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Sequences

Sequences

• Mathematics is the science of patterns and homogeneity
• Definition
• Special Sequences
  • Arithmetic Progression (AP)
  • Geometric Progression (GP)
  • Harmonic Progression (HP)
• Problem Solving Techniques
• Glossary of Formulas
• Questions with answers

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Mathematics is the science of patterns and homogeneity.

Sequences are all about them.

Sequence has a classic meaning in mathematics similar to that in ordinary language.  A sequence is any regular pattern exhibiting some characteristic property throughout.  To be precise, any sequence is characterized by a property.

Let us take a simple example of the most common sequence of all, the sequence of natural numbers 1, 2, 3, … , n.  This sequence is characterized by a property where every term is one more than the previous term.  Therefore, sequence is bounded by a property.  We may find an infinite number of sequences.  You may note that to generate a sequence we need two things.  It requires a starting term, and a defined rule to get to the next term, or a general term.  This general term is often called the generating term.

The sequences we are dealing with here are called finite sequences for the obvious reason that they are bounded, i.e., they have a last term.  There are also sequences that are unbounded and may extend infinitely on either side.  These are termed infinite sequences, which we will discuss afterwards.

We are now in a position to define a finite sequence.  We start with the bound of a sequence. The bounds of a sequence are the values which the terms or members of a sequence can take.  You may understand them as the limits of the sequence.  For example, if it was specified to write down all natural numbers less than 10, then we have specified that our sequence has a limit of n varying from 1 to 9, because we have to find all natural numbers less than 10.

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A sequence may be generated by a rule alone, along with the set or bound in which it is defined.  We call this generating term the general term of a sequence.  You may now guess the general or generating term of our sequence of natural numbers.  Yes, it would be t_n = n; here t_n is the symbol of the general term.  A sequence is, therefore, a function of a general term.  For each value of n we get a particular term of the sequence.  We are now in a position to define a few finite sequences ourselves.

We start with a general term t_n = 2n+1;

0 < n ≤ 10            n ∈ N        

 ∈ - is element of...
 N - is set of natural numbers 1, 2, 3, ...

You can compute a few values yourself by putting n = 1,  2,  3, etc.

∴     t₁ = 2·1+1 = 3

       t₂ = 2·2+1 = 5

and  t₃ = 2·3+1 = 7, etc.

This sequence is a bounded finite sequence, hence we would substitute only values in the range of  0 < n ≤ 10;  i.e., we have only 9 terms in the sequence.

Now we have an idea of a finite sequence.  The summary of discussion done till now is that a finite sequence has the following properties:

Starting term or the initial term.

The generating term, or general term, or the rule to get the next term.

Please note that a finite sequence is always bounded, but the reverse is not always true.

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There may be a sequence which is bounded but may not be finite. Consider the sequence 1, .1, .01, .001, …  Here the general term is given by t_n = 1 10 n .  You can see that all the values of this sequence lie in the range of 0 to 1.  (It is a continuously decreasing sequence that will never reach zero since the exponents are always positive; but one can reach as near to zero as is possible.  We call this phenomenon as converging, which you will learn in higher grades).  Hence, this sequence is bounded in the semi-open set [0, 1], i.e.,  greater than zero and less than or equal to 1.

Let us take a few examples of finite sequences with a general term given.  You are required to find the first few terms.

1. t_n = 3n+1   0 < n ≤ 3      n ∈ N
2. t_n = 2ⁿ−1   0 < n ≤ 13      n ∈ N
3. t_n = 1 n 3      0 < n ≤ 19      n ∈ N

Note that n can only take values of positive integers, although it may have a range.

You can easily get terms of the sequences defined above by putting the values of

n = 1,  2,  3,  etc.  You can check your answers below:

1. The sequence is 4,  7,  10
2. The sequence is 1,  3,  7, …,  2¹³−1
3. The sequence is 1, 1 8 , 1 27 , 1 19 3

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Definition

A sequence, or a suite, can be defined as an array of numbers = X₁,  X₂,  X₃ … X_n, arranged in a definite order, following a definite rule throughout.

