sn = n [2a + (n – 1) d] / 2
Or
Sn = na1 + n (n – 1) d / 2
This formula is suitable for any arithmetic series. The sum of a arithmetic series to n terms can be found by knowing the first term, the n terms and the common difference of the series.
Example 10
Find the sum of the first 20 terms of the arithmetic series 1 + 4 + 7 + 10 + 13 ……
Use the formula Sn = n [2a + (n – 1) d] / 2, with n = 20, a = 1, and d = −3
This gives:
S20 = 20 [2 + (20 – 1) (− 3)] / 2
= 20 ( 2 – 57) / 2
= – 110 / 2
= – 55
The sum of the first 22 terms is – 55.
Example 11
The sum of the first n terms of a series is n²+2n
- Find the nth term.
- Show the series of n term and find the common different.
-
the nth term series = Sn-Sn-1
= n²+2n-(n-1)²-2(n-1)
= n²+2n- n² +2n - 2n + 3
= 2n + 3
- Substituting n=1,2,3… in turn in the nth term, 2n + 3gives the series
+ 5, + 7, + 9…
The common different of this arithmetic is 2.
Example 12
Given that in an arithmetic series, a4 = 9, a9 = -6. Find the value of n in Sn = 54.
Method 1
a4 = a1 + 3d = 9
a9 = a1 + 8d = -6
Solve them and get a1 = 18, d = -3
Using the formula Sn = na1 + n (n – 1) d / 2
18n + n (n – 1) (– 3) / 2 = 54
n = 4 or n = 9
Method 2
a9 – a4 = 5d = –15
d = – 3
a1 = a4 – 3d = 18
So, Sn = (d/2) n2 + (a1 – d/2) n = 54
(– 3/2) n2 + (18 + 3/2) n – 54 = 0
n = 4 or n = 9
-
In arithmetic series there are 5 components: first term a1, nth term an, n, common difference d, and the sum of the series.
- Sum of the first n natural numbers
Natural number is an arithmetic series which has the first term 1 and common difference 1. This set of number can be written as
1 + 2 + 3 + 4 + …… + n
for the first n natural numbers.
The series can also be written as
Using the formula Sn = n [2a + (n – 1) d] / 2
= n (n + 1) / 2
The sum of the first n natural numbers is n (n + 1) / 2
Example 13
Find the sum of all those integers between 1 and 100, which are multiple of 3.
Here you require the sum S of the series
S=3 + 6 + 9 +…+ 99
S=3[1 + 2 + 3 +…+ 33]
=3 × 33 × 34 ÷2
=1683
Notice that we could also use
S= 3 +6 +9+…+ 99 and use the formula
Sn = n [2a+ (n-1) d] / 2
With a=3, d=3 and n=33 to give
S33 =33[2(3) + 32(3)] / 2 =1683
As the same result before we got.
A sequence is geometric if the ratio of any two consecutive terms is the same. Thus a sequence is geometric if there is a number r different from 0, such that an / an-1 = r
for any n positive integer greater than 1. The number r is the common ratio of the geometric sequence.
a1 = a1
a2 = a1d
a3 = a2d = a1d2
a4 = a3d = a2d2 = a1d3
……
Therefore the general term for any geometric series is an = a1dn-1
Example 14
Find the common ratio and the nth term of the geometric series
1 + 5 + 25 + ……
To find the common ratio we use the second term divide by first term
5 / 1= 5
So the common ratio is 5.
The series can be written as
1 + 5 + 5² + ……
The nth term is 5ⁿ־¹
The nth term of the series a + ar + ar² +… is arⁿ־¹
- The sum of a geometric series
The sum of a geometric series can be written:
S = a + ar + ar² +… + arⁿ־¹
Multiplying by r gives r S = ar+ ar²+… + arⁿ ־¹+ arⁿ
Subtracting the expression rS from the expression for S gives:
S – r S = a – a rⁿ
That is S (1 – r) = a (1 –rⁿ)
Dividing by 1-r gives:
S= a (1- rⁿ) / (1-r)
The above is the standard formula for the sum of the first nth terms of a geometric series. It can be also written as:
S=a (rⁿ-1) / (r-1)
Example 14
Find the sum of the first 10 terms of a geometric series 2 + 6 + 18 + 54 + ……
In this series, a1 = 2, r = 6 / 2 = 3, n = 10
Using the formula S= a (1- rⁿ) / (1-r)
S10 = 2 (1- 310) / (1-3) = 59048
So the sum of the first 10 terms of the series is 59048.
