##### Tallentex 2025 - ALLEN's Talent Encouragement Exam

ApplyRegister for Tallentex '25 - One of The Biggest Talent Encouragement Exam

Edited By Vishal kumar | Updated on Nov 08, 2023 10:28 AM IST

**NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.3-**In the previous exercise, you have already learned about the arithmetic progression, nth term and sum of the arithmetic progression, arithmetic mean, etc. In NCERT syllabus Class 11 Maths chapter 9 exercise 9.3, you will solve questions on geometric progression, nth term of the geometric progression, the sum of the geometric progression, geometric mean, relationship between the arithmetic mean and geometric mean, etc. You must have learned about the geometric progression, the nth term of the geometric progression, and the sum of the terms of the geometric progression in the previous classes. As you are familiar with these topics, it won't take much effort to understand these topics.

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

**Suggested: ****JEE Main: high scoring chapters | ****Past 10 year's papers**

This Story also Contains

- NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.3- Download Free PDF
- Download the PDF of NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.3
- Access Sequences And Series Class 11 Chapter 9 Exercise 9.3
- Question:1 Find the and terms of the G.P.
- More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3:-
- Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3:-
- NCERT Solutions of Class 11 Subject Wise
- Subject Wise NCERT Exampler Solutions

You will learn some new topics like arithmetic mean, geometric mean and the relationship between the arithmetic mean and geometric mean in the exercise 9.3 Class 11 Maths. Arithmetic mean and geometric mean are very similar to arithmetic progression and geometric progression respectively, so you will understand them very easily. Check NCERT Solutions if you are looking for NCERT solutions for Class 6 to Class 12 at one place. All the solutions are expertly crafted by subject experts at Careers360 in a user-friendly and detailed manner. Furthermore, PDF versions of these solutions are available for free, providing students with convenient access.

**Also, see**

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

****As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 8.**

Answer:

G.P :

first term = a

common ratio =r

the nth term of G.P

Question:2 Find the term of a G.P. whose term is 192 and the common ratio is 2.

Answer:

First term = a

common ratio =r=2

term is 192

is the term of a G.P.

Question:3 The terms of a G.P. are p, q and s, respectively. Show that

Answer:

To prove :

Let first term=a and common ratio = r

Dividing equation 2 by 1, we have

Dividing equation 3 by 2, we have

Equating values of , we have

Hence proved

Question:4 The term of a G.P. is square of its second term, and the first term is -3. Determine its term.

Answer:

First term =a= -3

term of a G.P. is square of its second term

Thus, seventh term is -2187.

Question:5(a) Which term of the following sequences:

Answer:

Given :

n th term is given as 128.

The, 13 th term is 128.

Question:5(b) Which term of the following sequences:

Answer:

Given :

n th term is given as 729.

The, 12 th term is 729.

Question:5(c) Which term of the following sequences:

Answer:

Given :

n th term is given as

Thus, n=9.

Question:6 For what values of x, the numbers are in G.P.?

Answer:

Common ratio=r.

Thus, for ,given numbers will be in GP.

Answer:

geometric progressions is 0.15, 0.015, 0.0015, ... .....

a=0.15 , r = 0.1 , n=20

Question:9 Find the sum to 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 first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

Answer:

Given : The sum of first three terms of a G.P. is and their product is 1.

Let three terms be .

Product of 3 terms is 1.

Put value of a in equation 1,

The three terms of AP are .

Question:13 How many terms of G.P. , … 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 value of n as 4 to get sum of 120.

Answer:

Let GP be

Given : The sum of first three terms of a G.P. is 16

Given : the sum of the next three terms is128.

Dividing equation (2) by (1), we have

Putting value of r =2 in equation 1,we have

Question:15 Given a G.P. with a = 729 and term 64, determine

Answer:

Given a G.P. with a = 729 and term 64.

(Answer)

Question:16 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

Answer:

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

If r=2, then

If r= - 2, then

Thus, required GP is or

Question:17 If the terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.

Answer:

Let x,y, z are in G.P.

Let first term=a and common ratio = r

Dividing equation 2 by 1, we have

Dividing equation 3 by 2, we have

Equating values of , we have

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 as

to n terms

Answer:

Here, is a GP.

first term =a=4

common ratio =r

Question:20 Show that the products of the corresponding terms of the sequences form a G.P, and find the common ratio.

Answer:

To prove : is a GP.

Thus, the above sequence is a GP with common ratio of rR.

