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In mathematics, a study of pattern's leads to significant generalization. A succession of numbers of which one number is designated as the first, another as the second, another as third and so on gives rise to what we call a sequence.The idea of a sequence originates in the process of counting in a very natural way. One of the conventional ways of representing the succession of counting numbers is to list the first few numbers separated by commas and then write dots at the end to indicate the remaining numbers. thus we can have the symbolic representation
1,2,3,4,5,6,…………..
in which the numbers are in given order: 1 is followed by 2, 2 by 3, 3 by 4 and so on. Hence , A sequence is an ordered set of numbers.In this chapter , we shall study the particular type of sequences called arithmetic sequences, geometric sequences and harmonic sequences and corresponding series.
Dipa deposited Rs.1000 in a bank at 10% interest per year compounded annually for 12 years. the amount at the end of first, second, third,……., twelfth year in rupees are 1100, 1210, 1331, ……., 3138.43. these amounts form what we call a sequence.
So from above example , we can see that the numbers are arranged in a definite order with a definite rule. Such patterns of numbers following certain rules is known as sequences. Each element in a sequence is called a term of a sequence. the terms of sequences are denoted by t_{1}, t_{2} , t_{3},….t_{n}, where t_{1} is the first term and t_{n}is the last term of the sequence.
Examples;
Let t_{1}, t_{2}, t_{3},…..,t_{n}…, be a given sequence. Then, the expressions t_{1}+t_{2}+t_{3}+…+t_{n}+… is called the series associated with the given sequence. If the terms of a sequence are connected with each other by a plus or minus sign, then the resulting expression is called a series. Thus, a series Is a well-ordered sum of a sequence .
The expression 2+4+6+8+10 is a finite series and 1-2+3-4+….. is an infinite series.The series is finite or infinite according to as given sequence is finite or infinite.So if the terms of a sequence are separated by using algebraic (+ or -) their it is called the series.
Some of the examples are as follows :
If a sequence of numbers is such that each term can be obtained from the preceding one by the operation of some law, the sequence is called a progression.
Some of the examples of progression are as follows;
There are three types of progression and they are listed below;
An arithmetic sequence has the property that the difference between two successive terms (any term subtracted from the following term) is always the same. This difference is called the common difference. Let us consider the following sequences;
In each of these above sequences, we note that each term except the first progresses in a definite manner.
How do these terms progress?
In (i) we get,
t_{1} = 2
t_{2} = t_{1}+3
t_{3} = t_{2}+3
t_{4} = t_{3}+3
Similarly , in (ii) we get;
t_{1} = 16
t_{2} = t_{1}+(-5)
t_{3} = t_{2}+(-5)
t_{4} = t_{3}+(-5)
Finally in (iii)
t_{1} = x-3b
t_{2} = t_{1}+4b
t_{3} = t_{2}+4b
t_{4} = t_{3}+4b
It is observed that in each term except the first is obtained by adding a fixed number (positive or negative) to the preceding term. In (i), fixed number is 3; in (ii) the fixed number is -5 and in (iii) the fixed number is 4b.
Such sequences are called arithmetic sequence or arithmetic progression abbreviated as A.P.
The sequence 2,5,8,11…. is an infinite progression (A.P) while 2+5+8+11+…. is an infinite arithmetics series, whose first term is 2 and the common difference is 3. IS one of the examples is arithmetic progression.
The nth term or general term of an A.P.
Let us consider an A.P. with first term 'a' and the common difference 'd' , i.e a, a+d, a+2d, a+3d….
then,
First term = t_{1} = a+ (1-1)d
second term = t_{2} = a+d = a+ (2-1)d
third term = t_{3} = a+2d = a+ (3-1)d
forth term = t_{4} = a+3d = a + (4-1)d
n^{th} term = t_{n} = a + (n-1)d
The sum to n terms of an A.P.
Let the term of an A.P. be 'a' and the common difference be 'd'. Let us denoted by Sn , the sum to n terms of the A.P. Then,
Sn = a + (a+d) + (a-2d) + ……,…… + [ a+ (n-2) d] + [a+ (n-1)d]…… (i)
Writing the term in reverse order we have
S_{n} = [a+(n-1)d] + [a+(n-2)d] + [a+(n-3)d] +……+ (a+d) +a ……. (ii)
Adding (i) and (ii) term be term,
We have;
2S_{n} = [2a+(n-1)d] + [2a+(n-1)d] +……+ [2a+ (n-1)d] + [2a+(n-1)d] in which [2a+(n-1)d] appear as a term n times. Thus,
2S_{n} = n[2a+(n-1)d]
\(\therefore\) S_{n} = \(\frac{n}{2}\) [ a+ {a+ (n-1)d]
=\(\frac{n}{2}\)[a+l].
