vNatural numbers are the product of human spirit. – DEDEKIND v
9.1 Introduction
In mathematics, the word, “sequence” is used in much the
same way as it is in ordinary English. When we say that a
collection of objects is listed in a sequence, we usually mean
that the collection is ordered in such a way that it has an
identified first member, second member, third member and
so on. For example, population of human beings or bacteria
at different times form a sequence. The amount of money
deposited in a bank, over a number of years form a sequence.
Depreciated values of certain commodity occur in a
sequence. Sequences have important applications in several
spheres of human activities.
Sequences, following specific patterns are called progressions. In previous class,
we have studied about arithmetic progression (A.P). In this Chapter, besides discussing
more about A.P.; arithmetic mean, geometric mean, relationship between A.M.
and G.M., special series in forms of sum to n terms of consecutive natural numbers,
sum to n terms of squares of natural numbers and sum to n terms of cubes of
natural numbers will also be studied.
9.2 Sequences
Let us consider the following examples:
Assume that there is a generation gap of 30 years, we are asked to find the
number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person
might have over 300 years.
Here, the total number of generations
=
300
10
30
=
Fibonacci
(1175-1250)
Chapter
SEQUENCES AND SERIES
9
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178 MATHEMATICS
The number of person’s ancestors for the first, second, third, …, tenth generations are
2, 4, 8, 16, 32, …, 1024. These numbers form what we call a sequence.
Consider the successive quotients that we obtain in the division of 10 by 3 at
different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These
quotients also form a sequence. The various numbers occurring in a sequence are
called its
terms. We denote the terms of a sequence by a
1
, a
2
, a
3
, …, a
n
, …, etc., the
subscripts denote the position of the term. The n
th
term is the number at the n
th
position
of the sequence and is denoted by a
n.
The
n
th
term is also called the general
term
of the
sequence.
Thus, the terms of the sequence of person’s ancestors mentioned above are:
a
1
= 2, a
2
= 4, a
3
= 8, …, a
10
= 1024.
Similarly, in the example of successive quotients
a
1
= 3, a
2
= 3.3, a
3
= 3.33, …, a
6
= 3.33333, etc.
A sequence containing finite number of terms is called a finite sequence. For
example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed
number).
A sequence is called infinite, if it is not a finite sequence. For example, the
sequence of successive quotients mentioned above is an infinite sequence, infinite in
the sense that it never ends.
Often, it is possible to express the rule, which yields the various terms of a sequence
in terms of algebraic formula. Consider for instance, the sequence of even natural
numbers 2, 4, 6, …
Here a
1
= 2 = 2 × 1 a
2
= 4 = 2 × 2
a
3
= 6 = 2 × 3 a
4
= 8 = 2 × 4
.... .... ....
.... .... ....
.... .... ....
.... .... ....
a
23
= 46 = 2 × 23, a
24
= 48 = 2 × 24, and so on.
In fact, we see that the n
th
term of this sequence can be written as a
n
=
2n,
where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, …,
the
n
th
term is given by the formula, a
n
= 2n – 1, where n is a natural number.
In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible
pattern, but the sequence is generated by the recurrence relation given by
a
1
= a
2
= 1
a
3
= a
1
+ a
2
a
n
= a
n – 2
+ a
n – 1
, n > 2
This sequence is called Fibonacci sequence.
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SEQUENCES AND SERIES 179
In the sequence of primes 2,3,5,7,…, we find that there is no formula for the n
th
prime. Such sequence can only be described by verbal description.
In every sequence, we should not expect that its terms will necessarily be given
by a specific formula. However, we expect a theoretical scheme or a rule for generating
the terms
a
1
, a
2
,
a
3
,…,a
n
,… in succession.
In view of the above, a sequence can be regarded as a function whose domain
is the set of natural numbers or some subset of it. Sometimes, we use the functional
notation a(n) for a
n
.
9.3 Series
Let
a
1
, a
2
,
a
3
,…,a
n
, be a given sequence. Then, the expression
a
1
+
a
2
+
a
3
+,…+ a
n
+
...
is called the
series associated with the given sequence .The series is finite or infinite
according as the given sequence is finite or infinite. Series are often represented in
compact form, called sigma notation, using the Greek letter
∑
(sigma) as means of
indicating the summation involved. Thus, the series a
1
+ a
2
+ a
3
+
... + a
n
is abbreviated
as
1
n
k
k
a
=
∑
.
Remark When the series is used, it refers to the indicated sum not to the sum itself.
For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase
“sum of a series,” we will mean the number that results from adding the terms, the
sum of the series is 16.
We now consider some examples.
Example 1 Write the first three terms in each of the following sequences defined by
the following:
(i) a
n
= 2n + 5, (ii) a
n
=
3
4
n
−
.
Solution (i) Here a
n
= 2n + 5
Substituting n = 1, 2, 3, we get
a
1
= 2(1) + 5 = 7, a
2
= 9, a
3
= 11
Therefore, the required terms are 7, 9 and 11.
(ii) Here
a
n
=
3
4
n
−
. Thus,
1 2 3
1 3 1 1
0
4 2 4
a , a , a
−
= = − = − =
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180 MATHEMATICS
Hence, the first three terms are
1 1
2 4
– , –
and 0.
Example 2 What is the 20
th
term of the sequence defined by
a
n
= (n – 1) (2 – n) (3 + n) ?
Solution
Putting n = 20 , we obtain
a
20
= (20 – 1) (2 – 20) (3 + 20)
= 19 × (– 18) × (23) = – 7866.
Example 3
Let the sequence a
n
be defined as follows:
a
1
= 1, a
n
= a
n – 1
+ 2 for n ≥ 2.
Find first five terms and write corresponding series.