Answer:

Let first term be a and 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 18.

Dividing equation 2 by 1 , we get

Putting value of r , we get

Thus, four terms of GP are

Question:22 If the terms of a G.P. are a, b and c, respectively. Prove that

Answer:

To prove :

Let A be the first term and R be common ratio.

According to the given information, we have

L.H.S :

=RHS

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, is a AP.

Put in equation (2),

Hence proved .

Question:24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from term is

Answer:

Let first term =a and common ratio = r.

Since there are n terms from (n+1) to 2n term.

Sum of terms from (n+1) to 2n.

Thus, the required ratio =

Thus, the common ratio of the sum of first n terms of a G.P. to the sum of terms from term is .

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),

= LHS

Hence proved

Question:26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer:

Let A, B be two numbers between 3 and 81 such that series 3, A, B,81 forms 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 may be the geometric mean between a and b.

Answer:

M of a and b is

Given :

Squaring both sides ,

Question:28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio

Answer:

Let there be two numbers a and b

geometric mean

According to the given condition,

.............................................................(1)

Also,

.......................................................(2)

From (1) and (2), we get

Putting the value of 'a' in (1),

Thus, the ratio is

Question:29 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are

Answer:

If A and G be A.M. and G.M., respectively between two positive numbers,

Two numbers be a and b.

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

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

We know

Put values from equation 1 and 2,

..................................................................3

From 1 and 3 , we have

Put value of a in equation 1, 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 and nth hour are 120,480 and respectively.

Answer:

Given: 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 roots of the quadratic equation be a and b.

According to given condition,

We know that

Thus, the quadratic equation =

Class 11 Maths chapter 9 exercise 9.3 consists of questions related to finding the nth term of the geometric progression, the sum of the terms of geometric progression, arithmetic mean, geometric mean, etc. The Class 11 Maths chapter 9 exercise 9.3 is quite lengthy as compared to other exercises of this chapter. Also, exercise 9.3 Class 11 Maths is the most important exercise of this chapter as it covers topics such as geometric mean, arithmetic mean.

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

- NCERT book Class 11 maths chapter 9 exercise 9.3 is more lengthy and many concepts of progression are covered in it.
- Most of the questions in exercise 9.3 Class 11 Maths are conceptually based on the geometric progression and arithmetic progression concepts.
- You must try to solve all the problems of this exercise on your own before moving to the NCERT solutions for Class 11 Maths chapter 9 exercise 9.3.

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook**Comprehensive Coverage:**The answers encompass all the ex 9.3 class 11 problem of the NCERT 11th Class Mathematics textbook.**Step-by-Step Solutions:**The class 11 maths ex 9.3 solutions are presented in a step-by-step format, making it easier for students to understand and follow the problem-solving process.**Clarity and Precision:**The class 11 ex 9.3 answers are written with clarity and precision, ensuring that students can grasp the mathematical concepts and methods required for effective problem-solving.**Conceptual Understanding:**The 11th class maths exercise 9.3 answers aim to promote a deep understanding of mathematical concepts rather than rote memorization, encouraging critical thinking and problem-solving skills.**Free Accessibility:**Typically, these class 11 maths ex 9.3 are available free of charge, allowing students to access them without any cost, making it a valuable resource for self-study.**Supplementary Learning Resource:**These ex 9.3 class 11 answers can be used as a supplementary learning resource to reinforce classroom instruction and aid in exam preparation.**Homework and Practice:**Students can utilize these answers to verify their work, practice problem-solving, and enhance their overall performance in mathematics.

**Happy learning!!!**

1. Find the 5th term of G.P. 2,4,8...

Given a = 2

r = 4/2 = 2

a_5 = ar^4 = 2(2)^4 = 32

2. If the first term and third term of the G.P. are 3, 48 respectively than find the common ratio of the G.P. ?

Given a = 3

ar^2 = 48

3(r^2) = 48

r^2 = 16

Common ratio (r) = 4

3. Find the arithmetic mean of two positive integers a and b ?

A.M. = (a+b)/2

4. Find the geometric mean of two positive integers a and b ?

G.M. = (ab)^(1/2)

5. What is the sum of first n natural numbers?

Sum of first n natural numbers = n(n+1)/2

6. What is the sum of square of first n natural numbers?

Sum of square of first n natural numbers = n(n+1)(2n+1)/6

Get answers from students and experts

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

Accepted by more than 11,000 universities in over 150 countries worldwide

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer!

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

News and Notifications

Back to top