1.1 Introduction to Series & Sequences
In Mathematical Analysis, a sequence is defined to be an ordered collection of terms/objects. This colelction or list can either be finite, or infinite. Thus we have both finite & infinite sequences.
Often we define sequences by a recursive formula with some base conditions. Thus let’s consider \(a_n=a_{n-1}+a_{n-2}\) with \(n \ge 3\) is the recursive formula that defines a sequence, with base conditions \(a_1=a_2=1\). Thus we can derive the terms of the seuence starting from the base condition. The sequence that we spoke of is the famous Fibonacci Sequence. So \(a_3=a_2+a_1=2\), \(a_4=a_3+a_2=2+1=3\), \(a_5=5\) and so on and so forth.
A series on the other hand is some mathematical operation like sum or product applied on the terms of the sequence. Thus \(1+2+3+…\) is a series. We’ll cover a few of the famous series’ in our next modules.
Progressions, is a sequence which is bounded by a strict or direct mathematical relation. So, given a positive integer \(n\), we should be able to directly calculate the nth term of the progression. We’ll cover the aspect of deriving this direct formula for sequences like the Fibonacci Sequence in later parts. For now, let’s get this clear – all prime numbers less than 100000 is a sequence (since there is no fixed mathematical formula to derive primes – in short, given \(n\), we cannot calculate directly the nth prime number). All numbers of the form \(2n\), with \(n \le 50000\) is a progression.
We’ll deal primarily with 3 types of progressions – Arithmetic Progression (AP), Geomteric Progression (GP) & Harmonic Progression (HP).
2.1 Arithmetic Progressions
A sequence \(A=\{a_1,a_2, a_3,…\}\) is said to be in AP iff the difference between any of its successive terms is a constant. So we can say for an AP \(a_2-a_1=a_3-a_2=a_4-a_3..\). We call this difference as the common difference \(d\). Thus, we can say, \(a_2=a_1+d\), \(a_3=a_2+d=a_1+2d\). In general we can say \(a_n=a_1+(n-1)d\). Thus given any positive integer \(n\), we are able to directly calculate the nth term of this sequence, and hence we call this a progression.
For 3 terms of an AP \(a,b,c\), we define \(b\) as the Arithmetic Mean(AM) of \(a\) & \(c\). Also we must note that \(b-a=c-b\thinspace \Rightarrow b=\frac{a+c}{2} \).
Ex-1
Insert 3 Arithmetic Means between \(1\) & \(10\)
Solution:
If we insert 3 Arithmetic Means, let’s say \(a_1,a_2,a_3\) between \(1\) & \(10\), then we have \(1,a_1,a_2,a_3,10\) in AP.
Thus \(10\) is the 5th term of this AP with first term \(1\). So \(10=1+4d\thinspace \Rightarrow d=2.25\). So we have the AP as \(1,3.25, 5.50, 7.75,10\). The means are \(3.25,5.5,7.75\).
We shall now explore the sum of an AP of n terms.
\(S=a+(a+d)+(a+2d)+…+a+(n-1)d=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+a+(n-1)d]=\frac{n}{2}(a+l)\), where \(a\) is the first term of an AP and \(l\) is the nth term of the sequence.
Key Points for an AP
1. If \(a,b,c\) are in AP
- \(ak,bk,ck\) with \(k \ne 0\) are in AP
- \(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\) are in AP with \(k \ne 0\)
- \(a \pm k, b\pm k, c\pm k\) are in AP
2. If we have to deal with 3 terms of an AP, the best substitution is \(a_1=a-d\), \(a_2=a\) & \(a_3=a+d\)
For four terms of an AP : \(a – 3d, a – d, a + d, a + 3d\).
3. If \(a_1,a_2,a_3..a_n\) be an AP, then \(a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=..=a_r+a_{n-r+1}\). (Try proving this)
4. AM of \(a_1,a_2,…a_n\) is defined as \(\frac{a_1+a_2+..a_n}{n}\)
Test Your Concepts:
- Show that the sum of any 2 terms of an AP equidistant from the beginning and the end is a constant
- Show that the sum of \(n\) Arithmetic Means between 2 numbers is \(n\) times the single AM between them
2.2 Geometric Progressions
We say three numbers \(a,b,c\) are in GP iff \(\frac{b}{a}=\frac{c}{b}\thinspace \iff b^2=ac\). Needless to say if one number of a GP is \(0\), then all numbers of the GP are \(0\). We define this ratio between 2 successive terms of a GP to be the common ratio \(r\).