Solution
We have
a
1
= 1, a
2
= a
1
+ 2 = 1 + 2 = 3, a
3
= a
2
+ 2 = 3 + 2 = 5,
a
4
= a
3
+ 2 = 5 + 2 = 7, a
5
= a
4
+ 2 = 7 + 2 = 9.
Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series
is 1 + 3 + 5 + 7 + 9 +...
EXERCISE 9.1
Write the first five terms of each of the sequences in Exercises 1 to 6 whose n
th
terms are:
1. a
n
= n (n + 2) 2. a
n
=
1
n
n
+
3. a
n
= 2
n
4.
a
n
=
2 3
6
n
−
5.
a
n
= (–1)
n–1
5
n+1
6. a
n
2
5
4
n
n
+
=
.
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n
th
terms are:
7. a
n
= 4n – 3; a
17
, a
24
8. a
n
=
2
7
;
2
n
n
a
9. a
n
= (–1)
n – 1
n
3
; a
9
10.
20
( – 2)
;
3
n
n n
a a
n
=
+
.
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SEQUENCES AND SERIES 181
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the
corresponding series:
11. a
1
= 3, a
n
= 3a
n – 1
+ 2 for all
n > 1 12. a
1
= – 1, a
n
=
1
n
a
n
−
, n ≥ 2
13. a
1
= a
2
= 2, a
n
= a
n – 1
–1, n > 2
14. The Fibonacci sequence is defined by
1 = a
1
= a
2
and a
n
= a
n – 1
+ a
n – 2
, n > 2.
Find
1
n
n
a
a
+
, for n = 1, 2, 3, 4, 5
9.4 Arithmetic Progression (A.P.)
Let us recall some formulae and properties studied earlier.
A sequence a
1
, a
2
, a
3
,…, a
n
,… is called arithmetic sequence or arithmetic
progression if a
n + 1
= a
n
+ d, n ∈ N, where a
1
is called the first term and the constant
term d is called the common difference of the A.P.
Let us consider an A.P. (in its standard form) with first term a and common
difference d, i.e., a, a + d, a + 2d, ...
Then the n
th
term (general term) of the A.P. is a
n
= a + (n – 1) d.
We can verify the following simple properties of an A.P. :
(i) If a constant is added to each term of an A.P., the resulting sequence is
also an A.P.
(ii) If a constant is subtracted from each term of an A.P., the resulting
sequence is also an A.P.
(iii) If each term of an A.P. is multiplied by a constant, then the resulting
sequence is also an A.P.
(iv) If each term of an A.P. is divided by a non-zero constant then the
resulting sequence is also an A.P.
Here, we shall use the following notations for an arithmetic progression:
a = the first term, l = the last term, d = common difference,
n = the number of terms.
S
n
= the sum to n terms of A.P.
Let a, a + d, a + 2d, …, a + (n – 1) d be an A.P. Then
l = a + (n – 1) d
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182 MATHEMATICS
[
]
S 2 ( 1)
2
n
n
a n d
= + −
We can also write,
[
]
S
2
n
n
a l
= +
Let us consider some examples.
Example 4 In an A.P. if m
th
term is n and the n
th
term is m, where m
≠
n, find the pth
term.
Solution We have a
m
= a + (m – 1) d = n, ... (1)
and a
n
= a + (n – 1) d = m ... (2)
Solving (1) and (2), we get
(m – n) d = n – m, or d = – 1, ... (3)
and a = n + m – 1 ... (4)
Therefore a
p
= a + (p – 1)d
= n + m – 1 + ( p – 1) (–1) = n + m
– p
Hence, the p
th
term is n + m – p.
Example 5 If the sum of n terms of an A.P. is
1
P ( – 1)Q
2
n n n+
, where P and Q
are constants, find the common difference.
Solution Let a
1
, a
2
, … a
n
be the given A.P. Then
S
n
= a
1
+ a
2
+ a
3
+...+ a
n–1
+ a
n
= nP +
1
2
n (n – 1) Q
Therefore S
1
= a
1
= P, S
2
= a
1
+ a
2
= 2P + Q
So that a
2
= S
2
– S
1
= P + Q
Hence, the common difference is given by d = a
2
– a
1
= (P + Q) – P = Q.
Example 6 The sum of n terms of two arithmetic progressions are in the ratio
(3n + 8) : (7n + 15). Find the ratio of their 12
th
terms.
Solution Let a
1
, a
2
and d
1
, d
2
be the first terms and common difference of the first
and second arithmetic progression, respectively. According to the given condition, we
have
3 8
7 15
Sum to termsof first A.P.
Sum to termsof second A.P.
n
n
n
n
+
+
=
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SEQUENCES AND SERIES 183
or
[ ]
[ ]
1 1
2 2
2 1
3 8
2
7 15
2 1
2
n
a ( n )d
n
n
n
a ( n )d
+ −
+
=
+
+ −
or
1 1
2 2
2 ( 1)
3 8
2 ( 1) 7 15
a n d n
a n d n
+ −
+
=
+ − +
... (1)
Now
th
1 1
th
2 2
11
12 term of first A.P.
11
12 term of second A.P
a d
a d
+
=
+
1 1
2 2
2 22
3 23 8
2 22 7 23 15
a d
a d
+
× +
=
+ × +
[By putting n = 23 in (1)]
Therefore
th
1 1
th
2 2
11
12 term of first A.P. 7
11 16
12 term of second A.P.
a d
a d
+
= =
+
Hence, the required ratio is 7 : 16.
Example 7 The income of a person is Rs. 3,00,000, in the first year and he receives an
increase of Rs.10,000 to his income per year for the next 19 years. Find the total
amount, he received in 20 years.
Solution Here, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20.
Using the sum formula, we get,
20
20
S [600000 19 10000]
2
= + ×
= 10 (790000) = 79,00,000.
Hence, the person received Rs. 79,00,000 as the total amount at the end of 20 years.
