Sequence And Series

3.1 Introduction

            In the previous unit we discussed set in which the order of the objects is of no significance. Thus, the set 1, , , , is the same as the set  , ,1, , .

            This unit is developed to the study of a set in which the order of the objects is of importance. Such a set is called a sequence. We shall also study series and progressions, specially arithmetic progress and geometric progression.

3.2 Sequence

            A sequence is not a just a set of numbers; it is an order set of numbers. There is therefore, a first number, a second, a third and so on. Thus,

(i) 1, , , , ,……….  or simply    1, , , , ,……….

(ii) {-1, 1, -1, 1 , -1, …….} or simply -1, 1, -1,1, -1,………..

Are examples of the sequences. Here, the numbers are in definite orders.

            A sequence, of course, is a special kind of function. It is defined as a f function whose domain is the set N = { 1, 2, 3, 4,……..} of natural numbers.

            A sequence is usually denoted by its values or images. The images of 1 , 2 , 3,….., n under a sequence f are f(1), f(2), f(3),……….. f(n). It is customary to denote them by f₁ , f₂ , f₃,……….fn or briefly by {f_n}.

Here, f₁, f₂, f₃,. ……, fn are also called the terms of the sequence. F₁ is the first term, f₂ the second term, f₃ the third term and so on. The term fn is called the n^th term or the general term of the sequence. It is sometimes denoted by t_n. Other conventional notations for the sequence are {x_n} and{ a_n}, where x_n and a_n denoted the respective general terms.

Finite and Infinite Sequence

             A sequence is said to be finite if the domain consists of a finite number of terms; otherwise it is said to be infinite.

Examples

(i) Let f: R be a function from N to R defined by

                        F(n) = , n N.

Then f(1) = , f(2) = , f(3) =  and so on.

 Images of 1, 2, 3,………. Under f are , , ,…………..

Thus, , , , ,………..from a sequence.

(ii) Let f:N R be defined by

            F(n) = (-1)n

Then, f(1) = -1, f(2) = 1 , f(3) = -1, f(4) = 1 and so on.

Image of 1, 2, 3, 4 …….under f are -1, 1, -1, 1,………….

(iii) The first 5 terms of a sequence

Let {an} be a sequence where

A_n = 4 + (-1)ⁿ

Then, a₁ = 4 + (-1) = 3

a₂ = 4 + 1         = 5

a₃ = 4 - 1          = 3

a₄ = 4 + 1         = 5

a₅ = 4 - 1          = 3

Thus, the first 5 terms of a sequence are 3 , 5, 3, 5, 3

3.3 Series

            Given a sequence a₁, a₂, a₃, a₄, …………,a_n, an expression of the form a₁ + a₂ + a₃ + a₄ + ……………..+ a_n is called a series. It is denoted by  .

Thus,  = a₁ + a₂ + a₃ + ……………..+ a_n.

Here, a1 is the first term, a2 the second term, a3 the third term,  ……….. and an the nth term of general term of the series.

            We also note that the series is finite; because the number of terms of term is   finite.

            An infinite series is given by

= a₁ + a₂ + a_{3…………}

K = 1

            The symbol is a short hand notation for the sum of the numbers. It is called sigma, a Greek letter.

Example:

(i) Express the series without using sigma notation.

Here, tk = 2k - 1

Replace k by 1, 2, 3, 4 and 5 respectively.

t₁ = 2.1 - 1 = 1

t₂ = 2.2 - 1 = 3

t₃ = 2.3 - 1 = 5

t₄ = 2.4 - 1 = 7

t₅ = 2.5 - 1 = 9

Thus ,  = 1 + 3 +5 +7 +9

            K = 1

Remarks

Operations like addition, subtraction, multiplication cannot be always applicable to an infinite series. For example, consider the series.

S = 4 + 8 + 16 + 32 + 64 + ……… (1)

 2S = 8 + 16 + 32 +64 + ………..(2)

Subtracting (1) from (2), we obtain S = -4, which is absurd.

3.4 Progressions

It is not necessary that the terms of a sequence always follow a certain pattern. Those sequence whose terms follows certain patterns are called progressions. All progressions are sequence but all sequence may not be progressions are series but all series mau not be progressions.

3.5 Type of Progressions

The various types of progressions are:

            (i) Arithmetic progression       (ii) Geometric progressions

And (iii) Harmonic progressions

Here we discuss Arithmetic and Geometric progressions only.

3.6 Arithmetic Progressions (A.P.)

            A sequence is said to be in arithmetic progressions if the difference of any term and its preceding term is always same; i.e.

t_n - t_n-1 = a constant, for all n

The constant quantity is called the common difference. The common difference is denoted by d.

Examples:

            (i)         The sequence 1 , 3 , 5, 7 , 9, ………..