Each term in the sequence can be assigned a value for each n.  A finite sequence has a fixed number of terms, while an infinite sequence, on the contrary, has endless terms.  Please note that the rule, or law, of generating terms must be consistent throughout.  For example,  consider the sequence 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, …  We have first three terms that are consecutive natural numbers, next come three consecutive even numbers, and so on.  You see that it is a sequence having no definite order or rule throughout, though there are a few fragmented sequences in between.  Consistency is a must, done to provide universality to the formula or rule so that everybody can understand it, not just the creator! This fact is equivalent to the grammar of any language, which is consistent throughout.

Note that because every sequence has definite rules to find the n^th term, this does not mean that it has a definite formula to find the n^th term.  As an example consider the sequence of prime numbers,  2,  3,  5,  7, etc.

This qualifies for being a sequence as it has a definite rule to find n^th prime number, but it does not have a formula to find the n^th term.  Another example would be n^th term in the decimal expansion of 2 .  Surely, we can find all the terms, but can’t have a formula to get the n^th term.  We should be clear in our mind that a rule is different from a formula.

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We now examine a few examples of infinite sequences.  They are the same as a finite sequence except for the fact that the terms are endless. You are again required to find first few terms, not all!

The generating terms are given by

1. t_n = log 1 n ,              n ∈ N
2. t_n = sin(n+1)

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   ,       n ∈ N
3. t_n = t_n1+ t_n2,   t₁ = t₂ = 1,   n > 2,     n ∈ N

It is easy to compute the terms by putting the values of n = 1, 2, 3, 4, …, etc., consecutively.

We have done that for you; you can check your answers.

1. 0,  log 1 2 ,  log 1 3 , …
2. sin 2x,  sin 3x,  sin 4x, …
3. 1,  2,  3,  5,  8, …

The third infinite sequence is the famous Fibonacci sequence (1175-1250).

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We can also find the n^th term of a sequence if a few terms are given.  We now employ a method of trial and error to find the n^th term.  Later, in higher grade, we will learn the method of differences to find these general terms in a more sophisticated mathematical way.

For example, try to find the n^th term of the following sequence;

1,  5,  9,  13, …

We can at once figure out that each preceding term is 4 less than next term.

A little algebra shows the n^th term could be 4n+1.

We will try to find n^th terms of the other sequences.

Let us find the n^th term of the following sequence 3,  5,  9,  17,  33,  65, …

It is a little difficult to find the n^th term in this case.  We will start with our standard procedure of finding the n^th term, by finding the difference of successive terms.  The successive differences are 2,  4,  8,  16,  32, etc.  Now we can guess that in the n^th term there must be some power of 2, because 2,  4,  8,  16,  32, … are successive exponents of 2.

Now the first term is 3 which becomes 2¹+1 = 3

Similarly,   2² + 1 = 5

                2³ + 1 = 9

                2⁴ + 1 = 17

                2⁵ + 1 = 33

                2⁶ + 1 = 65

Hence we get the n^th term of the sequence as t_n = 2ⁿ + 1.  Note that the base of the method of differences is similar to what we have done just before, with a few more steps, of course.

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Special Sequences

Arithmetic Progression (AP)

Consider a few special sequences exhibiting peculiar properties:

1,  3,  5,  7, …

1,  5,  9,  13, …

3, −5, −13, …

• Do they have anything common?

• Looking closely on the successive terms one can see that each successive term is differing by its next term by a constant.  We call this a common difference.  The common differences of above sequences are 2,  4 and −8 respectively.

• These special sequences are called arithmetic progression.  The name indicates that the difference between any two consecutive terms is an arithmetic constant.  In an arithmetic progression, AP for short, we have two important characteristics of sequences, viz.; it has a starting term and a rule to get the next terms.  We can say that an AP has a first term and a common difference (rule) to get next terms.

• An arithmetic progression is a sequence consisting of terms each one of which differs by a constant.

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We are now in a position to mathematically define a general AP.

A general AP is given by a,  a+d,  a+2d, …, a+(n-1)d

In the above description, ‘a’ is called the first term and ‘d‘ is called the common difference.  Note that a+(n-1)d is the general term or last term. We will learn about how to sum these AP in next chapter of series.

Let us take a few examples of AP.