9.4.1 Arithmetic mean Given two numbers a and b. We can insert a number A
between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean
(A.M.) of the numbers a and b. Note that, in this case, we have
A – a = b – A, i.e., A =
2
a b
+
We may also interpret the A.M. between two numbers a and b as their
average
2
a b
+
. For example, the A.M. of two numbers 4 and 16 is 10. We have, thus
constructed an A.P. 4, 10, 16 by inserting a number 10 between 4 and 16. The natural
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184 MATHEMATICS
question now arises : Can we insert two or more numbers between given two numbers
so that the resulting sequence comes out to be an A.P. ? Observe that two numbers 8
and 12 can be inserted between 4 and 16 so that the resulting sequence 4, 8, 12, 16
becomes an A.P.
More generally, given any two numbers a and b, we can insert as many numbers
as we like between them such that the resulting sequence is an A.P.
Let A
1
, A
2
, A
3
, …, A
n
be n numbers between a and b such that a, A
1
, A
2
, A
3
, …,
A
n
, b is an A.P.
Here, b is the (n + 2)
th
term, i.e.,
b = a + [(n + 2) – 1]d = a + (n + 1) d.
This gives
1
b a
d
n
−
=
+
.
Thus, n numbers between a and b are as follows:
A
1
= a + d = a +
1
b a
n
−
+
A
2
= a + 2d = a +
2( )
1
b a
n
−
+
A
3
= a + 3d = a +
3( )
1
b a
n
−
+
..... ..... ..... .....
..... ..... ..... .....
A
n
= a + nd = a +
( )
1
n b a
n
−
+
.
Example 8 Insert 6 numbers between 3 and 24 such that the resulting sequence is
an A.P.
Solution Let A
1
, A
2
, A
3
, A
4
, A
5
and A
6
be six numbers between 3 and 24 such that
3, A
1
, A
2
, A
3
, A
4
, A
5
, A
6
,
24 are in A.P. Here, a = 3, b = 24, n = 8.
Therefore, 24 = 3 + (8 –1) d, so that d = 3.
Thus A
1
= a + d = 3 + 3 = 6; A
2
= a + 2d = 3 + 2 × 3 = 9;
A
3
= a + 3d = 3 + 3 × 3 = 12; A
4
= a + 4d = 3 + 4 × 3 = 15;
A
5
= a + 5d
= 3 + 5 × 3 = 18; A
6
= a + 6d = 3 + 6 × 3 = 21.
Hence, six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21.
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SEQUENCES AND SERIES 185
EXERCISE 9.2
1. Find the sum of odd integers from 1 to 2001.
2. Find the sum of all natural numbers lying between 100 and 1000, which are
multiples of 5.
3. In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of
the next five terms. Show that 20
th
term is –112.
4. How many terms of the A.P. – 6,
11
2
−
, – 5, … are needed to give the sum –25?
5. In an A.P., if p
th
term is
1
q
and q
th
term is
1
p
, prove that the sum of first pq
terms is
1
2
(pq +1), where p ≠ q.
6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the
last term.
7. Find the sum to n terms of the A.P., whose k
th
term is 5k + 1.
8. If the sum of n terms of an A.P. is (pn + qn
2
), where p and q are constants,
find the common difference.
9. The sums of n terms of two arithmetic progressions are in the ratio
5n + 4 : 9n + 6. Find the ratio of their 18
th
terms.
10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then
find the sum of the first (p + q) terms.
11. Sum of the first p, q and r terms of an A.P. are a, b
and c, respectively.
Prove that
( ) ( ) ( ) 0
a b c
q r r p p q
p q r
− + − + − =
12. The ratio of the sums of m and n terms of an A.P. is m
2
: n
2
. Show that the ratio
of m
th
and n
th
term is (2m – 1) : (2n – 1).
13. If the sum of n terms of an A.P. is 3n
2
+ 5n and its m
th
term is 164, find the value
of m.
14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
15. If
1 1
n n
n n
a b
a b
− −
+
+
is the A.M. between a and b, then find the value of n.
16. Between 1 and 31, m numbers have been inserted in such a way that the resulting
sequence is an A. P. and the ratio of 7
th
and (m – 1)
th
numbers is 5 : 9. Find the
value of m.
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186 MATHEMATICS
17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the
instalment by Rs 5 every month, what amount he will pay in the 30
th
instalment?
18. The difference between any two consecutive interior angles of a polygon is 5°.
If the smallest angle is 120°
, find the
number of the sides of the polygon.
9.5 Geometric Progression (G . P.)
Let us consider the following sequences:
(i) 2,4,8,16,..., (ii)
1 1 1 1
9 27 81 243
– –
, , ,
...
(iii)
.
,
.
,
.
,
.
.
.
01
0001
000001
In each of these sequences, how their terms progress? We note that each term, except
the first progresses in a definite order.
In (i), we have
a
a
a
a
a
a
a
1
2
1
3
2
4
3
2 2 2 2= = = =, , ,
and so on.
In (ii), we observe,
a
a
a
a
a
a
a
1
2
1
3
2
4
3
1
9
1
3
1
3
1
3
= = = =, , ,
and so on.
Similarly, state how do the terms in (iii) progress? It is observed that in each case,
every term except the first term bears a constant ratio to the term immediately preceding
it. In (i), this constant ratio is 2; in (ii), it is
–
1
3
and in (iii), the constant ratio is 0.01.
Such sequences are called geometric sequence or geometric progression abbreviated
as G.P.
A sequence a
1
, a
2
, a
3
, …, a
n
, … is called geometric progression, if each term is
non-zero and
a
a
k
k
+ 1
= r (constant), for k ≥ 1.