                        Is in A.P; because d = 2.

                        [3 - 1 = 5 - 3 = 7 - 5 = 9 - 7 = 2]

            (ii)        The sequence 3, 2 , 1 , 0 , -1 , -2 , -3, ……… is also in A.P; because the common difference is - 1.

3.7 Technique for determining whether a given sequence is in A.P. or not when its nth term is given.

(i) Obtain tn

(ii) Replace n by n -1 to get t_n-1

(iii) Calculate t_n - t_n-1

(iv) If t_n - t_n-1 is independent of n, the given sequence is in A.P. otherwise it is not an A.P.

Examples:

(i) Show  that the sequence {an} , where an = 2n + 1 is in A.P., Also, fine commom difference.

            We have

a_n = 2n + 1

Replacing n by (n - 1) , we get

a_{n - 1} = 2(n - 1) + 1

Or, a_n-1 = 2n - 1

Now, an - an-1 = 2n + 1 - 2n + 1 = 2, which independent of n.

 The given sequence is in A.P; and its common difference is clearly 2.

            [Note : Here , t_n is denoted by a_n]

(ii) The nth term of a sequence is 2n + 3. Is the sequence an A.P. ? If so, find its 7th term.

            Let t_n = 2n + 3………………..(i)

Replacing  n by (n - 1), we obtain

            t_{n -1} = 2(n-1) + 3 = 2n +1

Now, t_n - t_{n -1} = 2n + 3 -2n - 1 = 2, which is independent of n.

So, the sequence is in A.P. with common difference 2.

Again, put n = 7 in (i). Then, t₇ = 2.7 + 3 = 17

Thus, the 7th term of A.P. is 17.

3.8 Nth term (or general term) of an A.P.

            Let a be the first term and d the common difference a an A.P. Then the sequence becomes

            a, a+d, a+2d, a+3d,………….

Clearly , t1 = a

T₂ = a+d = a + (2-1)d

T₃ = a+2d = a + (3-1)d

T₄ = a+ 3d = a + (4-1) d

 t_n = a + (n -1)d

Thus, the n^th term of an A.P. is t_n = a + (n-1) d, where a is the first term, d the common difference and n the number of terms. If the nth term is also the last term, then

= a + (n - 1) d, where is the last term.

Examples:

(i) Show the that sequence 29, 25, 21, 17, …….. is in A.P. Find is 10th term and general term.

Here, 25 - 29 = -4

            21 - 25 = -4

            17 - 21 = -4

The sequence is in A.P. with common difference -4.

Also, a = 29, n = 10 , 10^th term = t_10'

Now, (i) t_{10 =} a + (10-1)d = 29 + 9 (-4) = -7

(ii) t_n = a + (n -1)d = 29 + (n-1).(-4) = 29- 4n +4

Or, t_{n = 33 -4n}

(ii) Which term of the sequence 4, -1, -6 , …..is -66?

            Here, -1 -4 = -5

-6 -(-1) = -6 +1 = -5

 The given sequence is in A.P. with common difference -5.

Also, a = 4       d= -5

Let the n^th term be -66; i.e. t_n = -66

Now, t_{n =} a + (n-1)d

-66 = 4 + (n -1).(5)

-70 = -5n + 5

 n = 15

Hence, the 15th term of the given sequence is -66.

3.9Sum of n Terms of an A.P.

Let a be the first term, d the common difference, n the number terms and the last term of an A.P. Then the A.P. is

a ,+ a +d , a + 2d, ……………., 1 - 2d, 1 -d , 1.

Let S_n = a + (a + d) + (a + 2d) + …………..+ (1 - 2d) + (1 -d)  + 1

Writing the series in the reverse order, we have

S_n = 1 + (1 - d) + (1- 2d) + …………..+ (a + 2d) + (1+ d)  + a

By addition, we get

2S_n =  (a + 1) + (a+1)+ (a +1) + …………………+ (a+1) + (a +1) + (a+ )

Or, 2 S_n = n (a +1)

Or, S_n = (a+1)                      (1)

Since l = a + (n - 1) d ,we have

S_n = [ 2a + (n - 1)d]             (2)

The formula for obtaining the sum of the first n terms are

(i) S_n = (a+1) Use when the last term is given.

(ii) S_n = [ 2a + (n - 1)d] Use when last term is not given.

Examples:

Find the sum of the following A.S.

            (i) 1 + 3 + 5 + ……………to 21 terms

            (ii) 1 + 3 +5 +…………….+ 21.

Solution:

(i) Here, a = 1 , d = 2 , n = 21 , S₂₁ = ?

