JEE Mathematics Notes: Sequence and Series

JEE Mathematics Notes on Sequences and Series

Sequence

A sequence is a set of values which are in a particular order.

Mathematically, a sequence is defined as a map whose domain is the set of natural numbers (which may be finite or infinite) and the range may be the set of real numbers or complex numbers. We can represent it as

f: N → X, where N is the set of naturals. The sequence is finite or infinite depending on whether the set ‘N’ is finite or infinite. If ‘X’ is the set of real numbers then f is said to be a real valued sequence while if the set X is of complex numbers, then f is termed as a sequence of complex numbers.

Sequence

Difference between sequence and set….

A sequence is often confused with a set. Though they both appear to be same yet they are different. A sequence is almost the same as a set except for the fact that in a set the elements cannot repeat while it is not so in case of a sequence. Moreover, there is no importance of order in a set, while the order matters a lot in sequence. The following example will further clear the difference between the two:

{2, 4, 2, 4, ….} is an alternating sequence of 2s and 4s.

The corresponding set would be just {2, 4}.

Arithmetic Progression

A sequence is said to be in Arithmetic Progression when they increase or decrease by a constant number. This constant number is called the common difference (d) of the arithmetic progression.

1, 5, 9, 13, ……………             d = 4

–7, –2, 3, 8, 13 ………             d = 5

8, 5, 2, –1, –4 ………             d = –3

nth term of an A.P. will be given by

tn = a + (n–1)d.

Sum of the first n term of an A.P.

Let sum be Sn

Sn = {a} + {a + d} + {a + 2d} +…+ {a + (n–1)d}

we can write the above series in reverse way also.

Sn = {a+(n–1)d} + {a+(n–2)d} + {a+(n–3)d} +……+ {a}

Adding (1) and (2)

2Sn = {2a(n–1)d} + {2a(n–1)d} + (2a+(n–1)d}+…+ {2a+(n–1)d}

or 2 Sn = n {2a+(n–1)d}

Sn =  {2a + (n–1)d}

or,      Sn = first term+last term/2 × number of terms

Arithmetic Mean

When three numbers are in A.P., the middle term is called arithmetic mean (AM) of the two others.

A.M. = Sum of the numbers/2

Similarly we can find the two arithmetic means between two number. A1 and A2 are two arithmetic mean between two member a and ‘b’ then a, A1, A2 b are in A.P.

Now using the formula,

tn = a+(n–1)d

or,    d = b–a/3 [ here n = 4]

so,    A1 = a + d = a +b–a/3 = 2a+b/3

and   A2 = a + 2d = a + 2(b–a)/3 = a+2b/3

Arithmetic Mean of mth Power

The concept of arithmetic mean has been already discussed in the previous sections. This concept can be extended till the mth power for terms and such mean is termed as the arithmetic mean of mth power. Mathematically, the arithmetic mean for mth power is defined as

If a1, a2, …, an be n positive real numbers (not all equal) and let m be a real number.

  • However if m ∈ (0, 1), then a1+ a2m+…+ anm/n < (a1+a2+…+an/n)m.
  • Obviously if m ∈ {0, 1}, then a1+ a2m+…+anm/n = (a1+a2+…+an/n)m.

Illustration:

If a, b, c are positive real numbers such that a + b + c = 1, then prove that a/b + c + b/c+a + c/a+b > 3/2.

Solution: We have to show that (a/b+c+1)+(b/c+a + 1)+(c/a+b + 1) > 3/2 + 3.

i.e. 1/b+c + 1/c+1 + 1/a+b > 9/2.

Now A.M. of mth power > mth power arithmetic mean (m = – 1 here)

⇒ 1/3 [(b+c)–1 + (c+a)–1 + (a+b)–1] > [(b+c) + (c+a) + (a+b)/3]–1

⇒ 1/b+c + 1/c+a + 1/a+b > 3.3/2(a+b+c) = 9/2.

Geometric Progression

A geometric progression may be defined in the form

a, ar, ar2, ar3, ……, a rn–1.

n-th term of GP will be given by
tn = arn–1
The value of the sum of n terms in a G.P depends on the value of r. We have different values according to whether r is equal to 1 or is not equal to 1.

Sn = a (rn – 1)/ (r-1), when r ≠1

     = na                    ,  if r =1

If -1 < x < 1, then lim xn = 0, as n →∞.

Hence, the sum of an infinite G.P. is

1+x+x2+…. = 1/ (1-x).

 Geometric mean between two numbers

 Geometric mean of a,b
        G =√ab
Two geometric means between two given numbers a and b.
Let a, G1, G2, b are in G.P.
tn = a rn–1
or b/a = r3
r =(b/a)1/3
G1 = ar2 = a (b/a)1/3 = a1/3 b2/3

Harmonic Progression

The simplest way to define a harmonic progression is that if the inverse of a sequence follows the rule of an arithmetic progression then it is said to be in harmonic progression. This simply means that if a, a+d, a+2d, …..   is an A.P. then 1/a, 1/(a+d), 1/(a+2d), …… is an H.P.

For example, the series 1 +1/4 +1/7 +1/10 +….. is an example of harmonic progression, since the series obtained by taking reciprocals of its corresponding terms i.e. 1 +4 +7 +10 +… is an A.P.