By letting a
1
= a, we obtain a geometric progression, a, ar, ar
2
, ar
3
,…., where a
is called the first term and r
is called the common ratio of the G.P. Common ratio in
geometric progression (i), (ii) and (iii) above are 2,
–
1
3
and 0.01, respectively.
As in case of arithmetic progression, the problem of finding the n
th
term or sum of n
terms of a geometric progression containing a large number of terms would be difficult
without the use of the formulae which we shall develop in the next Section. We shall
use the following notations with these formulae:
a = the first term, r = the common ratio, l = the last term,
n = the numbers of terms,
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SEQUENCES AND SERIES 187
n = the numbers of terms,
S
n
= the sum of first
n terms.
9.5.1 General term of a G .P. Let us consider a G.P. with first non-zero term ‘a’ and
common ratio ‘
r’. Write a few terms of it. The second term is obtained by multiplying
a by r, thus a
2
= ar. Similarly, third term is obtained by multiplying a
2
by r. Thus,
a
3
= a
2
r = ar
2
, and so on.
We write below these and few more terms.
1
st
term = a
1
= a = ar
1–1
, 2
nd
term = a
2
= ar = ar
2–1
, 3
rd
term = a
3
= ar
2
= ar
3–1
4
th
term =
a
4
=
ar
3
= ar
4–1
, 5
th
term =
a
5
= ar
4
= ar
5–1
Do you see a pattern? What will be 16
th
term?
a
16
= ar
16–1
= ar
15
Therefore, the pattern suggests that the n
th
term of a G.P. is given by
a ar
n
n
=
−
1
.
Thus, a, G.P. can be written as a, ar, ar
2
, ar
3
, … ar
n – 1
; a, ar, ar
2
,...,ar
n – 1
...
;according
as G.P. is finite or infinite, respectively.
The series a + ar + ar
2
+ ... + ar
n–1
or a + ar + ar
2
+ ... + ar
n–1
+...are called
finite or infinite geometric series, respectively.
9.5.2. Sum to n terms of a G .P. Let the first term of a G.P. be a and the common
ratio be r. Let us denote by S
n
the sum to first n terms of G.P. Then
S
n
=
a + ar + ar
2
+...+ ar
n–1
... (1)
Case 1 If r = 1, we have S
n
= a + a + a + ... + a (n terms) = na
Case 2 If r ≠ 1, multiplying (1) by r, we have
rS
n
= ar + ar
2
+ ar
3
+ ... + ar
n
... (2)
Subtracting (2) from (1), we get (1 – r) S
n
= a – ar
n
= a(1 – r
n
)
This gives
(1 )
S
1
n
n
a r
r
−
=
−
or
( 1)
S
1
n
n
a r
r
−
=
−
Example 9 Find the 10
th
and n
th
terms of the G.P. 5, 25,125,… .
Solution Here a = 5 and r = 5. Thus, a
10
= 5(5)
10–1
= 5(5)
9
= 5
10
and a
n
= ar
n–1
= 5(5)
n–1
= 5
n
.
Example10 Which term of the G.P., 2,8,32, ... up to n terms is 131072?
Solution Let 131072 be the n
th
term of the given G.P. Here a = 2 and r = 4.
Therefore 131072 =
a
n
= 2(4)
n – 1
or 65536 = 4
n – 1
This gives 4
8
= 4
n – 1.
So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9
th
term of the G.P.
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188 MATHEMATICS
Example11 In a G.P., the 3
rd
term is 24 and the 6
th
term is 192.Find the 10
th
term.
Solution
Here,
a ar
3
2
24= =
... (1)
and
a ar
6
5
192= =
... (2)
Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6.
Hence a
10
= 6 (2)
9
= 3072.
Example12 Find the sum of first n terms and the sum of first 5 terms of the geometric
series
2 4
1
3 9
...
+ + +
Solution Here a = 1 and r =
2
3
. Therefore
S
n
=
2
1
3
(1 )
2
1
1
3
n
n
a r
r
−
−
=
−
−
=
2
3 1
3
n
−
In particular,
5
5
2
S 3 1
3
= −
=
211
3
243
×
=
211
81
.
Example 13 How many terms of the G.P.
3 3
3
2 4
, ,
,... are needed to give the
sum
3069
512
?
Solution Let n be the number of terms needed. Given that a = 3, r =
1
2
and
3069
S
512
n
=
Since
(1 )
S
1
n
n
a – r
r
=
−
Therefore
1
3(1 )
3069 1
2
6 1
1
512
2
1
2
n
n
−
= = −
−
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SEQUENCES AND SERIES 189
or
3069
3072
=
1
1
2
n
−
or
1
2
n
=
3069
1
3072
−
3 1
3072 1024
= =
or 2
n
= 1024 = 2
10
, which gives n = 10.
Example 14 The sum of first three terms of a G.P. is
13
12
and their product is – 1.
Find the common ratio and the terms.
Solution Let
a
r
, a, ar be the first three terms of the G.P. Then
a
ar a
r
+ +
=
13
12
... (1)
and
( ) ( ) 1
a
a ar –
r
=
... (2)
From (2), we get a
3
= – 1, i.e., a = – 1 (considering only real roots)
Substituting a = –1 in (1), we have
1 13
1
12
– – – r
r
=
or 12r
2
+ 25r + 12 = 0.
This is a quadratic in r, solving, we get
3 4
or
4 3
r – –
=
.
Thus, the three terms of G.P. are :
4 3 –3 3 4 –4
, 1, for = and , 1, for =
3 4 4 4 3 3
– r – r
,
Example15 Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.
Solution This is not a G.P., however, we can relate it to a G.P. by writing the terms as
S
n
= 7 + 77 + 777 + 7777 + ... to n terms
=
7
[9 99 999 9999 to term]
9
... n+ + + +
=
2 3 4
7
[(10 1) (10 1) (10 1) (10 1) terms]
9
...n− + − + − + − +
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190 MATHEMATICS
=
2 3
7
[(10 10 10 terms) (1+1+1+... terms)]
9
...n – n+ + +
=
7 10(10 1) 7 10 (10 1)
9 10 1 9 9
n n
n n
− −
− = −
−
.