            Sn = [ 2a + (n - 1)d]

Or, S_{21 =} [ 2.1 + (21 - 1)2] = [ 2 + 40] = 441

(ii) Here, l = 21 , a = 1 , d = 2

First of all, we find the number of terms we know that,

L = a + (n-1)d

21 = 1 +(n-1)2

Or, 20 = 2n - 2             n = 11

Now ,S_n = (a+1)

=   (1+21) = 121

3.10 Arithmetic Mean

            If three quantities are in A.P., then the middle term is called the arithmetic mean between the other two. The arithmetic mean is shortly written as A.M.

            Let a, m, b be the three numbers in A.P.

            Then, m - a = b -m

            Or, 2m = a + b

Or, m =

Thus, the arithmetic mean between a and b is ; ie, A.M. =

3.11 n arithmetic means between a and b

Let the n arithmetic means between  the given quantities a and b be

m₁, m₂, m₃, …………..m_n. Then

a , m₁, m₂ , m₃, …………….., m_n be are in A.P.

Here, the number of term is ( n +2).

Also, a is the first term and b the last term.

Suppose that d is the common difference. Then

b = a + (n +2 - 1)d       [ l = a + (n - 1)d],

Or, b - a = (n +1) d  or d =  

Now, m₁ = a + d = a+

m₂ = a + 2d = a + 2

m₃ = a + 3d = a + 3

m_n = a + nd = a + n

Example :

(i)Insert 5 A.M.'s between 3 and 15.

Here , a = 3

            l = 15

            n = 7

            d = ?

We know that

 l  = a + (n -1)d

Or, 15 = 3 + 6d or d = 2

 Now, m₁ = a + d =  3 +2 = 5

m₂ = a + 2d = 3 + 2.2 = 7

m₃ = a + 3d = 3 + 3.2 = 9

m₂ = a + 2d = 3 + 2.2 = 7

m₃ = a +3d = 3 +3.2 = 9

m₄ = a + 4d = 3 + 4.2 = 11

m₅ = a + 5d = 3 +5.2 = 13

Thus, the 5 arithmetic mans are

5, 7, 9, 11 and 13.

3.12 Selection of different numbers (or terms) in A.P.

            Sometimes, we have to select a finite number of terms in A.P. It will be easier to solve the problems if we select the terms (or numbers) in the following manner :

            No. of terms    Terms of numbers                   Common difference

Odd:                3          a -d, a , a + d                                       d

                         5         a - 2d , a -d , a , a + d, a + 2d             d

Even:                4                     a -3d, a-d, a+d, a+ 3d             2d

                         6         a-5d, a-3d, a-d, a+d, a+3d, a+5d        2d

                                               

                                                Illustrative Examples

1. Find the three numbers in A.P. such that their sum is 30 and the sum of their squares is 350.

Solution :

            Let the three numbers in A.P. be

                         a-d, a, a+d

By the question,

(i) a-d + a + a + d = 30

Or,       3a = 30

Or,       a = 10

(ii) (a-d)² + a² + (a+d)² = 350

Or, (10-d)² + 10² + (10 +d)² = 350

Or, 100-20d +d² + 100 + 100 + 20d + d² = 350

Or, 2d² = 50

Or d² = 25

Or, d = 5

(a) When d = 5, them the three numbers in A.P. are

            10 - 5, 10, 10 + 5

Or, 5, 10 , 15

(b) when d = -5, the three numbers are

            10 +5 + 10, 10 -5

Or, 15, 10, 5

Thus the three numbers in A.P. are

5, 10, 15 or 15, 10, 5

2. Divide 52 into 4 parts which are in A.P. such that the product of the second and third part is 165.

Solution:

            Let the four parts in A.P. be

            a-3d, a-d, a+d, a+3d.

By the question

(i) a - 3d + a - d + a +d + a + 3d = 52

            4a = 52

Or, a = 13

(ii) (a -d) (a+d) = 165

Or, a₂ - d₂ = 165

Or, 169 - d₂ = 165

Or d₂ = 4    d = 2

(a) when d = 2, then

a - 3d = 13 - 3.2 = 7

a-d = 13 -2 = 11

a +d = 13 +2 = 15

a +3d = 13 +3.2 = 19

(b) When d = -2, then a -3d = 13 - 3.(-2) = 19

                         a - d = 13 - (-2) = 15

                         a + d = 13 + (-2) = 11

                         a + 3d = 13 + 3. (-2) = 7

Thus, the four parts are

7, 11, 15, 19 or 19, 15, 11, 9

3. How many terms of the series 27+ 24 + 21 + ………… must be taken in order that the sum be 132? Explain the double answer.

