 Hello and welcome to the session. In this session first we will discuss about sequences. By sequence we mean an arrangement of number in definite order according to some rule. To redefine a sequence as a function whose domain is the set of natural numbers, some subsets of the type 1, 2, 3 and so on up to k. Then a sequence containing finite number of terms is called a finite sequence and a sequence which is not finite is infinite sequence. Next we discuss series. Consider a1, a2, a3 and so on be a sequence then the sum expressed as a1 plus a2 plus a3 plus and so on is called a series. Series is finite if the given sequence is finite and a series is infinite if the given sequence is infinite. A series a1 plus a2 plus and so on up to an is abbreviated as summation ak where k goes from 1 to n. Now let's discuss arithmetic progression or ap and ap or arithmetic progression is a sequence in which terms increase or decrease regularly by the same constant that is we can say that a sequence a1, a2, a3 so on up to an and so on is called an arithmetic sequence or arithmetic progression if we have an n plus 1 is equal to an plus d where n belongs to n. Let us consider an ap in its standard form given by a, a plus d, a plus 2d and so on. So here we have a is the first term of the ap, d is called the common difference and the nth term or you can say the general term of the ap is given by an and that is equal to a plus n minus 1 into d where n is the number of terms in the sequence. Now let a, a plus d, a plus 2d and so on a plus n minus 1 into d be an ap then here we have the last term of the ap is denoted by l and that is equal to a plus n minus 1 into d then sum of n terms of an ap is denoted by sn is equal to n upon 2 multiplied by 2a plus n minus 1 into d or we can also write this as sn that is the sum of n terms of an ap is equal to n upon 2 into a plus l. Usually we denote the first term of an ap by a and the last term by l. Let us consider the series 1, 3, 5, 7, 9, 11 it has 6 terms so we have n is equal to 6. Now here a that is the first term is 1 common difference d is 2. Now sum of the 6 terms of this series given by s6 is equal to n upon 2 multiplied by 2 into a plus n minus 1 into d. So this is equal to 3 into 2 plus 10 which is equal to 3 into 12 and that is 36. So sum of the 6 terms of the series is 36. Next we discuss arithmetic mean suppose we are given two numbers a and b we can insert a number a between these two numbers a and b so that this is an ap this number a is called the arithmetic mean or you can say am of the given two numbers a and b that is we have a is equal to a plus b by 2. So you can see that am that is the arithmetic mean between two numbers is their average. Generally we can insert as many numbers as we like between the given two numbers such that the resulting sequence is an ap. Suppose we have two numbers a and b and we insert n numbers between them a1 a2 and so on up to an such that this resulting sequence is an ap. So the total terms in this ap would be n plus 2 terms where we get this b that is the last term is equal to a plus n plus 1 into b. So from here we get b that is the common difference is equal to b minus a upon n plus 1 now a1 would be equal to a plus d. So on substituting this value of d in this we get this is equal to a plus b minus a upon n plus 1 then a2 equal to a plus 2d is equal to a plus 2 into b minus a upon n plus 1 then a3 equal to a plus 3d is equal to a plus 3 into b minus a upon n plus 1 and so on like this we get an is equal to a plus nd is equal to a plus n into b minus a upon n plus 1. Suppose that we are given two numbers 1 and 11 we need to insert four numbers between them that is a1 a2 a3 a4 so that the resulting sequence is an ap. So from this we see that a is equal to 1 b is equal to 11 b that is the last term is equal to 11 is equal to a that is 1 plus n that is the total terms of this ap which would be 6 minus 1 into d. So from here we get d is equal to 11 minus 1 upon 5 so we get d is equal to 2. Thus we have the first number a1 is equal to a plus d that is equal to 1 plus 2 is equal to 3 then a2 is equal to a plus 2d that is equal to 1 plus 2 into 2 which is equal to 5 then a3 is equal to a plus 3d and that is equal to 1 plus 2 into into 3 and that is equal to 7 then a4 is equal to a plus 4d is equal to 1 plus 2 into 4 that is equal to 9. So we get the resulting ap is 1, 3, 5, 7, 9, 11. This is how we can insert any number of terms between given two numbers. This completes the session. Hope you have understood what is an ap, how we find the sum of an ap and how we can insert any number of terms between given two numbers.