The following are an AP with the given general term

(i)   t_n = 2n+1

(ii)  t_n = − 2 3 n + 3 1 2

(iii) t_n = 3 + 2(n-2)

By putting n = 1, 2, 3, …, etc., we get

(i)  3, 5, 7, …

(ii) − 2 3 + 3 1 2 , − 2 3 ·2 + 3 1 2 , ....     or     17 6 , 13 6 , 9 6 ....     etc.

(iii) 1, 3, 5, …, etc.

The common differences for the three sequences are 2, - 2 3 , and 2 respectively.  Note that all mathematical tables are in AP. Take for example 2, 4, 6, 8, 10, …, etc., are in AP with a common difference equal to ‘2’.

We can also find the n^th term of an AP if the first few terms are given, or any term and a common difference is given.

We work out a few examples here and you will encounter more in the Question and Answer section later.

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Example: Find the n^th term of the following sequences ?

(i)   11, 3, -5, -13, …

(ii)  6, 8, 10, 12, …

(iii) Second term = 3 and the common difference ‘d’ = − 1 2

Solutions

(i) We have first term ‘a’ = 11; also common difference ‘d’ = common difference = difference of any two consecutive terms = 3 - 11 = -8.  Using the formula for the general term of an AP, we get t_n = a+(n-1)d = 11+(n-1)(-8) = 11-8n+8 = -8n+19 = 19-8n

(ii) Here we have first term ‘a’ = 6; also common difference ‘d’ = common difference = difference of any two consecutive terms = 8 - 6 = 2. Using the formula for general term of an AP we get t_n = a+(n-1)d = 6+(n-1)(2) = 6+2n-2 = 2n+4

(iii) Let the first term be ‘a’. Also, the common difference ‘d’ = - 1 2 .  Using the formula for general term of an AP we get t₂ = a+(2-1)d

⇒ a + (2−1)(− 1 2 ) = a − 3 − 1 2 = 3     (given)

∴ a = 3 + 1 2 = 7 2

Hence the n^th term of AP t_n = 7 2 + (n−1)(− 1 2 ) = − n + 8 2

Now we look at another type of special sequences.

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Geometric Progression (GP)

Consider the following array on numbers and like before, try to figure out any property common with all these sequences.

    3,  9,  27,  81, …

    1 2 , 1 10 , 1 50 , 1 250 , ...

    sin x, sin x·cos x, sin x·cos²X, …

Notice that all the terms are a constant multiple of the preceding terms. We can say that each term and its preceding term have a common ratio. The rule or property of this sequence is the common ratio. The common ratios are 3,  1 5 ,  and cos x, respectively.  These types of progressions are called geometric progressions.

The label geometric has an interesting significance.  The terms are a constant multiple of previous term.  For example, if next term is twice the previous term and the next term is again twice of previous term; we can attribute all these properties geometrically to these terms.  Consider a building, then consider another building twice its shape, and consider a sequence of these buildings.  You can at once see that properties exhibited by these geometric structures are characteristic of a geometric progression.  Hence, the name geometric is attributed because the properties of geometric progressions can be explained by geometric structures.

We are now in a position to find a general form of geometric progressions.

a,  ar,  ar²,  ar³,  ar⁴ … ar^n-1

Here ‘a’ is called the first term, ‘r’ is called the common ratio and ‘arⁿ’ is called the general term or the last term. Hence in a geometric progression each successive term is a constant multiple of the preceding term.

We will learn how to find the sum of these progressions in the next chapter of Series.

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Let us compute a few examples of GP.  The general terms are given and you are required to find first few terms.

(i) t_n = 3·2^n-1

(ii) t_n = − 1 3 ·3³⁻ⁿ

(iii) t_n = e^−nπ

Solutions

(i) Putting n = 1, 2, 3, 4 … into the given generating term we get

t₁ = 3,  t₂ = 3·2 = 6,   and t₃ = 3·2² = 12, etc.

(ii) Putting n = 1, 2, 3, … in n^th term t_n = − 1 3 ·3³⁻ⁿ, we get

t₁ = − 1 3 ·3^3-1 = − 1 3 · 9 = −3

t₂ = − 1 3 ·3^3-2 = − 1 3 · 3 = −1, and

t₃ = − 1 3 ·3^3-3 = − 1 3 , respectively.