Harmonic Mean

H = 2/ (1/a + 1/b)

or H = 2ab/(a+b) 

The harmonic mean of n positive numbers x1, x2, …. , xn is defined to be

The harmonic mean of two numbers is in fact the reciprocal of arithmetic mean of the reciprocal of the numbers. This simply means that if H is the harmonic mean between two numbers say a and b then 1/a, 1/H and 1/b are in A.P.

Formula for Harmonic mean

Let H be the harmonic mean between two numbers a and b.

So, a, H, b are in H.P.

This means that 1/a, 1/H, 1/b are in A.P.

or, 1/H – 1/a = 1/b – 1/H.

or, 2/H = 1/a + 1/b

= a+b/ab

∴     H = 2ab/a+b

On the same lines, we may also find two harmonic means between two numbers.

Let us assume that Ha and Hb are two harmonic means between a and b.

Then it follows that a, Ha, Hand b are in H.P.

Then 1/a, 1/Ha, 1/Hb, 1/b are in A.P.

Hence, using the formula we have

tn = a + (n-1)d

So, 1/b – 1/a = 3d, where‘d’ denotes the common difference of the A.P.

This implies that 3d = (a-b)/b

Hence, d = (a-b)/3ab

So, 1/H1 = 1/a + d = 1/a + a–b/3ab = a+2b/3ab

and 1/H2 = 1/a + 2d = 1/a + 2(a–b)/3ab = 2a+2b/3ab

Hence, these are the two harmonic means between a and b.

Weighted Harmonic Mean

Another concept closely related to harmonic mean is that of weighted harmonic mean. If we have a set of weights w1, w2, …. , wn associated with the set of values x1, x2, …. , xn, then the weighted harmonic mean is defined as

Harmonic mean is in fact a special case of weighted harmonic mean where all the weights are equal to 1 and is equal to any weighted harmonic mean having all equal weights.

Some Important Remarks:

• If a and b are two non-zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P. We have H = 1/H = 1/2 (1/a + 1/b) ⇒ H = 2ab/a+b.

• If a1, a2, ……, an are n non-zero numbers. then the harmonic mean H of these number is given by 1/H = 1/n (1/a1 + 1/a2 +…+ 1/an).

• The n numbers H1, H2, ……, Hn are said to be harmonic means between a and b, if a, H1, H2 ……, Hn, b are in H.P. i.e. if 1/a, 1/H1, 1/H2, …, 1/Hn, 1/b are in A.P. Let d be the common difference of the A.P., Then 1/b = 1/a + (n+1) d ⇒ d = a–b/(n+1)ab.

Thus 1/H1 = 1/a + a–b/(n+1)ab,

1/H2 = 1/a + 2(a–n)/(n+1)ab,

………..     ……….

1/Hn = 1/a + n(a–b)/(n+1)ab.

Relation between A.M., G.M. and H.M.

AM > GM > HM

Method of Differences

Suppose a1, a2, a3, …… is a sequence such that the sequence a2 – a1, 
a3 – a3, … is either an. A.P. or a G.P. The nth term, of this sequence is obtained as follows:

S = a1 + a2 + a3 +…+ an–1 + an

S = a1 + a2 +…+ an–2 + an–1 + an

Subtracting (2) from (1), we get, an = a1 + [(a2–a1) + (a3–a2)+…+(an–an–1)].

Since the terms within the brackets are either in an A.P. or a G.P., we can find the value of an, the nth term. We can now find the sum of the n terms of the sequence as S = Σnk=1 ak.

F corresponding to the sequence a1, a2, a3, ……, an, there exists a sequence b0, b1, b2, ……, bn such that ak = bk – bk–1, then sum of n terms of the sequence a1, a2, ……, an is bn – b0.

Illustration:

 Find the sum of 1st n terms of the series 5, 7, 11, 17, 25, …
Solution:
        Let    S = 5 + 7 + 11 + 17 + 25 +…+ tn.
        Also   S = 5 + 7 + 11 + 17 +… + tn–1 + tn.
        Subtracting we get
        0 = 5 + 2 + 4 + 6 + 8 +…+ nth term – tn
        ⇒ tn = 5 + 2 {n(n–1)/2}.
        or tn = n2 – n + 5
         Sn = Σn2 – Σn + 5Σ 1 = n(n+1)(2n+1)/6 – n(n+1)/2 + 5n
                = n/6 {(n + 1)(2n + 1) –3(n + 1) + 30}
                = n/6 (2n2 + 28).
Illustrations based on Vn method
Illustration:
        Find the sum of the series 1/1.2.3.4 + 1/2.3.4.5 +…… n terms.
Solution:
Let Tr =1/r(r+1)(r+2)(r+3) = 1/3 – [(r+3)–r]/r(r+1)(r+2)(r+3)
        =1/3 [1/r(r+1)(r+2) – 1/(r+1)(r+2)(r+3)] = 1/3 [Vr–1 – Vr]
 Sn = Σnr=1 Tr = 1/3Σnr=1 [Vr–1 – Vr] = 1/3 [V0 – Vn] = 1/3 [1/1.2.3 – 1/(n+1)(n+2)(n+3)]

 

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