Example
Find the sum of the series 19+17+15+….. to 15 terms.
solution:
Here, First term (a) = 19
Common difference (d) = -2
Number of terms (n) = 15
We know that,
S_{n} = \(\frac{n}{2}\) [2a + (n-1)d]
=\(\frac{15}{2}\) [ 2×19 + (15-1)(-2)]
= \(\frac{15}{2}\) [38-28]
=\(\frac{15}{2}\)(10)
=75. ans
Arithmetic Mean (A.M.)
When three numbers a, A and b are in A.P. then A is called the arithmetic mean of numbers a and b.
Given that a, A , b is an A.P. Then,
A-a = b-A
i.e. A= \(\frac{a+b}{2}\)
Thus , the required A.M. of two numbers a and b is
\(\frac{a+b}{2}\)
Example:
Find the A.M. between 15 and 31.
Here, let A be A.M. between 15 and 31,
where, a=15 , b=31 and A =?
We know that,
A= \(\frac{a+b}{2}\)
= \(\frac{15+31}{2}\)
= 23.
A geometric sequence has the property that the ratio of any term ( except the first) to its preceding term is always the same. this ratio is called the common ratio and is denoted by r,. We follow the same notations as used in the case of an arithmetic sequence and series.
Let us consider the following sequences.
In each of these sequences , we note that each term , except the first term , progresses in a definite order. How do these terms progress ?
In (1) we have,
t_{1} = 2 , \(\frac{t_2}{t_1}\) =2, \(\frac{t_3}{t_2}\)= 2, \(\frac{t_4}{t_3}\)= 2 and so on.
In (2) we have;
t_{1}= \(\frac{1}{9}\), \(\frac{t_2}{t_1}\) = \(\frac{-1}{3}\) , \(\frac{t_3}{t_2}\) = \(\frac{-1}{3}\) , \(\frac{t_4}{t_3}\) = \(\frac{-1}{3}\) and so on.
In (3) we have ;
t_{1} = 0.01 , \(\frac{t_2}{t_1}\) = (0.0001/ 0.01)=0.01 , \(\frac{t_3}{t_2}\) = (0.00001/0.0001)= 0.01 and so on.
It is observed that in each case , every term except the first term bears a constant ratio to the term immediately preceding it. In (1) , this constant ratio is 2 , in (2) it is -1/3 and in (3) , the constant ratio is 0.01. Such sequence are called geometric sequence or geometric progression abbreviated as G.P.
n^{th} term in G.P.
Let us consider a G.P. with first non-zero term 'a' and common ratio 'r'. Write a few term of it. The 2^{nd} term is obtained by multiplying a by r, thus t_{2}= ar. Similarly, thirs term is obtained by multiplying t_{2} by r. Thus, t_{3}=t_{2,}r=ar^{2} , and so on. We write below these a few more terms.
First term = t_{1}= a= a×r^{(1-1)}
second term = t_{2} = ar = a×r^{(2-1)}
third term = t_{3}= a×r^{2}= a×r^{(3-1)}
Therefore, the pattern suggests that n^{th} term of G.P. is given by;
tn=a×r(n-1)
EXAMPLE
Find the 10^{th} term of the series;
4-8+16-32+…….
solution:
Here, First term (a) = 4
Common ratio (r)= -2
Now,
10^{th} term (t_{10}) = a×r^{(10-1)}
=4×(-2)^{9}
= -2048.
A set of numbers t_{1}, t_{2}, t_{3}…. t_{n},…. is said to be in H.P. when their reciprocals \(\frac{1}{t_1}\) , \(\frac{1}{t_2}\), \(\frac{1}{t_3}\),……. \(\frac{1}{t_n}\) ,….. are in A.P. and vice-versa. For example;
\(\frac{1}{2}\), \(\frac{1}{4}\) , \(\frac{1}{6}\) … is H.P. since,
2,4,6,…. is A.P.
EXAMPLE;
Find the 20^{th} and nth terms of G.P.
\(\frac{5}{2}\),\(\frac{5}{4}\),\(\frac{5}{8}\)……
Solution:
Here, First term (a) = \(\frac{5}{2}\)
Common ratio (r) = \(\frac{1}{2}\)
Now, 20^{th} term (t_{20}) = a×r^{(20-1)}
=(\(\frac{5}{2}\)) (\(\frac{1}{2}\))^{19} = \(\frac{5}{20}\)^{20}
and nth terms (tn) = a×r (n-1)
= (\(\frac{5}{2}\)) (\(\frac{1}{2}\))^{(n-1)}
=(\(\frac{5}{2}\))^{n}.
References:
(Joshi, Adhikari and Aryal)
Joshi, Amba Datt, et al. "Measures of Dispersion." Business Maths. Kathmandu: Dreamland Publication, 2013 AD. 311 - 316.
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