(iii) Given t_n = e^−nπ.  Now putting n = 1, 2, 3, … into the n^th term t_n = e^−nπ we get

t₁ = e^-π,  t₂ = e^-2π and t₃ = e⁻³.

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Similarly we can find n^th term if the first few terms are given, or any term and the common ratio is given.  Let us take a few examples.  You are required to find the general term of the following geometric sequences:

(i) 1, 1 3 , 1 9 , …

(ii) -2, 6, -18, …

(ii) 3^rd term = -7  and the common ratio = 1 7

Solutions

(i) Given first term = a = 1
and the common ratio of any two consecutive terms = r = 1/3 1 = 1/3

⇒ ∴ n^th term = (First term)(common ratio)^n-1 = t_n = a·r^n-1 = 1 · ( 1 3 )^n-1 = 1 3 n-1

(ii) We have ‘a’ = first term = -2 and the common ratio of any two consecutive terms = r = 6 -2 = −3 ⇒ n^th term t_n = a·r^n-1 = (First term)(common ratio)^n-1 = (-2)·(-3)^n-1 = 6^n-1.

(iii) Given the 3rd term = t₃ = -7  and the common ratio = 1 7 .

Now, if the first term = a and the Common ratio = r then we have as the n^th term t_n = (First term)(common ratio)^n-1 = a·r^n-1

⇒ 3^rd term = t₃ = a·r^3-1 = (a)( 1 7 )^3-1 = −7,         (Given)

or   (a)( 1 7 )² = −7³( 1 7 )^n-1 = −7^-n+2, ⇒ a = −7³.

∴ the n^th term t_n = −7³( 1 7 )^n-1 = −7^-n+2

You will see other examples in the Question and Answer section.

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Harmonic Progression (HP)

We end our discussion by considering one more important class of progressions, harmonic progressions.

Consider the following sequences:

1, 1 2 , 1 3 , 1 4 , …

1 7 , 1 4 , 1 1 , 1 -2 , …

1 cos X , 1 cos X + tan X , 1 cos x + 2tan x , …

Do you notice any peculiar properties?

Yes, you again guessed it right! They all are terms of an inverted AP. This just means that when we invert all the terms (i.e. find the multiplicative inverse), we get an arithmetic progression.  The common differences are 1, -3, and tanx respectively, for the three given sequences.  We may call them arithmetic progression of reciprocal numbers or quantities: a sequence of numbers or quantities whose reciprocals form an arithmetic progression, e.g. 1/3, 1/5, 1/7, 1/9.  These sequences have been given a special name, harmonic progressions; HP for short.

We will take general form of these sequences.

1 a , 1 a+d , 1 a+2d , …, 1 a+(n-1)d .

Proceeding as before, the general term, or last term, is 1 a+(n-1)d .

The common difference is ‘d’ and the first term is 1 a .

An important point to note is that there is no formula to find the sum for the ‘n’ terms of a harmonic series.  We can find the HP if the n^th term is given.  Let us take a few examples.

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You are required to find the HP whose n^th term is given as follows:

(i)   t_n = 1 1-2n

(ii)  t_n = 1 3n + 2

(iii) t_n = 1 sin x + (n-1)cos x

Solutions

(i) Using n = 1, 2, 3, 4 …, in the given generating, or n^th, term we get

t₁ = 1 1−2·1 = -1,  t₂ = 1 1−2·2 = -1/3,  t₃ = 1 1−2·3 = -1/5, etc.

(ii) Using n = 1, 2, 3, … in general term we get

t₁ = 1 3·1 + 2 = 1/5,  t₂ = 1 3·2 + 2 = 1/8, and t₃ = 1 3·3 + 2 = 1/11 respectively.

(iii) Given t_n = 1 sin x + (n-1)cos x .   Using for n = 1, 2, 3, … in the n^th term we get

t₁ = 1 sinx ,  t₂ = 1 sin x + cos x ,  and t₃ = 1 sin x + 2cos x .