Example 16 A person has 2 parents, 4 grandparents, 8 great grandparents, and so on.
Find the number of his ancestors during the ten generations preceding his own.
Solution Here a = 2, r = 2 and n = 10
Using the sum formula S
n
=
( 1)
1
n
a r
r
−
−
We have S
10
= 2(2
10
– 1) = 2046
Hence, the number of ancestors preceding the person is 2046.
9.5.3 Geometric Mean (G .M.) The geometric mean of two positive numbers a
and b is the number
ab
. Therefore, the geometric mean of 2 and 8 is 4. We
observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a
generalisation of the concept of geometric means of two numbers.
Given any two positive numbers a and b, we can insert as many numbers as
we like between them to make the resulting sequence in a G.P.
Let G
1
, G
2
,…, G
n
be n numbers between positive numbers a and b such that
a,G
1
,G
2
,G
3
,…,G
n
,b is a G.P. Thus, b being the (n + 2)
th
term,we have
1
n
b ar
+
=
,
or
1
1
n
b
r
a
+
=
.
Hence
1
1
1
G
n
b
ar a
a
+
= =
,
2
1
2
2
G
n
b
ar a
a
+
= =
,
3
1
3
3
G
n
b
ar a
a
+
= =
,
1
G
n
n
n
n
b
ar a
a
+
= =
Example17 Insert three numbers between 1 and 256 so that the resulting sequence
is a G.P.
Solution Let G
1
, G
2
,G
3
be three numbers between 1 and 256 such that
1, G
1
,G
2
,G
3
,256 is a G.P.
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SEQUENCES AND SERIES 191
Therefore 256 = r
4
giving r = ± 4 (Taking real roots only)
For r = 4, we have G
1
= ar = 4, G
2
= ar
2
= 16, G
3
= ar
3
= 64
Similarly, for r = – 4, numbers are – 4,16 and – 64.
Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are
in G.P.
9.6 Relationship Between A.M. and G.M.
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Then
A and G
2
a b
ab
+
= =
Thus, we have
A – G =
2
a b
ab
+
−
=
2
2
a b ab
+ −
=
(
)
2
0
2
a b−
≥
... (1)
From (1), we obtain the relationship A ≥ G.
Example 18 If A.M. and G.M. of two positive numbers a and b are 10 and 8,
respectively, find the numbers.
Solution Given that
A.M. 10
2
a b
+
= =
... (1)
and
G.M. 8
ab
= =
... (2)
From (1) and (2), we get
a + b = 20 ... (3)
ab = 64 ... (4)
Putting the value of a and b from (3), (4) in the identity (a – b)
2
= (a + b)
2
– 4ab,
we get
(a – b)
2
= 400 – 256 = 144
or a – b = ± 12
... (5)
Solving (3) and (5), we obtain
a = 4, b = 16 or a = 16, b = 4
Thus, the numbers a and b are 4, 16 or 16, 4 respectively.
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192 MATHEMATICS
EXERCISE 9.3
1. Find the 20
th
and n
th
terms of the G.P.
5 5 5
2 4 8
, ,
, ...
2. Find the 12
th
term of a G.P. whose 8
th
term is 192 and the common ratio is 2.
3. The 5
th
, 8
th
and 11
th
terms of a G.P. are p, q and s, respectively. Show
that q
2
= ps.
4. The 4
th
term of a G.P. is square of its second term, and the first term is – 3.
Determine its 7
th
term.
5. Which term of the following sequences:
(a)
2 2 2 4 is 128 ?
, , ,...
(b)
3 3 3 3 is729 ?
, , ,...
(c)
1 1 1 1
is
3 9 27 19683
, , ,...
?
6. For what values of x, the numbers
2 7
7 2
– , x, –
are in G.P.?
Find the sum to indicated number of terms in each of the geometric progressions in
Exercises 7 to 10:
7. 0.15, 0.015, 0.0015, ... 20 terms.
8.
7
,
21
, 3
7
, ... n terms.
9. 1, – a, a
2
, – a
3
, ... n terms (if a ≠ – 1).
10. x
3
, x
5
, x
7
, ... n terms (if x ≠ ± 1).
11. Evaluate
11
1
(2 3 )
k
k =
+
∑
.
12. The sum of first three terms of a G.P. is
39
10
and their product is 1. Find the
common ratio and the terms.
13. How many terms of G.P. 3, 3
2
, 3
3
, … are needed to give the sum 120?
14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is
128. Determine the first term, the common ratio and the sum to n terms of the G.P.
15. Given a G.P. with a = 729 and 7
th
term 64, determine S
7
.
16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is
4 times the third term.
17. If the 4
th
, 10
th
and 16
th
terms of a G.P. are x, y and z, respectively. Prove that x,
y, z are in G.P.
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SEQUENCES AND SERIES 193
18. Find the sum to
n terms of the sequence, 8, 88, 888, 8888… .
19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,
16, 32 and 128, 32, 8, 2,
1
2
.
20. Show that the products of the corresponding terms of the sequences a, ar, ar
2
,
…ar
n – 1
and A, AR, AR
2
, … AR
n – 1
form a G.P, and find the common ratio.
21. Find four numbers forming a geometric progression in which the third term is
greater than the first term by 9, and the second term is greater than the 4
th
by 18.
22. If the p
th
, q
th
and r
th
terms of a G.P. are a, b and c, respectively. Prove that
a
q – r
b
r – p
c
P – q
= 1.
23. If the first and the n
th
term of a G.P. are a and b, respectively, and if P is the
product of n terms, prove that P
2
= (ab)
n
.