- The sum to infinity of a convergent geometric series
As we discussed the formula of the sum of geometric series is: a (1- rⁿ) / (1-r)
a (1- rⁿ) / (1-r) = a / (1 – r) – arn / (1 – r)
If r lies between -1 and 0 or between 0 and 1, then as n increases rⁿ gets smaller and smaller. For example, (1/2)²=1/4, (1/2)³= 1/8… this means that you can make the term arⁿ / (1-r) as small as you like provided you take a large enough value of n. as n approaches infinity so arⁿ(1-r) approaches zero. In other words the values approaches infinity of arⁿ (1-r) from a decreasing sequence, which has the limiting values zero as n, approach infinity.
The two statement -1< r < 0 and 0 < r < 1, can be combined by writing ⏐r⏐< 1. This reads ‘mod r is less than 1’. mod r, written ⏐r⏐, means the absolute value of r, or the actual numerical value of r. for example, if r = 4, ⏐r⏐ =4 if r = - 4 then ⏐r⏐= 4 as well.
If ⏐r⏐<1, the sum to n terms of the geometrical series is
S= a (1- rⁿ) / (1-r)
This converges to the value a / (1 – r) as n approaches infinity.
So the series has a sum to infinity of a/ (1-r), provided ⏐r⏐ <1.
Example 15
For a geometric series with first term 7 and common ration 0.5. Find the 5th term, the sum of the first 10 terms and the sum to infinity. Give answer to 3 significant figures where necessary.
a = 7, r = 0.5
5th term is: ar4 = 7 (0.5)4 = 0.438 (3 s. f.)
The sum to 10 terms is: 7 (1 – 0.510) / 1 – 0.5 = 13.986 (3 s. f.)
The sum to infinity is: 7 / (1 – 0.5) = 14
Example 16
Given that ⏐p⏐< 1, ⏐q⏐< 1, calculate the sum of
1 + (1 + p) q + (1 + p + p2) q2 + ……+ (1 + p + p2 + ……+ pn) q n+….
The nth term of this series is an = (1 – pn+1) / (1 – p)
Because ⏐p⏐< 1, ⏐q⏐< 1,
So ⏐p q⏐< 1,
So:
S = 1 + lim n→∞ (a1 + a2 + a3 +……+ an)
= 1 + [1 / (1 – p) ] × [q / (1 – q) – p2q / (1 – p q)]
= 1 / (1 – q) (1 – p q)
- Further examples on arithmetic and geometric series
Example 17
Given an arithmetic series which has the sum of first n terms sum 25, the sum of first 2n terms are 100, find its sum of first 3n terms.
Suppose the first term of the series is a1, common difference is d.
[2a1 + (n – 1) d ] n / 2 = 25 (1)
[2a1 + (2n – 1) d ] 2n / 2 = 100 (2)
(2) – 2 (1) dn2 = 50
From (2): [2a1 + (2n – 1) d] n / 2 = 50,
So S3n = [2a1 + (3n – 1) d] 3n / 2
= 3 (50 + 50 / 2)
= 225
Example 18
Known that f (x) is a linear vector function, f (10) = 21, and f (2), f (7), f (22) are geometric series, find f (1) + f (2) + f (3) +……+ f (n)
Suppose that f (x) = k x + b (k is not equal to 0)
f (10) = 10 k + b = 21 (1)
(7k + b)2 = (2k + b) ( 22k + b) (2)
From (2): 5k2 = 10kb ⇒ k = 26 (3)
Substitute k = 26 into (1)
f (x) = 2x + 1
So
f (1) + f (2) + f (3) + …… + f (n)
= 3 + 5 + 7+……+ (2n + 1)
= [3 + (2n + 1) ] / 2
= n2 + 2n
Example 19
A factory has the production values last year is 1 million pounds, planning the production values increase 10% than the previous year in the next 3 years. Find that from this year, the production values of the third year. The total production values in the 3 years.
Suppose the last year’s production value is a1, this year’s is a2 and so on.
This is a geometric sequence a1 = 100, q = 1 + 10% = 1.1
So a4 = a1 q3 = 100 × 1.13 = 133.1 million
The total production values of these 3 years are
S4 – a1 = a1 (q4 – 1) / (q – 1) – a1
= 100 (1.14 – 1) / (1.1 – 1) – 100
= 364 million
Thus the third year’s production value is 133.1 million and the total values of these 3 years are 364 million.
Having seen the sequences and series, there are 6 points I can make in the list follow:
-
The nth term of the arithmetic series an = a1 + (n – 1) d
-
The sum of the arithmetic series Sn = n [2a + (n – 1) d] / 2
- The sum of the first n natural numbers n (n + 1) / 2
-
The nth term of the geometric series an = a1rn-1
- The sum of the finite geometric series S=a (rⁿ-1) / (r-1)
-
If ⏐r⏐• 1, then the infinite geometric series has sum a / (1 – r)