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Sequence and Series Maths Chapter 8 Class 11 are important topics in the Class 11 CBSE Syllabus of Maths. It is a branch of class 11th that deals with arranged numbers, known as sequences, and their sum, known as series. In this chapter, we study sequences and series ncert solutions, which include Geometric Progression (GP), answers to FAQs of finding general terms, sum of terms, and properties. This article helps you to make the concept of Sequence And Series Notes easier for Class 11 students. With the help of sequences and series class 11 solutions, we will discuss questions such as, 'What is a sequence?', 'How do we find the nth term of a geometric sequence?', and 'What is the sum of the first n terms in a geometric series?'
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NCERT Solutions for Class 11 Maths Chapter 8 Sequence and Series, crafted by experts of Careers360, provide detailed study material for students preparing for the Class 11 CBSE exam. In this article, we cover all Sequences and series class 11 NCERT solutions and important topics of the chapter, with brief explanations and detailed methods for solving problems by applying suitable formulae. NCERT Solutions for Class 11 provide us understanding of the concepts and build confidence in handling sequence and series questions. For the best results for class 11 maths chapter 8 question answer, students are encouraged to refer to the NCERT Class 11 Maths book, and for more practice, go through NCERT Exemplar Solutions For Sequence And Series.
Progression: A sequence whose terms follow certain patterns is known as a progression.
Geometric Progression (GP):
A sequence in which the ratio of two consecutive terms is constant is called a geometric progression (GP).
The constant ratio is called the common ratio (r), i.e.,
The general term or nth term of GP is
The nth term of a GP from the end:
If a, b, and c are three consecutive terms of a GP, then
Geometric Mean (GM):
If a, G, and b are in a GP, then G is called the geometric mean of a and b and is given by
If a, G1, G2, G3, …, Gn, b are in GP, then G1, G2, G3, …, Gn are in GMs between a and b.
The common ratio r is given as:
The GM of a1, a2, a3,…, an is given by:
Product of n GMs is
Relationship between A.M. and G.M.
Thus, we have
Thus, we get the relationship
Sequences and Series class 11 questions and answers: Exercise: 8.1 |
Question:1 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms = 3, 8, 15, 24, 35
Question:2 Write the first five terms of each of the sequences in Exercises 1 to 6 whose terms are:
Answer:
Given :
Therefore, the required number of terms
Question:3 Write the first five terms of each of the sequences in Exercises 1 to 6 whose terms are:
Answer:
Given :
Therefore, the required number of terms
Question:4 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:5 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:6 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:7 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Put
Put n = 24,
Hence, we have
Question:8 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n=7,
Hence, we have
Question:9 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n = 9,
The value of
Question:10 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n = 20,
Hence, value of
Answer:
Given :
Hence, the five terms of the series are
Series
Answer:
Given :
Hence, five terms of series are
Series
Answer:
Given :
Hence, the five terms of the series are
Series
Sequences and Series class 11 questions and answers: Exercise: 8.2 |
Question:1 Find the
Answer:
G.P :
first term = a
common ratio = r
Question:2 Find the
Answer:
First term = a
common ratio = r = 2
Question:3 The
Answer:
To prove :
Let first term and common ratio = r
Dividing equation 2 by 1, we have
Dividing equation 3 by 2, we have
Equating values of
Hence proved.
Question:4 The
Answer:
First term =a= -3
Thus, the seventh term is -2187.
Question:5 Which term of the following sequences:
(b) Which term of the following sequences:
(c) Which term of the following sequences:
Answer:
(a) Given :
nth term is given as 128.
The 13th term is 128.
(b) Given :
n The term is given as 729.
The 12th term is 729.
(c) Given :
nth term is given as
Thus, n = 9.
Question:6 For what values of x, the numbers
Answer:
Common ratio = r.
Thus, for
Answer:
Geometric progressions are 0.15, 0.015, 0.0015, ... .....
a = 0.15 , r = 0.1 , n = 20
Question:9 Find the sum to the indicated number of terms in each of the geometric progressions in
Answer:
The sum to the indicated number of terms in each of the geometric progressions is:
Question:12 The sum of the first three terms of a G.P. is
Answer:
Given: The sum of the first three terms of a G.P. is
Let three terms be
The product of 3 terms is 1.
Put the value of an in equation 1,
The three terms of AP are
Question:13 How many terms of G.P. $3, 3 ^ 2 3 ^ 3 … are needed to give the sum 120?
Answer:
G.P.=
Sum = 120
These terms are GP with a = 3 and r = 3.
Hence, we have a value of n as 4 to get a sum of 120.
Answer:
Let GP be
Given: The sum of the first three terms of a G.P. is 16
Given: the sum of the next three terms is 128.