24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
(n + 1)
th
to (2n)
th
term is
1
n
r
.
25. If a, b, c and d are in G.P. show that
(a
2
+ b
2
+ c
2
) (b
2
+ c
2
+ d
2
) = (ab + bc + cd)
2
.
26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
27. Find the value of n so that
a b
a b
n n
n n
+ +
+
+
1 1
may be the geometric mean between
a and b.
28. The sum of two numbers is 6 times their geometric mean, show that numbers
are in the ratio
(
)
(
)
3 2 2 : 3 2 2
+ −
.
29. If A and G be A.M. and G.M., respectively between two positive numbers,
prove that the numbers are
A A G A G
( )( )
± + −
.
30. The number of bacteria in a certain culture doubles every hour. If there were 30
bacteria present in the culture originally, how many bacteria will be present at the
end of 2
nd
hour, 4
th
hour and n
th
hour ?
31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays
annual interest rate of 10% compounded annually?
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then
obtain the quadratic equation.
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194 MATHEMATICS
9.7 Sum to n
Terms of Special Series
We shall now find the sum of first n terms of some special series, namely;
(i) 1 + 2 + 3 +… + n (sum of first n natural numbers)
(ii) 1
2
+ 2
2
+ 3
2
+… +
n
2
(sum of squares of the first n natural numbers)
(iii) 1
3
+ 2
3
+ 3
3
+… +
n
3
(sum of cubes of the first n natural numbers).
Let us take them one by one.
(i) S
n
=1 + 2 + 3 + … +
n, then
S
n
=
( 1)
2
n n
+
(See Section 9.4)
(ii) Here S
n
= 1
2
+ 2
2
+ 3
2
+ … + n
2
We consider the identity k
3
– (k – 1)
3
= 3k
2
– 3k + 1
Putting k = 1, 2…, n successively, we obtain
1
3
– 0
3
= 3 (1)
2
– 3 (1) + 1
2
3
– 1
3
= 3 (2)
2
– 3 (2) + 1
3
3
– 2
3
= 3(3)
2
– 3 (3) + 1
.......................................
.......................................
......................................
n
3
– (n – 1)
3
= 3 (n)
2
– 3 (n) + 1
Adding both sides, we get
n
3
– 0
3
= 3 (1
2
+ 2
2
+ 3
2
+ ... + n
2
) – 3 (1 + 2 + 3 + ... + n) + n
3 2
1 1
3 3
n n
k k
n k – k n
= =
= +
∑ ∑
By (i), we know that
1
( 1)
1 2 3
2
n
k
n n
k ... n
=
+
= + + + + =
∑
Hence S
n
=
2 3
1
1 3 ( 1)
3 2
n
k
n n
k n n
=
+
= + −
∑
=
3 2
1
(2 3 )
6
n n n
+ +
=
( 1)(2 1)
6
n n n
+ +
(iii) Here S
n
= 1
3
+ 2
3
+ ...+n
3
We consider the identity, (k + 1)
4
– k
4
= 4k
3
+ 6k
2
+ 4k + 1
Putting k = 1, 2, 3… n, we get
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SEQUENCES AND SERIES 195
2
4
– 1
4
= 4(1)
3
+ 6(1)
2
+ 4(1) + 1
3
4
– 2
4
= 4(2)
3
+ 6(2)
2
+ 4(2) + 1
4
4
– 3
4
= 4(3)
3
+ 6(3)
2
+ 4(3) + 1
..................................................
..................................................
..................................................
(n – 1)
4
– (n – 2)
4
= 4(n – 2)
3
+ 6(n – 2)
2
+ 4(n – 2) + 1
n
4
– (n – 1)
4
= 4(n – 1)
3
+ 6(n – 1)
2
+ 4(n – 1) + 1
(n + 1)
4
–
n
4
= 4n
3
+ 6n
2
+ 4n + 1
Adding both sides, we get
(n + 1)
4
– 1
4
= 4(1
3
+ 2
3
+ 3
3
+...+
n
3
) + 6(1
2
+ 2
2
+ 3
2
+ ...+ n
2
) +
4(1 + 2 + 3 +...+ n) + n
3 2
1 1 1
4 6 4
n n n
k k k
k k k n
= = =
= + + +
∑ ∑ ∑
... (1)
From parts (i) and (ii), we know that
2
1 1
( 1) ( 1) (2 1)
and
2 6
n n
k k
n n n n n
k k
= =
+ + +
= =
∑ ∑
Putting these values in equation (1), we obtain
3 4 3 2
1
6 ( 1) (2 1) 4 ( 1)
4 4 6 4
6 2
n
k
n n n n n
k n n n n – – – n
=
+ + +
= + + +
∑
or 4S
n
= n
4
+ 4n
3
+ 6n
2
+ 4n – n (2n
2
+ 3n + 1) – 2n (n + 1) – n
= n
4
+ 2n
3
+ n
2
= n
2
(n + 1)
2
.
Hence, S
n
=
[ ]
2
2 2
( 1)
( 1)
4 4
n n
n n
+
+
=
Example 19 Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41…
Solution Let us write
S
n
= 5 + 11 + 19 + 29 + ... + a
n–1
+ a
n
or S
n
= 5 + 11 + 19 + ... + a
n–2
+ a
n–1
+ a
n
On subtraction, we get
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196 MATHEMATICS
0 = 5 + [6 + 8 + 10 + 12 + ...(n – 1) terms] – a
n
or a
n
= 5 +
( 1)[12 ( 2) 2]
2
n – n
+ − ×
= 5 + (n – 1) (n + 4) = n
2
+ 3n + 1
Hence
S
n
=
2 2
1 1 1 1
( 3 1) 3
n n n n
k
k k k
a k k k k n
= = =
= + + = + +
∑ ∑ ∑ ∑
=
( 1)(2 1) 3 ( 1)
6 2
n n n n n
n
+ + +
+ +
( 2) ( 4)
3
n n n
+ +
=
.