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In the following examples we will find the n^th term when a few terms are given, or any term and the common difference is given.  You are required to find the n^th term.

(i)   1 2 , 1 7 , 1 12 , …

(ii)  7 3 , 7 5 , 1, …

(iii) t₂ = -3 5 , and the common difference = d = 2.

Solutions

(i)   Let the HP be 1 a , 1 a + d , 1 a + 2d , …, 1 a + (n-1)d .

Here the first term = 1/a = 1/2,  and the 2^nd term = 1 a + d = 1/7.

⇒ a = 2 and a + d = 7  ⇒  d = 7 - 2 = 5.

∴  The n^th term of HP is 1 2 + (n-1)5 = 1 5n -3 .

(ii)  We will solve this problem a bit differently.  We know that HP is the reciprocal of AP.  And if we revert all the terms we will get an AP.  Or 3 7 , 5 7 , 1, …, are in AP.  Here the first term = a = 3 7 .

Also, the common difference d = difference of any two consecutive terms = 5 7 - 3 7 = 2 7 .

Hence, the n^th term of AP = a+(n-1)d = 3 7 + (n-1) 2 7 = 3n + 1 7

⇒  The n^th term of the corresponding HP = 1 nth term of AP = 7 3n + 1 .

You will learn more of these methods in the Question and Answer section.

(iii) Let the HP be 1 a , 1 a + d , 1 a + 2d , … 1 a + (n-1)d

where a = first term and d = common difference.

Accordingly, the problem’s 2nd term = t₂ = 1 a + d = −3 5     ⇒ a + d = − 5 3 .

But d = 2 ⇒ a + 2 = - 5 3 ⇒ a = -2 - 5 3 = − 11 3

∴  The n^th term = 1 -11/3 + (n-1)2 = 3 -11+6n-6 = 3 6n - 17

We will work other examples in the Question and Answer section.

In the next chapter we will learn how to find the means of these important sequences and following those we will try to find the sum for the n terms of AP and GP series.  But let us first learn a few problem-solving techniques for Sequences.

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Problem Solving Techniques

In the present section we propose to discuss various methods that may be employed to solve problems with Sequences.

The basis of mathematical problem solving is observation.  Mathematics is the science of observing patterns in nature.  We will discuss a few peculiar patterns here and see how they can be resolved.

Let us start with an example of A.P..

Example 1.  Find the n^th term of the following A.P.:   1, 4, 7, 10… ?

Approach: To approach all problems of A.P. in general we try to figure out two basic things which virtually define an A.P.  Those two things are the first term and the common difference (C.D.).  In our case the first term is ‘1’ and the common difference is 4-1 = 3.  The common difference is always the difference of any two consecutive terms taken in increasing order and subtracting the lower order term from the higher order term.  It is advisable to take the difference of the initial two terms if possible.

∴  t_n = a+(n-1)d = 1+(n-1)3 = 3n-3+1 = 3n - 2

∴  We state the approach/step number (1) for problems concerning AP sequences.

A₁  —  Find ‘a’ & ‘d’

With some problems we have no need to find t_n.  Hence, we concentrate on our approach/step number (2).  After finding ‘a’ & ‘d’ we write the concerned formula.

A₂  —  Write the concerned formula

In our case the formula was t_n = a+(n-1)d

The next approach/step number (3) in our present problem could be ‘fill in the values’

A₃  —  Fill in the known values

The next approach/step should be to calculate the values

A₄  —  Calculate the unknown values

The last but step is to write the answer

A₅  —  Write the calculated values/answer

It always pays to write the answer in the end.  The teacher/examiner may not award you full credits if the answer alone is correct while a few of your steps are missing or they are not in the correct order.  On the other hand, if your answer is incorrect then it is also beneficial to write or mention the answer in the end.  It always plays a psychological trick for the examiner that the problem was attempted in full but a few steps might be wrong.

The basic approach to all the problems should be to observe any patterns, and if no approach seems possible, to return to the basics.  Also, if one is unable to find any possible approach then the golden rule is doing every possible intelligent mathematical operation [Refer problem solving techniques section of chapter 3] and you will arrive at the solution in most of the cases! All that may or may not be needed in basic problems, but for the advanced problems we can always use that tactic with great success.