Solution:

Here, a = 27 and d = -3

Let S_n = 132, then

S_n = [ 2a + (n - 1)d]

Or, 132 = [2.27 + (n -1) (-3)]

Or, 264 = n [54 - 3n + 3]         or, 264 = 54n - 3n² + 3n

Or, 3n² - 57n + 264 = 0           or, n² - 19n + 88 = 0

Or, (n - 8) (n- 11) =                 n = 8, 11

Now, t₉ = a + 8d = 27 + 8.(-3) = 3

            t₁₀ = a + 9d = 27 +9.(-3) = 0

t₁₁ = a + 10d = 27 + 10.(-3) = -3

Since the sum of the last three terms is zero; therefore the sum of 8 terms as well as that of 11 terms is 132.

4. Find the sum of all  natural numbers between 250 and 1000 which are exactly divisible by 3.

Solution:

            The numbers between 250 and 1000 which are exactly divisible by 3 are 252, 255, 258, …………. 999.

This is an A.P. with first term a = 252, common difference d = 3 and the last term l = 999.

            Using l = a + (n-1)d, we get

            999 = 252 + (n -1).3

Or, 747 = 3n - 3          or, 3n = 750     or, n = 250

Let S_n be the required sum. Then

S_n = (a+1) = (252 + 999) = 156375.

5. If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of A.P. is zero.

Solution:

            Let a and d be the first term and common difference respectively of an A.P. Then

            t_{10 =} a + 9d  and t₁₅ = a + 14d

By the question

10 t_{10  =} 15 t₁₅

Or, 10 (a +9d) = 15 (a + 14d)

Or, 10a +90d = 15a + 210d

Or, a = -24d

Now, t₂₅ = a + 24d = -24d + 24d = 0

i.e., 25th term of the A.P. is zero.

6. If the nth term of am A.P. is 5n + 1, find the first term and common difference.Also find the upto n terms.

Solution

            We have : t_n = 5n  + 1

Replacing by 1, 2, 3, ………, we get

t₁ = 6

t₂ = 11

t₃ = 16

Then A.P. is 6, 11, 16,………..

Clearly, the first term = 6

Common difference = 11 -6 = 5

Let sn be the sum of the first terms, Then

S_n = [ 2a + (n - 1)d]

= [2.6 + (n - 1).5]

= (5n + 7)

7.Insert 9 A.M.'s between -7 and 43.

Solution:

Let m₁, m₂, m₃, ……………..m₉ be the 9 A.M.'s between -7 and 43.

Then,

            -7, m₁, m₂, m₃, …………..m₉, 43 are is A.P.

Here,  a = -7,    l = 43; n = 11

Let d be the common difference. Then

L = a + (n - 1)d

43 = -7 + 10d    d = 5

Now, m₁ = a + d = -7 + 5 = -2

            m₂ = a + 2d = -7 +10 = 3

            m₃ = a + 3d = -7 + 15 = 8

            m₄ = a + 4d = -7 + 20 = 13

            m₉ = a + 9d = -7 + 45 = 38

 The required means are

            -2, 3, 8, 13,………….., 38.

8. Find the A.P. whose 5^th term is 11 and 9^th term is 7. Find also 14^th term.

Solution:

Let a be the first term and d the common difference. Then

            t₅ = a + 4d                   and t₉ = a + 8d

By the question

            a + 4d = 11                  (1)

            a + 8d = 7                    (2)

Subtracting (1) from (2), we get 4d = -4 or d = -1

From (1) : a = 11 - 4.(-1) = 15

Thus, a = 15 and d = -1

 The required A.P. is 15, 4, 13,12,……………

Also, t₁₄ = a + 13d = 15 + 13.(-1) = 2

9. A man takes a job at Rs. 70000 a year. He receives an annual increase in pay of Rs. 1000. What is his salary during the tenth year?

Solution:

Here, a = 70000,         d = 1000, n = 10

t_{10 = ?}

Now, t_n = a + (n-1)d

Or, t₁₀  = 70000 + (10-1) 1000

= 70000 + 9000 = 79000

The required salary = Rs. 79000

10. Eighty coins are placed in a line on the ground. The distance between any two consecutive coins is 10 meters. How far must a person travel to bring them one to a bag placed 10 meters behind the first coin?

Solution:

Distance traveled to bring the first coin in the bag is 10 + 1 0 = 20 meters. Similarly, the distance traveled to bring the 2nd coin is 40 meters. The distance traveled to bring the 3rd coin is 60 meters, and so on.

Clearly, these distance

20,40,60,………. Are in A.P.

Let S_n denote the total distance traveled.

Here, a = 20, d = 20, n = 80

Now, S_n= [ 2a + (n - 1)d]

= [2 20 + 79 20] = 40 (40 +1580) = 64800

So, the total distance traveled is 64800 meters.

