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MATHEMATICS The number of person’s ancestors for the first, second, third, …, tenth generations are , , , , , …, . These numbers form what we call a sequence . Consider the successive quotients that we obtain in the division of by at different steps of division. In this process we get , .

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 , a , a , …, 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 = , a = , a = , …, a = . Similarly, in the example of successive quotients a = , a = . , a = .

, …, a = .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 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 , , , … Here a = = × a = = × a = = × a = = × .... ....

.... .... a = = × , a = = × , and so on. In fact, we see that the n th term of this sequence can be written as a n = n , where n is a natural number.

Similarly, in the sequence of odd natural numbers , , , …, the n th term is given by the formula, a n = n – , where n is a natural number. In some cases, an arrangement of numbers such as , , , , , ,.. has no visible pattern, but the sequence is generated by the recurrence relation given by a = a = a = a + a a n = a n – + a n – , n > This sequence is called Fibonacci sequence . SEQUENCES AND SERIES In the sequence of primes , , , ,…, 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 , a , a ,…, 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 an . . Series Let a , a , a ,…, a n , be a given sequence. Then, the expression a + a + a +,… + 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 + a + a = ∑ Remark When the series is used, it refers to the indicated sum not to the sum itself. For example, + + + 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 . We now consider some examples. Example Write the first three terms in each of the following sequences defined by the following: a n = n + , a n = n − Solution (i) Here a n = n + Substituting n = , , , we get a = ( ) + = , a = , a = Therefore, the required terms are , and . Here a n = n − .

Thus, , a ,a = − = − MATHEMATICS Hence, the first three terms are – , – and . Example What is the th term of the sequence defined by an = ( n – ) ( – n ) ( + n ) ? Solution Putting n = , we obtain a = ( – ) ( – ) ( + ) = × (– ) × ( ) = – . Example Let the sequence an be defined as follows: a = , a n = a n – + for n ≥ .

Find first five terms and write corresponding series. Solution We have a = , a = a + = + = , a = a + = + = , a = a + = + = , a = a + = + = . Hence, the first five terms of the sequence are , , , and . The corresponding series is + + + + +...

EXERCISE . Write the first five terms of each of the sequences in Exercises to whose n th terms are: . a n = n ( n + ) . a n = n + .

a n = n . a n = n − . a n = (– ) n – n + . a n Find the indicated terms in each of the sequences in Exercises to whose n th terms are: .

a n = n – ; a , a . a n = ; n . a n = (– ) n – n ; a . ( – ) ; n n SEQUENCES AND SERIES Write the first five terms of each of the sequences in Exercises to and obtain the corresponding series: .

a = , a n = a n – + for all n > . a = – , a n = n a − , n ≥ . a = a = , a n = a n – – , n > . The Fibonacci sequence is defined by = a = a and a n = a n – + a n – , n > .

Find + , for n = , , , , . Arithmetic Progression (A.P.) Let us recall some formulae and properties studied earlier. A sequence a , a , a ,… , a n , … is called arithmetic sequence or arithmetic progression if a n + = a n + d , n ∈ N , where a 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 + d , ... Then the n th term ( general term ) of the A.P. is a n = a + ( n – ) d. We can verify the following simple properties of an A.P.

: If a constant is added to each term of an A.P., the resulting sequence is also an A.P. 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 + d, …, a + ( n – ) d be an A.P. Then l = a + ( n – ) d MATHEMATICS ] S n a l Let us consider some examples. Example In an A.P. if m th term is n and the n th term is m , where m ≠ n , find the p th term.

Solution We have a m = a + ( m – ) d = n , ... ( ) and a n = a + ( n – ) d = m ... ( ) Solving ( ) and ( ), we get ( m – n ) d = n – m , or d = – , ... ( ) and a = n + m – ...

( ) Therefore a p = a + ( p – ) d = n + m – + ( p – ) (– ) = n + m – p Hence, the p th term is n + m – p. Example If the sum of n terms of an A.P. is P ( – )Q n n , where P and Q are constants, find the common difference. Solution Let a , a , … a n be the given A.P.

Then S n = a + a + a +...+ a n – + a n = n P + n ( n – ) Q Therefore S = a = P, S = a + a = 2P + Q So that a = S – S = P + Q Hence, the common difference is given by d = a – a = (P + Q) – P = Q. Example The sum of n terms of two arithmetic progressions are in the ratio ( n + ) : ( n + ). Find the ratio of their th terms. Solution Let a , a and d , d be the first terms and common difference of the first and second arithmetic progression, respectively.