Dividing equation (2) by (1), we have
Putting the value of r =2 in equation 1, we have
Question:15 Given a G.P. with a = 729 and
Answer:
Given a G.P. with a = 729 and
Question:16 Find a G.P. for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term
Answer:
Given the sum of the first two terms is – 4 and the fifth term is 4 times the third term
Let the first term be a and the common ratio be r
If r = 2, then
If r = - 2, then
Thus, required GP is
Question:17 If the
Answer:
Let x,y, and z be in G.P.
Let first term and common ratio = r
Dividing equation 2 by 1, we have
Dividing equation 3 by 2, we have
Equating values of
Thus, x, y, z are in GP.
Question:18 Find the sum to n terms of the sequence, 8, 88, 888, 8888….
Answer:
8, 88, 888, 8888… is not a GP.
It can be changed in GP by writing terms such s
Answer:
Here,
first term = a = 4
common ratio =r
Question:20 Show that the products of the corresponding terms of the sequences
Answer:
To prove :
Thus, the above sequence is a GP with a common ratio of rR.
Answer:
Let the first term be a and the common ratio be r.
Given: the third term is greater than the first term by 9, and the second term is greater than the
By dividing equation 2 by 1 we get
Putting the value of r, we get
Thus, four terms of GP are
Question:22 If the
Answer:
To prove :
Let A be the first term and R be the cthemon ratio.
According to the given information, we have
L.H.S :
Thus, LHS = RHS.
Hence proved.
Answer:
Given First term =a and n th term = b.
Common ratio = r.
To prove :
Then ,
P = product of n terms
Here,
Put in equation (2),
Hence proved.
Question:24 Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms from
Answer:
Let the first term =a and common ratio = r.
Since there are n terms from (n+1) to 2n term.
The sum of terms from (n+1) to 2n.
Thus, the required ratio =
Thus, the common ratio of the sum of the first n terms of a G.P. to the sum of terms from
Question:25 If a, b, c, and d are in G.P. show that
Answer:
If a, b, c, and d are in G.P.
To prove :
RHS :
Using equation (1) and (2),
Hence proved.
Question:26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Answer:
Let A, and B be two numbers between 3 and 81 such that series 3, A, B, and 81 form a GP.
Let a=first term and common ratio =r.
For
The required numbers are 9,27.
Question:27 Find the value of n so that
Answer:
M of a and b is
Given :
Squaring both sides
Question:28 The sum of two numbers is 6 times their geometric mean, showing numbers are in the ratio
Answer:
Let there be two numbers a and b
Geometric mean
According to the given condition,
Also,
From (1) and (2), we get
Putting the value of 'a' in (1),
Thus, the ratio is
Question:29 If A and G are A.M. and G.M., respectively between two positive numbers, prove that the numbers are
Answer:
If A and Garee A.M. and G.M., respectively between two positive numbers,
Two numberarebe a and b.
We know
Put values from equation 1 and 2,
From 1 and 3, we have
Put the value of an in equation 1, and we get
Thus, numbers are
Answer:
The number of bacteria in a certain culture doubles every hour. It forms GP.
Given a = 30 and r = 2.
Thus, bacteria present at the end of the 2nd hour, 4th hour, r, and nth hour are 120,480 and
Answer:
Given: The Bank pays an annual interest rate of 10% compounded annually.
Rs 500 amounts are deposited in the bank.
At the end of the first year, the amount
At the end of the second year, the amount
At the end of the third year, the amount
At the end of 10 years, the amount
Thus, at the end of 10 years, amount
Answer:
Let the roots of the quadratic equation be a and b.
According to the given conditions,
We know that
Thus, the quadratic equation =
Sequences and Series class 11 NCERT solutions: Miscellaneous Exercise |
Question:1 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y
Answer:
Given : f (x +y) = f(x) f(y) for all x, y
Taking
Therefore,
Thus, the value of n is 4.
Answer:
Let the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2
Therefore,
Thus, the value of n is 6.
Last term of GP=6th term
The last term of GP =160
Question:3 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Answer:
Given: The first term of a G.P. is 1. The sum of the third term and fifth term is 90.
Thus, the common ratio of GP is
Answer:
Let three terms of GP be
Then, we have
From equation 1 and 2, we get
If r = 2, GP = 8,16,32
If r=0.2, GP= 32,16,8.
Thus, the numbers required are 8,16,32.
Answer:
Let GP be
Number of terms = 2n
According to the given condition,
Lettheme be GP as
Thus, the common ratio is 4.
Question:6 If
Answer:
Given :
Takin,
Taking,
From equation 1 and 2, we have
Thus, a,b,c,d and are in GP.
Question:7 Let S be the sum, P the product, and R the sum of reciprocals of n terms in a G.P. Prove that
Answer:
Let there be a GP
According to the given information,
To prove :
LHS :
Hence proved.