                                                Exercise 3 (A)

1. If the n^th term of a sequence is given by an = n2 + 1, write down its five terms.

2. The n^th term of a sequence is given by an = 2n + 3. Show that it is an A.P; also find its 7^th term and common difference.

3. Show that the sequence 3, -1, -5, -9, -13, -17……..is an A.P.Also find its common difference.

4. find (i) 11th term of the A.P. 25, 23, 21,……….

            (ii) 18th term of the A.P. ,3 , 5 ,………

5. The 11th term of am A.P. is 30 and the common difference is 2. Find its respectively; obtain the 12th term.

7. (i) Which term of the A.P. 84, 80,76,……….is 0?

     (ii) Is -447 a term of the A.P. 8, 5,2,………. ?

8. (i)  If x +2, 3x and 4x + 1 are in A.P. find x.

   (ii) Find x and y if 3, 10, x, y, 31 are in A.P.

9. If the nth term of an A.P. 23, 26, 29, 32,………..is equal to the n^th term of an A.P. 59, 58, 57, 56…………., find the number of terms.

10. The first term of an A.P. is 20, the common difference is 5 and the last term is 205; Find the number of terms.

11. (i) What is the 21st term of an A.P. whose 11th term is 23 and 15th term is 3 ? Is 40  a term of the series?

(ii) Find the A.P. whose 5^th and 8^th terms are 11 and 17 respectively. Find also its 15^th term.

12. How many terms of an A.P. 2 , 2 , 2,……….. must be taken so their sum mayu be 13 ? Explain the double answer.

13. If the 9^th term of an A.P. is zero, prove that its 29th term is double the 19th term.

14. Find the sum of the following series.

(i) 20 + 18 + 16 + ……………to 12 terms.

(ii)  +  +  + ……………to 9 terms.

(iii) 5 + 17 +29 + …………… + 233.

15. If the nth term of an A.P. is 2n + 3. Find the first term and common difference. Also find the  sum upto n terms.

16. In an A.P. (i) if t₁₃ = 8,    find S₂₅

                _­(ii) if S₁₅ = 300,  find t₈

            (iii) if t₂ : t₄ = 3:7, find the value of t₅ : t₉

17. (i) Find the sum of all integers between 50 and 500 what are divisible by 7.

     (ii) Find the sum of all odd numbers between 100 and 200.

     (iii) Find the sum of natural numbers from 4 to 100 inclusive which are multiples of 4.

18. The sum of the 7 terms of an A.P. is 10 and that of the nest 7 terms is 17. Find the first term and common difference.

19. The first term of an A.P. is 2 and the last term is 5. Then sum of all these terms is 442. Find the common difference.

20. Insert (i) 22 A.M. 's between 1 and 70.

            (ii) 12 A.M. 's between  4 and 43

            (iii) 6 A.m. 's between 15 and -13

21. Three numbers are in the ratio of 1:2:4. If 3 is added to the first and 8 is subtracted from the third, the new numbers will be the first and third term of an A.P. whose second term is the second number. Find the original numbers.

22. Find the three numbers in A.P. such that

(i) Their sum is 15 and the sum of their squares is 83.

(ii) their sum is 21 and their product is 280.

(iii) their sum is 15 ,and sum of the squares of its first and the third in 58.

23. divided 60 into three parts which are in A.P. such that the last one is three times the first.

24. Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

25. A man undertakes to pay off a debt of Rs. 65 by monthly ,installments; he pays Rs. 2 in the first month and continually increase the installments is every subsequent month by Re. 1. In what time will the debt be cleared up?

26. Shiva, a clerk in Nepal Bank is engaged in counting one rupee currency notes.For the first 10 minutes ,he counts the notes quickly at a speed of Rs. 150 per minute, Then he slows and begins to count at a speed of Rs. 2 less every minute then in the previous minute. Find what time he will take to count  a sum of Rs. 4500.

27. Small town whose populations was 18395 ten years ago, has lost 270 inhabitants each year since then. What is the present  population of small town?

28. A class consists of a certain number of boys whose age are in A.P. ; the common difference being 4 months. If the youngest boy is just 8 years old and the sum of the ages is  168 years;  find the number of boys in the class?

29. How much salary does a man receive in the 9th year and also find the total salary for Rs. 2000 with an incensement of Rs. 200 in each year ?