Example 20 Find the sum to n terms of the series whose n
th
term is n (n+3).
Solution Given that a
n
= n (n + 3) = n
2
+ 3n
Thus, the sum to n terms is given by
S
n
=
2
1 1 1
3
n n n
k
k k k
a k k
= = =
= +
∑ ∑ ∑
=
( 1) (2 1) 3 ( 1)
6 2
n n n n n
+ + +
+
( 1) ( 5)
3
n n n
+ +
=
.
EXERCISE 9.4
Find the sum to n terms of each of the series in Exercises 1 to 7.
1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +... 2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ...
3. 3 × 1
2
+ 5 × 2
2
+ 7 × 3
2
+ ... 4.
1 1 1
1 2 2 3 3 4
+ + +
× × ×
...
5. 5
2
+ 6
2
+ 7
2
+ ... + 20
2
6. 3 × 8 + 6 × 11 + 9 × 14 + ...
7. 1
2
+ (1
2
+ 2
2
) + (1
2
+ 2
2
+ 3
2
) + ...
Find the sum to n terms of the series in Exercises 8 to 10 whose n
th
terms is given by
8. n (n+1) (n+4). 9. n
2
+ 2
n
10.
2
(2 1)
n –
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SEQUENCES AND SERIES 197
Miscellaneous Examples
Example21 If p
th
, q
th
, r
th
and s
th
terms of an A.P. are in G.P, then show that
(p – q), (q – r), (r – s) are also in G.P.
Solution
Here
a
p
= a + (p –1) d ... (1)
a
q
= a + (q –1) d ... (2)
a
r
=
a + (r –1) d ... (3)
a
s
=
a + (s –1) d ... (4)
Given that a
p
, a
q
, a
r
and a
s
are in
G.P
.,
So
q q r
r
p q p q
a a a
a
q r
a a a a p q
−
−
= = =
− −
(why ?)
... (5)
Similarly
s r s
r
q r q r
a a aa
r s
a a a a q r
−
−
= = =
− −
(why ?)
... (6)
Hence, by (5) and (6)
q – r r – s
p – q q – r
=
, i.e., p – q, q – r and r – s are in G.P.
Example 22 If a, b, c are in G.P. and
1
1
1
y
x
z
a b c
= =
, prove that x, y, z are in A.P.
Solution Let
1
x
a
=
1
1
y
z
b c k
= =
Then
a = k
x
, b = k
y
and c = k
z
. ... (1)
Since a, b, c are in G.P., therefore,
b
2
= ac ... (2)
Using (1) in (2), we get
k
2y
= k
x + z
, which gives 2y = x + z.
Hence, x, y and z are in A.P.
Example 23 If a, b, c, d and p are different real numbers such that
(a
2
+ b
2
+ c
2
)p
2
– 2(ab + bc + cd) p + (b
2
+ c
2
+ d
2
) ≤ 0, then show that a, b, c and d
are in G.P.
Solution Given that
(a
2
+ b
2
+ c
2
) p
2
– 2 (ab + bc + cd) p + (b
2
+ c
2
+ d
2
) ≤ 0 ... (1)
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198 MATHEMATICS
But L.H.S.
= (a
2
p
2
– 2
abp + b
2
) + (b
2
p
2
– 2
bcp + c
2
) + (c
2
p
2
– 2cdp + d
2
),
which gives (ap – b)
2
+ (bp – c)
2
+ (cp –
d)
2
≥ 0 ...
(2)
Since the sum of squares of real numbers is non negative, therefore, from (1) and (2),
we have, (ap – b)
2
+ (
bp – c)
2
+ (
cp – d)
2
= 0
or ap – b = 0, bp – c = 0, cp – d = 0
This implies that
b c d
p
a b c
= = =
Hence a, b, c and d are in G.P.
Example 24 If p,q,r are in G.P. and the equations, px
2
+ 2qx + r = 0 and
dx
2
+ 2ex + f = 0 have a common root, then show that
d
p
e
q
f
r
, ,
are in A.P.
Solution The equation px
2
+ 2qx + r = 0 has roots given by
2
2 4 4
2
q q rp
x
p
− ± −
=
Since p ,q, r are in G.P. q
2
= pr. Thus
q
x
p
−
=
but
q
p
−
is also root of
dx
2
+ 2ex + f = 0 (Why ?). Therefore
2
2 0
q q
d e f ,
p p
− −
+ + =
or dq
2
– 2eqp + fp
2
= 0 ... (1)
Dividing (1) by pq
2
and using
q
2
= pr, we get
2
0
d e fp
,
p q pr
− + =
or
2
e d f
q p r
= +
Hence
d e f
, ,
p q r
are in A.P.
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Miscellaneous Exercise On Chapter 9
1. Show that the sum of (m + n)
th
and (m – n)
th
terms of an A.P. is equal to twice
the
m
th
term.
2. If the sum of three numbers in A.P., is 24 and their product is 440, find the
numbers.
3. Let the sum of n, 2n, 3n terms of an A.P. be
S
1
, S
2
and S
3
, respectively, show that
S
3
= 3(S
2
– S
1
)
4. Find the sum of all numbers between 200 and 400 which are divisible by 7.
5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
6. Find the sum of all two digit numbers which when divided by 4, yields 1 as
remainder.
7. If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈
N such that
f(1) = 3 and
1
( ) 120
n
x
f x
=
=
∑
, find the value of n.
8. The sum of some terms of G.P. is 315 whose first term and the common ratio are
5 and 2, respectively. Find the last term and the number of terms.
9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90.
Find the common ratio of G.P.
10. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers
in that order, we obtain an arithmetic progression. Find the numbers.