Question:8 If a, b, c, d are in G.P, prove that
Answer:
Given: a, b, c, d are in G.P.
To prove:
Then we can write,
Let
LHS:
Hence proved.
Thus,
Answer:
Given: a and b are the roots of
Then,
Also, c, d are roots of
Given: a, b, c, d form a G.P
Let,
From 1 and 2, we get
On dividing them,
When , r = 2,
When, r = -2,
Case (1) when r=2 and x=1,
i.e. (q + p) : (q – p) = 17:15.
Case (2) when r = -2 and x = -3,
i.e. (q + p) : (q – p) = 17 : 15.
Question:10 The ratio of the A.M. and G.M. of two positive numbers a and b, is m n. Show that
Answer:
Let two numbers be a and b.
According to the given condition,
We get,
From 1 and 2, we get
Putting the value of an ann equation 1, we have
Question:11 Find the sum of the following series up to n terms:
(ii) .6 +. 66 +. 666+…
Answer:
(i)
It can be changed in GP by writing terms such s
Thus, the sum is
(ii) The sum of 0.6 +0. 66 + 0. 666+….................
It can be written as
Question:12 Find the 20th term of the series
Answer:
The series =
Thus, the 20th term of the series is 1680
Answer:
Given Farmer pays Rs 6000 cash.
Therefore, unpaid amount = 12000-6000=Rs. 6000
According to the condition, interest paid annually is
12% of 6000,12% of 5500,12% of 5000,......................12% of 500.
Thus, the total interest to be paid
Here,
We know that
The sum of AP:
Thus, interest to be paid :
Thus, cost of tractor = Rs. 12000+ Rs. 4680 = Rs. 16680
Answer:
Given: Shamshad Ali buys a scooter for Rs 22000.
Therefore, the unpaid amount = 22000-4000=Rs. 18000
According to the given condition, interest paid annually is
10% of 18000,10% of 17000,10% of 16000,......................10% of 1000.
Thus, the total interest to be paid
Here,
We know that
The sum of AP:
Thus, interest to be paid :
Thus, cost of tractor = Rs. 22000 + Rs. 17100 = Rs. 39100
Answer:
The number of letters mailed from a GP:
First term = a=4
Common ratio=r=4
Number of terms = 8
We know that the sum of GP is
Costs to mail one letter are 50 paise.
Cost of mailing 87380 letters
Thus, the amount spent when the 8th set of the letter is mailed is Rs. 43690.
Answer:
Given: A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
Answer:
Cost of machine = Rs. 15625
Machines depreciate each year by 20%.
Therefore, its value every year is 80% of the original cost i.e.
Thus, the value of the machine at the end of 5 years is Rs. 5120
Answer:
Let x be the number of days in which 150 workers finish the work.
According to the given information, we have
Series
First term=a=150
Common difference= -4
Number of terms = x+8
Since x cannot be negative so x=17.
Thus, in 17 days 150 workers finish the work.
Thus, the required number of days = 17+8=25 days.
If you are interested in Sequence and Series Maths chapter 8 class 11 exercises then these are listed below.
Sequence and Series are core concepts of algebra, and having basic knowledge in class 11 will help students build a strong base for higher-level topics of mathematics. Solving NCERT questions helps in formula applications that are essential for solving progressions. These NCERT solutions are specially designed by subject experts at Careers360. The answers are precisely explained. Since the solutions are prepared in a lucid manner, students can use their time efficiently during exams for the BSE board by directly focusing on the most important concepts and practicing questions efficiently.
Here are the subject-wise links for the NCERT solutions of class 11:
NCERT solutions for class 11 biology |
NCERT solutions for class 11 maths |
NCERT solutions for class 11 chemistry |
NCERT Solutions for Class 11 Physics |
Here are some useful links for NCERT books and NCERT syllabus for class 11:
Happy Reading !!!
Chapter 8 of Class 11 Maths focuses on sequences and series, which includes Arithmetic Progression (AP) and Geometric Progression (GP). It covers the nth term of an AP and GP, the sum of the first n terms of both AP and GP and the concept of Arithmetic Mean and Geometric Mean. Understanding the relationship between terms in these progressions, as well as the formulas to find the sum and nth terms, are key parts of this chapter.
The sum of the first
The nth term of a geometric progression (GP) is given by the formula
The key difference between arithmetic and geometric sequences lies in how the terms are generated. In an Arithmetic Progression (AP), the difference between any two consecutive terms is constant, referred to as the common difference. In contrast, a Geometric Progression (GP) is defined by a constant ratio between consecutive terms, known as the common ratio. These differences in structure give rise to different formulas and methods for calculating terms and sums in each sequence.
The sum of an infinite geometric series, provided the common ratio
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