ANSWERS

1. 2, 5, 17, 26              2. 17; 2            3. -4                 4. (i) 5  (ii) 35         5. 10    6. 62   

7. (i) 22 (ii) No 8. (i) 3   (ii) 17 and 27                        9. 10    10. 38              11. (i) -27, No (ii) 3, 5, 7, 9… 31        12. 10 or 11            14. (i) 108    (ii) -1.5 (iii) 2380

15. 5; 2; n (n +4)         16. (i) 200 (ii) 20 (iii) 9:17       17.  (i) 17696 (ii) 7500 (iii) 1300

18.1; 19. 3               20. (i) 4, 7, 10….., 67 (ii) 7, 10, 13,……., 40 (iii) 11, 7, 3, -1, -5, -9

21. 5, 10, 20     22. 3, 5, 7 or 7, 5, 3    (ii) 4, 7, 10, or 10, 7, 4 (iii) 3, 5, 7 or 7, 5, 3

24. 2, 4, 6, 8     25. 10       26. 26.34 min   27. 15965      28. 16      29. Rs. 3600, Rs. 25200

3.13 Geometric Progression

             A sequence is said to be in geometric progression if the ratio of any term to the preceding one is always the same throughout the sequence.

            The ratio of any term to its preceding is called the common ratio. It is generally denoted by r.

For the sake of convenience the word geometric progression is abbreviated as G.P.

Examples:

1. The sequence 3,6, 12, 24,……… is in G.P. because =  =  = This is a sequence whose first term is 3 and the common ratio 2.

2. The sequence 1, , , ,…… is a G.P. whose first term is 1 and common ratio .

3. The sequence a,   ar,  ar² ,  ar³, ……….is in G.P. because  =  = ……..= r

The first term of this sequence is a and the common ratio is r.

3.14 General Term

            Let a be the first term and r be the common ratio of a G.P. Then

            t₁ = a

            t₂ = ar = ar^2-1

            t₃ = ar2 = ar^3-1

            t_n = ar ^n-1

Thus, the nth term of a G.P. is t_n = ar ^n-1

The nth term is also called the general term.

If nth term is the last term l, then

l = ar ^n-1

If the first term and common ratio are given, then we can write down any term of a G.P.

            Thus, if the first term, i.e. a = 1 and the common ratio r = 3, then the 7th term of the G.P. is  t₇ = 1.3^7-1 = 3⁶ = 729.

3.16 Sum of n Terms of a G.P.

Let S_n be the sum of the first n terms of a G.P. whose first term is a and common ratio r. Then S_n = a + ar + ar² + ……….+ ar^n-1

 rS_n = ar + ar² + …….+ ar^n-1 + arⁿ

By subtraction,

S_n (l-r) = a - arⁿ = a(l- rⁿ)

 Sn =  or Sn =

If  is the last term, then  = ar^n-1. Now (1) can be written as

S_n =  =   =  =

Thus, we have

S_{n =}    Use when r < 1

S_n =   Use when r > 1

S_n =  Use when the last term is given

Sum of an infinite geometric series. We know

S_n =              (r < l)

If r < l and the number of terms be infinite, then

rⁿ 0 as n

S  = 

3.17 Geometric Mean

If three quantities are in G.P. then the meddle one is called the geometric mean of the other two.

Let the numbers a, G, b be in G.P. Then

=   or G² = ab     G =

The geometric mean between two given quantities is the square root of their product.

3.18 n Geometric means between a and b

Let a and b be the two given numbers. If n numbers are inserted between a and b such that the sequence

a, G₁, G₂, G₃,………. G_n , b

is in G.P. Then the numbers G₁ , G₂, G₃,………, G_n intermediate. The sequence 1, , , ,  is in G.P. Therefore, , , , are three geometric means between 1 and .

Let ₁ , G₂, G₃,………, G_n  be the n geometric means between a and b. Then

            a, ₁ , G₂, G₃,………, G_n b are in G.P.

It consist of (n + 2) terms. The first term is a and the last term is b.

            Using               = ar^n-1, we obtain

b= a r^n+2-1 = arⁿ⁺¹

 r = ^{1/ (n+1)}

Now, G₁ = ar = a. ^{1/ (n+1)}

G₂ = ar² = a ^{2/ (n+1)}

G_n ⁼ arⁿ = a ^{n/ (n+1)}

3.19 Selection of Terms in G.P.

            Sometimes it is required to select a finite number in G.P.It will always be convenient if we select the terms in the following manner:

No. of terms	Terms	Common ratio
3	, a, ar	r
4	, ,ar, ar3	r2
5	, , a, ar , ar2	r

If the product of the numbers is not given, then the numbers are taken as a, ar, ar²……….

                                    Illustrative Examples

1. Examine whether the sequence 4, 12, 36, 108 ………. From a G.P. Also find the 10^th term.

Solution:

The given sequence is

            4, 12, 36, 108,………..

Here,  =3 ,  = 3 ,  = 3

Since the ratio of any term to its preceding is 3, therefore the sequence forms a G.P.

Here, a = 4,  r = 3, n = 10, t₁₀ = ?

