Focused on dealing with a list of objects and events (in a certain fashion), sequences and series teach students all about different functions that dominate a sequence chain. Further, sequences and sets also provide a description of different operations that are used for adding different quantities of a sum. The chapter Sequences and Series Class 11 is about mathematical concepts that are based around lines and series. Students can refer to the CBSE Class 11 Sequence and Series study resource on Vedantu to the to get a complete idea of the basic calculations of Sequence and Series at its core.
About Sequence and Series
In mathematics, a sequence is a compile list of objects or events that are ordered or placed in a sequential fashion, created in such a way that each member of a given sequence comes before or after every other member in the list of sequences.
More precisely, a sequence is a mathematical expression where the function paired to a domain is equal to the set of positive integers.
Key Terms in Sequences and Series
Sequence: An aggregative function whose domain is a subset of natural numbers. A sequence can often be denoted in a fashion: 1, 2, 3,… ,n, as f1, f2, f3, …., fn , where fn = f(n).
Real Sequence: Sequence where the range is a subset of R is called a real sequence.
Series: In the mathematical expression where a1, a2, a3 , … , it is a sequence and the expression a1 + a2 + a3 + … + an is a series.
Progression: Sequence where there are certain rules to follow, becomes a progression.
Finite Series: A series that is cued with a finite number of terms is called finite series.
Infinite Series: A series that comprises an infinite number of terms is called infinite series.
Solved Examples of Sequence and Series
Listed below is a compilation of different types of sums that you may come across in the chapter, Sequences and Series. It is a composition of different types and levels of sums from the chapter.
Q1. Between any two numbers ‘a’ and ‘b’, show how the ‘n’ numbers can be inserted, in such a formation that the resulting sequence is an Arithmetic Progression.
Let, the sequence start with the following progression:
B1, B2, B3,……, Bn. Let the suggested sequence have ‘n’ numbers, in such a manner that between b and c, C1, C2, C3,……, Cn, c is in A.P.
Here, b is the 1st term and b is (n+2)th term. Therefore,
c = b + d[(n + 2) – 1] = b + d (n + 1).
Hence, common difference (d) = (c – b)/(n + 1)
Now, A1 = b + d = b + ((c – b)/(n + 1))
B2 = b + 2d = b + ((2(c – b)/(n + 1))
Bn = b + nd= b + ((n(c – b)/(n + 1))}
Therefore, the nth term around the suggested geometric progression is highlighted with the help of
Sum of the nth term:
Sn = n/2 [2b + (n-1)d]
In the solution stated above, n denoted the number of terms, b denoted the first term and d is about the common difference.
Q2. Derive the first five terms of the given terms where:
an = n (n + 2)
nth term of a sequence an = n (n + 2)
By substituting n = 1, 2, 3, 4, and 5, we get the first five terms
a1 = 1(1 + 2) = 3
a2 = 2(2 + 2) = 8
a3 = 3(3 + 2) = 15
a4 = 4(4 + 2) = 24
a5 = 5(5 + 2) = 35
Therefore, the required terms are 3, 8, 15, 24, and 35.