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For problems in GP we take the following course of action:

A₁  —  Find ‘a’ & ‘r’

In some problems we have no need to find t_n.  Hence we advance to approach/step number (2).  After finding ‘a’ & ‘r’ we can write the concerned formula.

A₂  —  Write the concerned formula

In this case the formula is t_n = a·r^n-1

The next approach/step number (3) in our present problem could be ‘fill in the values’.

A₃  —  Fill in the known values

The next approach/step is to calculate the values.

A₄  —  Calculate the unknown values

The last, but never the least, step is writing the answer.

A₅  —  Write the calculated values/answer

It is always advisable to quote the formula since there are credits for steps taken and you may not want to miss the chance to get some.

In GP the basic items to look for are the first term and the common ratio.  If we know these two factors we can always find the GP.  Methods vary from problem to problem, but the basics remain the same.  Identify the known and unknown and also the relations concerning them.

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Here is another example.  The student is cautioned that the following examples are a little difficult and may not represent the exact level of problems in this chapter.  We have chosen them so we can include as many steps and explain the essence of the methods in detail.

Example 2.  Find the least positive term of the sequence

1 1024 , − 1 512 , 1 256 , …

Solution: We have first term = a = 1 1024

Also, the second term = − 1 512

∴ The common ratio = r = Ratio of any two consecutive terms = Second Term First Term = − 512 1024 = − 1 2

∴ Nth term = 1 1024 ( 1 2 )^n-1 a · r^n-1 = 2¹⁰ · (-1)^n-1 · 2^-n+1 = 2^-n+11 · (-1)^n-1

∴   T_n = 2^-n+11 · (-1)^n-1

Now t_n = Positive ⇒ n = odd, since an even exponent to the (-1) is positive and n-1 is even when n is odd.

The lowest positive exponent of 2 is 1, i.e., 2⁰.

⇒ t_n = 2^-n+11·(-1)^n-1 > 2⁰ = 1.

or  -n + 11 > 1  and  n is odd.

Or n < 10 and odd ⇒ n = 9

∴ The lowest value term = T₉ = 2^-9+11 · (-1)^9-1) = 1 4 .

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And here is another example for HP and then we end this section.  You will learn many other methods for solutions in the Question and Answer section.

Example 3. Which terms of the given sequences

− 1 4 , − 1 3 , − 1 2 , …   and − 1 729 , − 1 243 , − 1 81 , …  are equal, and
where the harmonic mean is not defined for a fifth term?

Solution: The n^th term of the HP is 1 nth term of corresponding AP

1 a + (n-1)d = − 1 4 −(n-1) = − 1 5 - n

First term of GP = a = 1 729 and common ratio r = Ratio of consecutive terms = 729 243 = −3.

∴ The N^th term of the GP = 1 729 · (-3)^n-1 = -3^-6 · (-3)^n-1 = −3^n-7 (-1)ⁿ.

According to the problem the n^th term of HP is equal to the n^th term of GP.

− 1 5 - m = −3^n-7(-1)ⁿ ……… (1)

We halt here and tell you another advanced method of problem solving which you will encounter in the future.  This method may be termed an indirect method of problem solving.  We try to solve problems by using properties that affect the solution indirectly.  We narrow down the range for a solution and in the process find the solution.

Now (1) is true since − 1 5 - m = −3^n-7(-1)ⁿ

⇒ n is even when LHS is positive, and odd when LHS is negative.

Also, you see that LHS can be numerically equal to all the inverse powers of 3, i.e., − 1 3 , −1, 1, 1 3 , 1 9 , etc., as LHS is an inverse AP with common difference = 1.

∴ The common terms are −1, 1 3 , 1 27 , 1 243 , which can be obtained by putting n = 6, 4, 2, 0 and 7.

The above examples are not meant to daunt you.  Rather, we have given them to inspire you.  The basic methods of problem solving are finding patterns through observation.  In the end we must add all wise men admonish that only practice makes a mathematician perfect.  You may memorize all the formulas, but you still need to practice so you can learn all the necessary steps and also gain speed and proficiency.

Good luck in the Question and Answer section.