Using t_n  = ar^n-1,  we get

t₁₀  = 3.4^10-1 = 4.3⁹ = 4 19683 = 78732

2. If the 5^th and 8^th terms of a G.P. are 81 and 2187 respectively; find the first term, common ratio and also determine the series.

Solution:

Let the first term and r be the common ratio of a G.P.

(i) The 5_th term is 81, i.e. t₅ = 81 , n = 5

            Using t_n = ar^n-1 we get

            t₅= a. r^5-1

            Or, 81 = ar⁴                 (1)

(ii) The 8^th term is 2187; i.e. t₈ = 2187 and n = 8

Now, tn = ar^n-1 = ar^8-1 = ar⁷

Or, 2187 = ar⁷

Dividing (2) by (1), we get

 =   or, r3 = 27  r = 3

Substituting the value of r in  (1), we obtain

81 =a.3⁴    a = 1

Thus, a = 1 and r = 3

The required series is 1 +3 + 9 + 27 + 81 +……..

3. Which term of the progression 18, -12, 8,……is  ?

Solution:

            The given progression is in G.P. whose first term is 18 and common ratio -

Let the nth term be . Then

tn = arn-1

or,  = 18 .

^or,  =      or, ⁸  =

 n -1 = 8  or, n = 9

4.The 7^th term of a G.P. is 8 times the fourth term and the 5^th term is 48. Find G.P

Solution:

Let a be the first term and r be the common ratio of a G.P.

By the question,          t₇ = 8t₄

                         Or, ar^{7 - 1} = 8. ar^{4 -1}

                         Or, a r⁶ = 8 ar³            or, r³ = 8  r = 2

Also, we have

            .t₅= 48

            Or, ar ^5-1 = 48              or, ar⁴ = 48      or, a2⁴ = 48

 a = 3

Thus, a = 3 and r = 2

 The G.P. is 3,6,12,24………..

5. How many terms of the series 2 + 6 +18 +……….must be taken to make the sum equal to 728?

Solution:

Here, a = 2 and r = 3

Let n be the number of terms so that S_n = 728. Then

S_n =

Or, 728 =  or,  728 = 3ⁿ- 1 or, 3ⁿ = 729

Or, 3+= 36  n = 6

6. Find the sum of the following series

            (i) 7 + 77+ 777 +…………. t n terms

            (ii) 0.5 +0.55 + 0.555 +……… to n terms

Solution:

(i) we have:

            7+ 77 + 777+………to n terms

= 7( 1 +11 +111 +……..to n terms)

= [9 +99 +999 + ……….to n terms]

= [(10-1) + (100-1) +(1000-1) + ……….to n terms]

= [(10 +100 +1000 +………to n terms]

= ,  Sn = ; and 1 +1 +1 +……to n terms = n]

=    -

(ii) 0.5 + 0.55 + 0.555+ ………..to n terms

= 5 (0. 1+0.11+ 0.111 + ……….to n terms)

= [ 0.9 + 0.99 +0.999 + ………..to nterms]

= [(1-0.1) +(1-0.01) + (1- 0.001)+…………..to n terms]

= [(1 +1+1+……to n terms) - (0.1 + 0.01 +0.001+……….to n terms)]

= , [  Sn = ]

=     =

                 0.9

= -  

7. find the three numbers in G.P. whose sum is 65 and whose products is 3375.

Solution:

Let the three numbers in G.P. be

, a, ar

By the question,

 + a +ar = 65            (1)

.a. ar = 3375

Or, a³ = 3375   or, a³ = 153                   a = 15

Subtracting the value of a in (1), we get

 + 15 + 15r = 65

Or, 15 + 15r + 15r² = 65r

Or, 15r² - 50r + 15 = 0            or, 3r² - 10r + 3 = 0

Or, 3r² - 9r - r +3 = 0               or, 3r(r -3)-1 (r - 3) = 0

Or, (r- 3) (3r - 1) = 0

 r = 3 ,

i) When r = 3 and a = 15, then the three numbers are

            , 15 , 15, 3

i.e. 5, 15, 45

ii) When r =  and a = 15, the three numbers are

45, 15, 5

Thus, the required three numbers in G.P. are

5, 15, 5            or, 45, 15, 5

8. find the three numbers inG.P. whose sum is 13 and the sum of whose squares is 91.

Solution

Let the three numbers in G.P. be

            a, ar, ar²

Then, a +  ar + ar² = 13           or, a (1 + r + r²) = 13               (1)

a2 + a²r² + a²r⁴ = 91                 or, a² (1 +r² + r4)                     (2)

Squaring (1) we get

a² + (1 +r + r²)² = 169                                                              (3)

Dividing (2) and (3) we get

=

Or,  =               [  a² + b² = (a+b)² - 2ab]

Or, =

Or,  =        or,   =

Or, 13 + 13r² -13r = 7 + 7r + 7r²

Or, 6r² - 20r + 6 = 0

Or, 3r² - 10r + 3 = 0

Or, (r-3) (3r- 1) = 0

 r = 3,

When r = 3, then a (1 + 3 +9) = 13

                        Or, a = 1

So, the numbers are 1, 3, 9.

