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**NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- **In this article, you will get NCERT solutions for Class 11 Maths Chapter 9 miscellaneous exercise. The miscellaneous exercise consists of mixed kinds of questions from all the topics covered in this chapter. You must have solved the previous exercise of this chapter. Now you can try to solve problems from the miscellaneous exercise chapter 9 Class 11.

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- NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise- Download Free PDF
- NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series Miscellaneous Exercise
- Access Sequences And Series Class 11 Chapter 9:Miscellaneous Exercise
- Question:1 Show that the sum of and terms of an A.P. is equal to twice the term.
- More About NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-
- Topic Covered in Class 11 Chapter 9 Maths Miscellaneous Solutions
- Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise:-
- Key Features of Class 11 Maths ch 9 Miscellaneous Exercise Solutions
- NCERT Solutions of Class 11 Subject Wise
- NCERT Solutions for Class 11 Maths
- Subject Wise NCERT Exampler Solutions

As you have already familiar with the NCERT syllabus Class 11 Maths chapter 9 concepts, you can try to solve problems from miscellaneous exercise chapter 9 Class 11. These problems are a bit lengthy and complex as compared to the other exercise of this chapter but these problems will check your understanding of this chapter. You don't need to worry if you are not able to solve these problems by yourself at first. Class 11 Maths Chapter 9 miscellaneous exercise solutions are here to help you where you will get detailed miscellaneous solutions. The class 11 maths ch 9 miscellaneous exercise solutions Maths Miscellaneous Exercises are meticulously crafted by subject experts from Careers360. Each step is explained in detail, providing comprehensive and well-explained solutions. Furthermore, the availability of a PDF version of class 11 chapter 9 maths miscellaneous solutions allows students to access the solutions offline, providing flexibility and convenience to study according to their preferences. You can check NCERT Solutions link if you are looking for NCERT solutions for all the Classes at one place.

**Also, see**

- Sequences and Series Exercise 9.1
- Sequences and Series Exercise 9.2
- Sequences and Series Exercise 9.3
- Sequences and Series Exercise 9.4

** As per the CBSE Syllabus for the academic year 2023-24, it has been noted that miscellaneous exercise class 11 chapter 9 has been renumbered and is now recognized as Chapter 8.

Answer:

Let a be first term and d be common difference of AP.

Kth term of a AP is given by,

Hence, the sum of and terms of an A.P. is equal to twice the term.

Question:2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Answer:

Let three numbers of AP are a-d, a, a+d.

According to given information ,

When d=3, AP= 5,8,11 also if d=-3 ,AP =11,8,5.

Thus, three numbers are 5,8,11.

Question:3 Let the sum of n, 2n, 3n terms of an A.P. be , respectively, show that

Answer:

Let a be first term and d be common difference of AP.

Subtract equation 1 from 2,

Hence, the result is proved.

Question:4 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Answer:

Numbers divisible by 7 from 200 to 400 are

This sequence is an A.P.

Here , first term =a =203

common difference = 7.

We know ,

The sum of numbers divisible by 7from 200 to 400 is 8729.

Question:5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer:

Numbers divisible by 2 from 1 to 100 are

This sequence is an A.P.

Here , first term =a =2

common difference = 2.

We know ,

Numbers divisible by 5 from 1 to 100 are

This sequence is an A.P.

Here , first term =a =5

common difference = 5.

We know ,

Numbers divisible by both 2 and 5 from 1 to 100 are

This sequence is an A.P.

Here , first term =a =10

common difference = 10

We know ,

Thus, the sum of integers from 1 to 100 that are divisible by 2 or 5 is 3050.

Question:6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer:

Numbers divisible by 4, yield remainder as 1 from 10 to 100 are

This sequence is an A.P.

Here , first term =a =13

common difference = 4.

We know ,

The sum of numbers divisible by 4 yield 1 as remainder from 10 to 100 is 1210.

Question:7 If f is a function satisfying f (x +y) = f(x) f(y) for all x, y N such that f(1) = 3 and

, find the value of n.

Answer:

Given : f (x +y) = f(x) f(y) for all x, y N such that f(1) = 3

Taking , we have

is forms a GP with first term=3 and common ratio = 3.

Therefore,

Thus, 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

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

...............................................1

from an AP.

....................................................................2

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,

Let the be GP as

Thus, the common ratio is 4.

Answer:

Given : first term =a=11

Let AP be

Given: The sum of the first four terms of an A.P. is 56.

Also, The sum of the last four terms is 112.

Thus, the number of terms of AP is 11.

Question:13 If then show that a, b, c and d are in G.P.

Answer:

Given :

Taking ,

Taking,

From equation 1 and 2 , we have

Thus, a,b,c,d are in GP.

Question:14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that

Answer:

Ler there be a GP

According to given information,

To prove :

LHS :

Hence proved

Question:15 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that

Answer:

Given: The pth, qth and rth terms of an A.P. are a, b, c, respectively.

To prove :

Let the first term of AP be 't' and common difference be d

Subtracting equation 2 from 1, we get

Subtracting equation 3 from 2, we get

Equating values of d, from equation 4 and 5, we have

Hence proved.

Question:16 If are in A.P., prove that a, b, c are in A.P.

Answer:

Given: are in A.P.

Thus, a,b,c are in AP.

Question:17 If a, b, c, d are in G.P, prove that are in G.P.

Answer:

Given: a, b, c, d are in G.P.

To prove: are in G.P.

Then we can write,

Let be in GP

LHS:

Hence proved

Thus, are in GP

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

and

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:19 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,

...................................................................1

We get,

.....................................................2

From 1 and 2, we get

Putting the value of a in equation 1, we have

Question:20 If a, b, c are in A.P.; b, c, d are in G.P. and 1/c , 1/d , 1/e are in A.P. prove that a, c, e are in G.P.

