Sequences and Series ISC Class-11 Maths ML Aggarwal

Sequences and Series ISC Class-11 Maths ML Aggarwal Solutions Chapter-10. Step by step Solutions of ML Aggarwal ISC Class-11 Mathematics with Exe-1,  Exe-2, Exe-3 Exe-4, Exe-5, Exe-6, Exe-7, Exe-8, Exe-9, Exe-10 and Chapter Test Questions. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Sequences and Series ISC Class-11 Maths ML Aggarwal

Sequences and Series ISC Class-11 Maths ML Aggarwal Solutions Chapter-10

Board   ISC
Class  11
Subject Mathematics
Chapter-10 Sequences and Series
Session  2024-25
Topics  Solutions of ML Aggarwal

Sequences and Series

Sequence: A succession of numbers arranged in a definite order according to a given certain rule is called sequence. A sequence is either finite or infinite depending upon the number of terms in a sequence.

Series:  If a1, a2, a3,…… an is a sequence, then the expression a1 + a2 + a3 + a4 + … + an is called series.

Progression : A sequence whose terms follow certain patterns are more often called progression

Arithmetic Progression (AP)

A sequence in which the difference of two consecutive terms is constant, is called Arithmetic progression (AP).

Properties of Arithmetic Progression (AP): If a sequence is an A.P. then its nth term is a linear expression in n i.e. its nth term is given by An + B, where A and S are constant and A is common difference.

nth term of an AP : If a is the first term, d is common difference and l is the last term of an AP then

  • nth term is given by an = a + (n – 1)d.
  • nth term of an AP from the last term is a’n =an – (n – 1)d.
  • an + a’n = constant
  • Common difference of an AP i.e. d = an – an-1,∀ n > 1

Note:  If a constant is added or subtracted from each term of an AR then the resulting sequence is an AP with same common difference.

Sum of n Terms of an AP

Sn = n/2 [2a + (n – 1)d]

Arithmetic Mean:  if a, and b are in A.P then A  = (a+b) /2 is called the arithmetic mean of a and b

Geometric Progression (GP)

A sequence in which the ratio of two consecutive terms is constant is called geometric progression. The constant ratio is called common ratio(r).

If a is the first term and r is the common ratio, then the general term or nth term of GP is an =arn-

nth term of a GP from the end is a’n = 1/ r (n1),  l = last term

If all the terms of GP be multiplied or divided by same non-zero constant, then the resulting sequence is a GP with the same common ratio.

The reciprocal terms of a given GP form a GP.

If each term of a GP be raised to some power, the resulting sequence also forms a GP

If a, b and c are three consecutive terms of a GP then b2 = ac


Exe-10.1

Sequences and Series ISC Class-11 Maths Chapter-10


Exe-10.2

Sequences and Series ISC Class-11 Maths ML Aggarwal Chapter-10


Exe-10.3

Sequences and Series ISC Class-11 Maths  Solutions Chapter-10


Exe-10.4

Sequences and Series ISC Class-11 Maths Solutions Chapter-10


Exe-10.5

 Solutions of Sequences and Series ISC Class-11 Maths Chapter-10


Exe-10.6

Solutions of Sequences and Series  for ISC Class-11 Maths Ch-10


Exe-10.7

ML Aggarwal Solutions for Class 11


Exe-10.8

 Sequences and Series ISC Class-11 Maths ML Aggarwal Ch-10


Exe-10.9

 Sequences and Series ISC Class-11 Maths Ch-10


Exe-10.10

Sequences and Series ISC Class-11 Maths Ch-10


Chapter Test

 Sequences and Series ISC Class-11 Maths Ch-10

-: End of Sequences and Series ISC Class-11 Maths ML Aggarwal  Solution :-

Return to :-  ML Aggrawal ISC Class-11 Maths Vol-1 Solutions

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18 thoughts on “Sequences and Series ISC Class-11 Maths ML Aggarwal”

    • dear student / well wisher / Teacher
      the previous version of 2020-21 has been removed because council has decided to start new session from 1st April Therefore we are upgrading the solutions of 2021-22 editions
      Sorry for inconvenience
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      team icsehelp

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  1. It is useless website I ever seen in my life I came here for ML Aggrawal class 11 solution but when i click on solution it says that this content is protected . What is this I request you to solve his problem as soon as possible

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