Sequences and Series Class 11
In Sequences and Series class 11 chapter 9 deals with the study of sequences which follow a specific pattern called progression. In this chapter, the concepts such as arithmetic progression (A.P), geometric mean, arithmetic mean, the relationship between A.M. and G.M., special series in forms of the sum to n terms of consecutive natural numbers, sum to n terms of squares and cubes of natural numbers will also be studied.
Sequence and Series Class 11 Concepts
The topics and subtopics covered in chapter 9 – Sequences and Series class 11 concepts are:
- Arithmetic Progression (A.P.)
- Arithmetic mean
- Geometric Progression (G.P.)
- The general term of a G.P.
- Sum to n terms of a G.P.
- Geometric Mean (G.M.)
- Relationship Between A.M. and G.M.
- Sum to n Terms of Special Series
Sequences and Series Class 11 Notes
The different numbers occurring in any particular sequence are known as terms. The terms of a sequence are denoted by
a1, a2, a3,….,an
If a sequence has a finite number of terms then it is known as a finite sequence. A sequence is termed as infinite if it is not having a definite number of terms. The nth term of an AP is given by
a + (n-1) d.
Between any two numbers ‘a’ and ‘b’, n numbers can be inserted such that the resulting sequence is an Arithmetic Progression. A1, A2, A3,……,An be n numbers between a and b such that a, A1 , A2 , A3,……,An, b is in A.P.
Here, a is the 1st term and b is (n+2)th term. Therefore,
b = a + d[(n + 2) – 1] = a + d (n + 1).
Hence, common difference (d) = (b-a)/(n+1)
Now, A1= a+d= a+((b-a)/(n+1))
A2= a+2d = a + ((2(b-a)/(n+1))
An = a+nd= a + ((n(b-a)/(n+1))}
The nth term of a geometric progression is given by an = arn-1
Sum of nth term:
Sn = n/2 [2a + (n-1)d]
where n = number of terms, a = first term and d = common difference
Q.1: If an = 2n, then find the first five terms of the series.
Solution: Given: an = 2n
On substituting n = 1, 2, 3, 4, 5, we get
a1 = 21 = 2
a2 = 22 = 4
a3 = 23 = 8
a4 = 24 = 16
a5 = 25 = 32
Therefore, the required terms are 2, 4, 8, 16, and 32.
Q.2: Find the sum of odd integers from 1 to 2001.
The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.
It clearly forms a sequence in A.P.
Where, the first term, a = 1
Common difference, d = 2
a + (n -1)d = 2001
1 + (n-1)(2) = 2001
2n – 2 = 2000
2n = 2000 + 2 = 2002
n = 1001
Sn = n/2 [2a + (n-1)d]
Sn = 1001/2[2×1+(1001-1)2]
Sn = 1001/2[2+1000×2]
Sn = 1001/2
Sn = 1001 x 1001
Sn = 1002001
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