11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times
the sum of terms occupying odd places, then find its common ratio.
12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is
112. If its first term is 11, then find the number of terms.
13. If
a
bx
a bx
b
cx
b cx
c
dx
c dx
x
+
−
=
+
−
=
+
−
≠( ) ,0
then show that a, b, c and d are in G.P.
14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P.
Prove that P
2
R
n
= S
n
.
15. The p
th
, q
th
and r
th
terms of an A.P. are a, b, c, respectively. Show that
(q – r )a + (r – p )b + (p – q )c = 0
16. If
1 1 1 1 1 1
a ,b ,c
b c c a a b
+ + +
are in A.P., prove that a, b, c are in A.P.
17. If a, b, c, d are in G.P, prove that (a
n
+ b
n
), (b
n
+ c
n
), (c
n
+ d
n
) are in G.P.
18. If a and b are the roots of x
2
– 3x + p = 0 and c, d
are roots of x
2
– 12x + q = 0,
where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.
2020-21
200 MATHEMATICS
19. The ratio of the A.M. and G.M. of two positive numbers a and
b, is m : n
. Show
that
(
)
(
)
2 2 2 2
:
a b m m – n : m – m – n
= +
.
20. If a, b, c are in A.P.; b, c, d are in G.P. and
1 1 1
, ,
c d e
are in A.P. prove that a, c, e
are in G.P.
21. Find the sum of the following series up to n terms:
(i) 5 + 55 +555 + … (ii) .6 +. 66 +. 666+…
22. Find the 20
th
term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.
23. Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…
24. If S
1
, S
2
, S
3
are the sum of first n natural numbers, their squares and their
cubes, respectively, show that 9
2
2
S
= S
3
(1 + 8S
1
).
25. Find the sum of the following series up to n terms:
3 3 3 3 3 3
1 1 2 1 2 3
1 1 3 1 3 5
...
+ + +
+ + +
+ + +
26. Show that
2 2 2
2 2 2
1 2 2 3 ( 1) 3 5
3 1
1 2 2 3 ( 1)
... n n n
n
... n n
× + × + + × + +
=
+
× + × + + × +
.
27. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to
pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid
amount. How much will the tractor cost him?
28. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to
pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid
amount. How much will the scooter cost him?
29. A person writes a letter to four of his friends. He asks each one of them to copy
the letter and mail to four different persons with instruction that they move the
chain similarly. Assuming that the chain is not broken and that it costs 50 paise to
mail one letter. Find the amount spent on the postage when 8
th
set of letter is
mailed.
30. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
Find the amount in 15
th
year since he deposited the amount and also calculate the
total amount after 20 years.
31. A manufacturer reckons that the value of a machine, which costs him Rs. 15625,
will depreciate each year by 20%. Find the estimated value at the end of 5 years.
32. 150 workers were engaged to finish a job in a certain number of days. 4 workers
dropped out on second day, 4 more workers dropped out on third day and so on.
2020-21
SEQUENCES AND SERIES 201
It took 8 more days to finish the work. Find the number of days in which the work
was completed.
Summary
®
By a sequence, we mean an arrangement of number in definite order according
to some rule. Also, we define a sequence as a function whose domain is the
set of natural numbers or some subsets of the type {1, 2, 3, ....k}. A sequence
containing a finite number of terms is called a finite sequence. A sequence is
called infinite if it is not a finite sequence.
®
Let a
1
, a
2
, a
3
, ... be the sequence, then the sum expressed as a
1
+ a
2
+ a
3
+ ...
is called
series. A series is called finite series if it has got finite number of
terms.
®
An arithmetic progression (A.P.) is a sequence in which terms increase or
decrease regularly by the same constant. This constant is called common
difference of the A.P. Usually, we denote the first term of A.P. by a, the
common difference by d and the last term by l. The general term or the n
th
term
of the A.P. is given by a
n
= a + (
n – 1)
d.
The sum S
n
of the first n terms of an A.P. is given by
( ) (
)
S 1
n
n n
= 2a+ n – d = a + l
2 2
.
®
The arithmetic mean A of any two numbers a and b is given by
2
a + b
i.e., the
sequence a, A, b is in A.P.
®
A sequence is said to be a geometric progression or G.P., if the ratio of any
term to its preceding term is same throughout. This constant factor is called
the common ratio. Usually, we denote the first term of a G.P. by a and its
common ratio by r. The general or the n
th
term of G.P. is given by a
n
= ar
n – 1
.
The sum S
n
of the first n terms of G.P. is given by
2020-21
202 MATHEMATICS
(
)
(
)
– 1 1–
S 1
1 1 –
n n
n
a r a r
= or , if r
r – r
≠
®
The geometric mean (G.M.) of any two positive numbers a and b is given by
ab
i.e., the sequence a, G, b is G.P.
Historical Note
Evidence is found that Babylonians, some 4000 years ago, knew of arithmetic and
geometric sequences. According to Boethius (510), arithmetic and geometric
sequences were known to early Greek writers. Among the Indian mathematician,
Aryabhatta (476) was the first to give the formula for the sum of squares and cubes
of natural numbers in his famous work Aryabhatiyam, written around
499. He also gave the formula for finding the sum to n terms of an arithmetic
sequence starting with p
th
term. Noted Indian mathematicians Brahmgupta
(598), Mahavira (850) and Bhaskara (1114-1185) also considered the sum of squares
and cubes. Another specific type of sequence having important applications in
mathematics, called Fibonacci sequence, was discovered by Italian mathematician
Leonardo Fibonacci (1170-1250). Seventeenth century witnessed the classification
of series into specific forms. In 1671 James Gregory used the term infinite series in
connection with infinite sequence. It was only through the rigorous development of
algebraic and set theoretic tools that the concepts related to sequence and series
could be formulated suitably.
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