According to the given condition, we have Sumto termsof firstA.P. Sumto termsof secondA.P. SEQUENCES AND SERIES or ] ( n )d ( n )d or ) ) d d ... ( ) Now th th termof first A.P.

termof secondA.P d d d d × × [By putting n = in ( )] Therefore th th term of first A.P. term of second A.P. d d Hence, the required ratio is : . Example The income of a person is Rs.

, , , in the first year and he receives an increase of Rs. , to his income per year for the next years. Find the total amount, he received in years. Solution Here, we have an A.P.

with a = , , , d = , , and n = . Using the sum formula, we get, S [600000 10000] × = (790000) = , , . Hence, the person received Rs. , , as the total amount at the end of years.

. . 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 = b We may also interpret the A.M. between two numbers a and b as their average b . For example, the A.M.

of two numbers and is . We have, thus constructed an A.P. , , by inserting a number between and . The natural 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 and can be inserted between and so that the resulting sequence , , , 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 , A , A , …, A n be n numbers between a and b such that a , A , A , A , …, A n , b is an A.P.

Here, b is the ( n + ) th term, i.e., b = a + [( n + ) – ] d = a + ( n + ) d. This gives b d Thus, n numbers between a and b are as follows: A = a + d = a + b A = a + d = a + ( b A = a + d = a + ( b ..... ..... .....

..... A n = a + nd = a + n b Example Insert numbers between and such that the resulting sequence is an A.P. Solution Let A , A , A , A , A and A be six numbers between and such that , A , A , A , A , A , A , are in A.P. Here, a = , b = , n = .

Therefore, = + ( – ) d , so that d = . Thus A = a + d = + = ; A = a + d = + × = ; A = a + d = + × = ; A = a + d = + × = ; A = a + d = + × = ; A = a + d = + × = . Hence, six numbers between and are , , , , and . SEQUENCES AND SERIES EXERCISE .

. Find the sum of odd integers from to . . Find the sum of all natural numbers lying between and , which are multiples of .

. In an A.P., the first term is and the sum of the first five terms is one-fourth of the next five terms. Show that th term is – . .

How many terms of the A.P. – , , – , … are needed to give the sum – ? . In an A.P., if p th term is q and q th term is p , prove that the sum of first pq terms is ( pq + ), where p ≠ q .

. If the sum of a certain number of terms of the A.P. , , , … is . Find the last term.

. Find the sum to n terms of the A.P., whose k th term is k + . . If the sum of n terms of an A.P.

is ( pn + qn ), where p and q are constants, find the common difference. . The sums of n terms of two arithmetic progressions are in the ratio n + : n + . Find the ratio of their th terms.

. 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. .

Sum of the first p , q and r terms of an A.P. are a , b and c , respectively. Prove that b c q p p q p q . The ratio of the sums of m and n terms of an A.P.

is m : n . Show that the ratio of m th and n th term is ( m – ) : ( n – ). . If the sum of n terms of an A.P.

is n + n and its m th term is , find the value of m . . Insert five numbers between and such that the resulting sequence is an A.P. .

If b b is the A.M. between a and b , then find the value of n . . Between and , m numbers have been inserted in such a way that the resulting sequence is an A.

P. and the ratio of th and ( m – ) th numbers is : . Find the value of m . MATHEMATICS .

A man starts repaying a loan as first instalment of Rs. . If he increases the instalment by Rs every month, what amount he will pay in the th instalment? .

The difference between any two consecutive interior angles of a polygon is °. If the smallest angle is ° , find the number of the sides of the polygon. . Geometric Progression (G .

P.) Let us consider the following sequences: (i) , , , ,..., (ii) – – , , , ... (iii) . ,. ,.

,... 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 , , , and so on.

In (ii), we observe, a , , , 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 ; in (ii), it is – and in (iii), the constant ratio is .

. Such sequences are called geometric sequence or geometric progression abbreviated as G.P. A sequence a , a , a , …, a n , … is called geometric progression , if each term is non-zero and k k + = r (constant), for k ≥ . By letting a = a , we obtain a geometric progression, a , ar , ar , ar ,…., 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 , – and . , 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, SEQUENCES AND SERIES n = the numbers of terms, S n = the sum of first n terms.

. . 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 = ar . Similarly, third term is obtained by multiplying a by r .

Thus, a = a r = ar , and so on. We write below these and few more terms. st term = a = a = ar – , nd term = a = ar = ar – , rd term = a = ar = ar – th term = a = ar = ar – , th term = a = ar = ar – Do you see a pattern? What will be th term?

a = ar – = ar Therefore, the pattern suggests that the n th term of a G.P. is given by ar − . Thus, a , G.P. can be written as a , ar , ar , ar , … ar n – ; a , ar , ar ,..., ar n – ...

;according as G.P. is finite or infinite , respectively. The series a + ar + ar + ... + ar n – or a + ar + ar + ...