We can define the terms of the GP as \(a,ar,ar^2,…ar^{n-1}\). Thus the nth term of a GP with first term \(a\) & common ratio \(r\) is \(ar^{n-1}\)
To find the sum of a GP of \(n\) terms we may need to evaluate the expression \(S_n=a+ax+ax^2+…ax^{n-1}\).
For this, let’s use our knowledge of secondary school Polynomial Division. Any Polynomial of the form \(x^n-1\) when divided by \(x-1\) gives the quotient as \(x^{n-1}+x^{n-2}+..x^2+x+1\). Thus we know \(1+x+x^2+..x^{n-1}=\frac{x^n-1}{x-1}\). Trivially, we can thus obtain from here
Geometric mean of 2 numbers \(a,c\) is \(\sqrt{ac}\). This comes from the fact that:
If \(b\) be the GM of \(a,c\), we must have \(a,b,c\) in GP. Thus, from what we studied above, \(b^2=ac \thinspace \Rightarrow b=\sqrt{ac}\). And similarly, we can say that for \(n\) terms \(a_1,a_2,…a_n\), the GM is \(\sqrt[n]{\prod a_i}\).
Ex-2
Insert 2 Geometric Means between \(1\) & \(27\).
Solution:
Let the Geometric Means be \(a,b\), thus we have \(1,a,b,27\) in GP. Let the common ratio of the GP be \(r\), thus the fourth term of this sequence is \(27\), which means \(r^{4-1}=27 \thinspace \Rightarrow r=3\). Thus the geometric means are \(3,9\), which gives the GP as \(1,3,9,27\) with common ratio \(3\).
Key Points of a GP:
1. If \(a,b,c\) are in GP
- \(ak,bk,ck\) with \(k \ne 0\) are in GP
- \(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\) are in GP with \(k \ne 0\)
2. For 3 terms of a GP, we can take them as \(\frac{a}{r}\), \(a\), \(ar\).
For four terms : \(ar^3,ar,\frac{a}{r}, \frac{a}{r^3}\)
3. If \(x_1,x_2,…x_n\) is a GP and \(y_1,y_2,y_3,…y_n\) is also a GP then \(x_1y_1,x_2y_2,…x_ny_n\) is a GP.
Test Your Concepts
If \(n\) GMs are inserted between 2 numbers then show that the product of the \(n\) GMs is the nth power of the single GM between the 2 numbers.
2.3 An Infinite Arithmetic & Geometric Series
In our earlier paragraphs, we had assumed the series’ to be finite. For an infinite AP it becomes difficult to accurately predict the sum when the number of terms \(n\) tends to infinity. The sum of a GP, however, converges when \(n\) tends to infinity under the condition that the common ratio \(r\) should satisfy \(|r|<1\).
We have \(S_n=\frac{a(1-r^n}{1-r}\), when \(|r|<1\) and \(n \rightarrow \infty\), \(r^n \rightarrow 0\), thus the sum of an infinite GP is
Ex-3
Find the number of tems \(n\) of a GP if \(a_1=3\), \(a_n=96\) & \(S_n=189\).
Solution:
We have \(a_1r^{n-1}=96\thinspace \Rightarrow r^{n-1}=32\).
Also \(S_n=\frac{3(r^n-1)}{r-1}=189\thinspace\Rightarrow \frac{32r-1}{r-1}=63\thinspace \Rightarrow r=2\). That gives \(2^{n-1}=32\thinspace \Rightarrow n=6\).
2.4 Harmonic Prgressions
The sequence \(a_1,a_2,…a_n\) is in Harmonic Progression iff \(\frac{1}{a_1},\frac{1}{a_2},.., \frac{1}{a_n}\) is in Arithmetic Progression. Thus if \(a,b,c\) is in HP, we have
\(\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\thinspace \Rightarrow b=\frac{2ac}{a+c}\). In this case \(b\) is the harmonic mean of \(a\) & \(c\).
In general the harmonic mean of a harmonic progression \(a_1,a_2,..a_n\) is given by
Thus, if we were to insert harmonic means between 2 numbers, we would insert arithmetic means between their inverses.
Thus, if we were to insert 2 HMs between \(1\) & \(4\), it’d be equivalent to insert 2 AMs between \(\frac{1}{4}\) & \(1\). Thus the HMs would be \(\frac{4}{3}\) & \(2\).
Relationship between AM-GM-HM
Try proving the following theorems:
If A,G,H be the AM, GM, HM of 2 numbers \(a,b\), then :
- \(A \ge G \ge H\)
- \(A,G,H\) form a GP, thus \(G^2=AH\)
- \(x^2-2Ax+G^2=0\) has roots \(a,b\).
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