When r = , then a  = 13

            Or, a = 9

So, the numbers are 9, 3, 1.

Hence the numbers are 1, 3, 9 and 9, 3, 1

9. A pingpong ball is dropped form a height of 16m. and always rebounds  of the height of the pervious fall. What height does it fall the 6th time.

Exercise 3 (b)

1. Show that each on of the following progression is a G.P., fi0nd also the common ratio in each case.

(i) 1, 3, 9, 27,………………

(ii) 27, 9, 3, 1,…………

(iii) , , , , …………..

2. If the n^th term of a sequence is given by a_n = 4ⁿ, n N, so that the sequence is in G.P.

3. find the 10th term of the G.P. , , ,……………..

4. If the 7^th term of a series inG.P. whose common ratio is 2, is 256; find the first term.

5.Find the 5th term of a series in G.P.if its 2^nd and 8^th terms are 9 and  respectively.

6. Find the first term and common ratio of a G.P. whose 4^th term and 6^th term are 54 and 486 respectively .

7. The fifth term of a certain G.P.is 81 whereas its second term is 24. find the series.

8. (i) find r if a = 27 and t₇ =

(ii) Find n if a = 2, r = 2 and t_n = 128

9. (i) Which term of the G.P. 5, 10, 20 , 40,………is 5120?

(ii) The 4th term of a G.P. is 54 and the 6th term is 24. Is 7  a term of the G.P.?

10. Find the  sum of the following series.

(i) 128 + 64 + 32 + ………… to 10 terms.

(ii) 1-3 +9 - 27 + …………to 9 terms.

(iii) 2 + 6 +18 + …………+ 4374.

11. How many terms of the G.P. 64. 32, 16, 8,………… must be taken so that the sum may be 127.5?

12.       (i) In a G.P. the first term is 7, the last term is 440 and the sum is 889. Find the      common ratio.

            (ii)The sum of  a G.P. whose common ratio is 2 and the last term is 768, is 1533; find the first term.

13. In a G.P. the sum of the first two terms is  and the sum of the nest two is 12. write down the sequence.

14. Insert

            (i) 3 G.M. between 1 and 81.

            (ii) 6 G.M. between 56 and -

15. (i) The sum of three numbers in G.P. is 38 and their product is 1728. find them.

            (ii) If the sum of the three numbers in G.P. is 7 and the sum of their squares is 21, find them.

            (iii) The product of three numbers in G.P. is 27 and the sum of their product in pair is 39. find the numbers.

            (iv) The ratio of the sum of the first three terms is to that of the first six terms of a G.P. is 125:152. Find the common ratio.

16. Sum to n terms

(i) 9 + 99 +999+ 9999 + …………

(ii) 5 + 55 +555+ 5555+ …………

(iii) 4 +44 + 444 + 4444 + ………….

(iv) 0.7 + 0.77 + 0.777 + 0.7777 + ……………

17. There are 8 varieties of monkeys in a certain zoo. The number of each variety forms a G.P. If the4th and 6th variety consist of 54 and 486, monkeys respectively, find the number in the first and the last variety.

18. A man was appointed to a post at a salary of Rs 1000 a year with an increasing each year of 10 % of his salary for the pervious year. How much does he receive during his fifth year?

19. A person borrows Rs. 6300 which the promises to pay in 6annual first and last installment.

2. Three numbers are in the rate 1:4:13. If 1 is added to the first number, the resulting number together with other two numbers from a G.P. find the original three numbers.

                                                ANSWERS

1. (i) 3   (ii)   (iii)               3. 2^-10               4. 4      5.     6. 2, 3 

7. 16 +24 +36 +54 +……….              8. (i)  (ii) 7               9. (i) 11th term

(ii) Yes, 9th term         10. (i) 255.75   (ii) 4921   (iii) 6560                 11. 8   

12. (i)    (ii) 3       13. , 1, 3 , 9,………. Or, - , 2, -6, 18, -54,……………

14. (i) 3, 9, 27      (ii) -28, 14, -7, , ,

15. (i) 8, 12, 18 or, 18, 12, 8   (ii) 4, 2, 1         (iii) 1, 3, 9        (iv)

16. (i)  (10ⁿ - 1) - n (ii)  (10n - 1) - n             (iii) (10n - 1) -

(iv) -  (1- ) 17. 2, 4347      18. Rs. 1464.10           19. Rs 100, Rs, 3200

20. 3, 12, 36