Answer:

Given: a, b, c are in A.P

Also, b, c, d are in G.P.

Also, 1/c, 1/d, 1/e are in A.P

To prove: a, c, e are in G.P. i.e.

From 1, we get

From 2, we get

Putting values of b and d, we get

Thus, a, c, e are in G.P.

Question:21(i) Find the sum of the following series up to n terms:

Answer:

is not a GP.

It can be changed in GP by writing terms as

to n terms

Thus, the sum is

Question:21(ii) Find the sum of the following series up to n terms: .6 +. 66 +. 666+…

Answer:

Sum of 0.6 +0. 66 + 0. 666+….................

It can be written as

to n terms

Question:22 Find the 20th term of the series terms.

Answer:

the series =

Thus, the 20th term of series is 1680

Question:23 Find the sum of the first n terms of the series: 3+ 7 +13 +21 +31 +…

Answer:

The series: 3+ 7 +13 +21 +31 +…..............

n th term =

Question:24 If are the sum of first n natural numbers, their squares and their cubes, respectively, show that

Answer:

To prove :

From given information,

Here ,

Also,

From equation 1 and 2 , we have

Hence proved .

Question:25 Find the sum of the following series up to n terms:

Answer:

n term of series :

Here, are in AP with first term =a=1 , last term = 2n-1, number of terms =n

Question:26 Show that

Answer:

To prove :

the nth term of numerator

nth term of the denominator

Numerator :

Denominator :

From equation 1,2,3,we have

Hence, the above expression is proved.

Answer:

Given : Farmer pays Rs 6000 cash.

Therefore , unpaid amount = 12000-6000=Rs. 6000

According to given condition, interest paid annually is

12% of 6000,12% of 5500,12% of 5000,......................12% of 500.

Thus, total interest to be paid

Here, is a AP with first term =a=500 and common difference =d = 500

We know that

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 , 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, total interest to be paid

Here, is a AP with first term =a=1000 and common difference =d = 1000

We know that

Sum of AP:

Thus, interest to be paid :

Thus, cost of tractor = Rs. 22000+ Rs. 17100 = Rs. 39100

Answer:

The numbers of letters mailed forms 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.

Interest in fifteen year 10000+ 14 times Rs. 500

Amount in 15 th year

Amount in 20 th year

Answer:

Cost of machine = Rs. 15625

Machine depreciate each year by 20%.

Therefore, its value every year is 80% of the original cost i.e. of the original cost.

Value at the end of 5 years

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 is a AP

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.

Class 11 Maths chapter 9 miscellaneous solutions consist of questions related to finding the nth term of the arithmetic progression, the sum of the terms of arithmetic progression, arithmetic mean, nth term of the geometric progression, the sum of the terms of geometric progression, geometric mean, etc. There are few solved examples given before the miscellaneous exercise chapter 9 Class 11 that you can try to solve.

The Miscellaneous Exercise in NCERT Solutions for Class 11 Maths Chapter 9 - Sequences and Series covers the following topics:

- Introduction
- Sequences
- Series
- Arithmetic Progression (A.P.)
- Geometric Progression (G.P.)
- Relationship Between A.M. and G.M.
- Sum to n terms of Special Series

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Download EBookBy engaging with the concepts presented in NCERT Solutions for Class 11 Maths, students can address any doubts related to these topics, establishing a strong foundation for their understanding of Class 12 Maths.

**Also Read| **Sequences And Series Class 11 Notes

- There are 32 questions in the miscellaneous exercise chapter 9 Class 11 which you can try to solve by yourself.
- You can use Class 11 Maths chapter 9 miscellaneous exercise solutions when you are finding it difficult to solve miscellaneous problems.
- Class 11 Maths chapter 9 miscellaneous solutions are designed by subject matter experts in a detailed manner that could be understood by an average student also.

**Comprehensive Approach:**The miscellaneous exercise class 11 chapter 9 offers a thorough explanation of topics, ensuring a well-rounded understanding of Sequences and Series.**Stepwise Clarity:**Each problem in Class 11 maths miscellaneous exercise chapter 9 is solved with step-by-step solutions, facilitating a logical and clear progression of concepts for Class 11 students.**Accessible Language:**The class 11 chapter 9 miscellaneous exercise solutions are presented in clear and concise language, making intricate mathematical concepts more accessible for students in Chapter 9 of Class 11 Maths.**Variety of Problems:**The Miscellaneous Exercise in Class 11 Maths Chapter 9 provides a diverse range of problems, catering to different difficulty levels and scenarios, allowing for a comprehensive understanding.**Free PDF Access:**Solutions for Chapter 9 of Class 11 Maths Miscellaneous Exercise are accessible in a free PDF format, providing students with easy access to study materials according to their convenience.

**Also see-**

**Happy learning!!!**

1. What is the meaning of sequence ?

Sequence means an arrangement of numbers in a definite order according to some rule.

2. What is the meaning of finite sequence ?

The finite sequence is s sequence containing a finite number of terms.

3. What is the meaning of infinite sequence ?

The infinite sequence is s sequence containing an infinite number of terms.

4. What is the meaning of series ?

If the terms of a sequence are expressed as the sum of terms then it is called a series.

5. What is the sum of A.P. if the first term is 'a' and the last term of the A.P. is 'l' and the number of terms are 'n' ?

Sum of A.P. = n(a+l)/2

6. What is the arithmetic mean of two numbers 5, 9 ?

Arithmetic mean (A.M.) = (5+9)/2 = 7

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