NCERT Solutions of Class 11 Maths Exercise 9.3 -Sequences and Series
NCERT Solutions of Class 11 Maths Exercise 9.3 of the chapter 9-Sequences and Series are created here by a step by step method by an expert teacher of maths.These NCERT solutions of of exercise 9.3 are the solutions of chapter 9-Sequences and series of class 11 NCERT text book. All questions of exercise are based on Geometric Progression.If you Study these NCERT solutions till the end you will clear your concept that enable you in solving all questions based on the Geometric Progressions in maths question paper of class 11 CBSE board exam of 2020-21.
Class 11 maths NCERT Solutions of chapter 9-Sequences and Series
exercise 9.1 -Sequence and Series
Exercise 9.2- Sequences and Series
Exercise 9.4-Sequence and Series
NCERT Solutions of Class 11 Maths Exercise 9.3 of the chapter 9-Sequences and Series
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Q1. Find the 20^{th }and n^{th }terms of the G.P. 5/2, 5/4,5/8…..
Ans. The given G.P is 5/2, 5/4, 5/8…..
In the G.P. first term, a = 5/2
Common ratio, r
n^{th }terms of the G.P,a_{n} =ar^{n-1}
Hence 20^{th }and n^{th }terms of the G.P are 5 × 2^{-20} and 5/2^{n }respectively
Q2. Find the 12^{th }terms of a G.P. whose 8^{th }term is 192 and the common ratio is 2.
Ans. We are given 8^{th }term(a_{8}) is 192 and the common ratio,r is 2
n^{th }terms of the G.P,a_{n} =ar^{n-1}
= 3×1024 =3072
Q3.The 5^{th }term, 8^{th }term, and 11^{th }terms of a G.P are p, q, and s, respectively. Show that q² = ps.
Ans. According to question
5^{th }term = p, 8^{th }term = q and 11^{th }terms = s
The n^{th }term of G.P
Squaring both sides of equation (ii)
Hence proved
Q4. The 4^{th }term of a G.P is square of its second term , and the first term is -3. Determine its 7^{th }term.
Ans. We are given the first term, of G.P, a = -3
n^{th }term of G.P,a_{n} = ar^{n-1}
The fourth term and and second term of the G.P is given by the (i) and (ii) equation
According to question
r = -3
The seventh term of G.P is given by
= -2187
Therefore the 7^{th }term of the series is -2187
Q5.Which term of the following sequences :
(a) 2, 2√2, 4……. is 128? (b) √3, 3, 3√3…..is 729
Ans.(a) The given G.P is 2, 2√2, 4……. is 128
Common ratio ,r of the G.P is = 2√2/2 = √2 and the first term is ,a = 2
n^{th }term of G.P,a_{n} = ar^{n-1}
n = 15
Therefore 15^{th }term of G.P is 1s 128
(b) The given G.P is √3, 3, 3√3…..is 729
Common ratio ,r of the G.P is = 3/√3 = √3 and the first term is ,a = √3
n^{th }term of G.P,a_{n} = ar^{n-1}
n = 12
Therefore 12^{th }term of G.P is 729
The first term of the G.P
Common ratio, r
n^{th }term of G.P,a_{n} = ar^{n-1}
n – 1 = 9
n = 10
There for 10^{th }term of G.P is
Q6. For what values of x, the numbers are in G.P ?
Ans. Since numbers are in G.P, therefore
14x² = 14
x = ±1
Hence for x = ±1, the given numbers are in G.P
Find the sum to indicated number of terms in each of the geometric progression in exercises 7 to 10.
Q7.0.15, 0.015, 0.0015…. 20 terms.
Ans. We are given the G.P 0.15, 0.015, 0.0015…. 20 terms
In the given G.P ,a = 0.15
The common ratio, r
In case r< 1,the sum of the G.P is given by
Hence the required sum of the G.P is
Q8.√7, √21, 3√7………n terms.
Ans. The given G.P is √7, √21, 3√7………n terms.
In the G.P ,a = √7 and common ratio, r = √21÷√7 = √3
Since, r > 1 so, the sum of the G.P is given by
Q9.Find the sum to n terms in the geometric progression 1, -a, a^{2}, -a^{3} …. (if a ≠ -1)
Ans.The given series is 1, -a, a^{2}, -a^{3} …. (if a ≠ -1)
Common ratio,r = -a/1=-a and first term,A=1
Since Since r < 1, therefore the sum of n terms is given as
Q10.Find the sum to n terms in the geometric progression x^{3}, x^{5}, x^{7}, … (if x ≠ ±1 )
Ans.First-term, a =x³, common ratio,r = x^{5}/x^{3}=x²
Since r <1 therefore the sum of n terms is given as
Ans. The sum given to us defined as follows
Expanding the sum by putting k=1.2.3…..11
⇒(2 +3) + (2 + 3²) +(2+ 3³) +…….(2 + 3^{11})
=(2+2+2….up to 11 terms)+(3 +3² +3³ +……3^{11})
= 2×11 + (3 +3² +3³ +……3^{11})
=22+ (3 +3² +3³ +……3^{11})….(i)
3 +3² +3³ +……3^{11} is a G.P
Where a = 3, r= 3²/3 = 3
Since r > 1, therefore the sum of n terms is given as
Substituting this sum in (i), we get the total required sum
Q12. The sum of the first three terms of a G.P is 39/10 and their product is 1. Find the common ratio and the terms.
Ans. Let the first term of the G.P is a and the common ratio of the sequence is r
Therefore the first three terms are = a, ar and ar²
According to first condition of the question
According to second condition of the question
(ar)³ = 1
ar =1
Substituting the value of a in equation (i)
39r = 10 +10r +10r²
10r² – 29r +10 = 0
10r² – 25 r-4r +10 =0
5r(2r -5) -2(2r – 5) = 0
(2r -5)(5r -2) = 0
r = 5/2, r = 2/5
If r = 5/2 then a = 2/5 ( a = 1/r)
ar = (2/5)(5/2) =1
ar² = (2/5)(5/2)²= 5/2
When r =2/5, then a = 5/2
ar =(5/2)(2/5) =1²
ar² = (5/2)(2/5)²= 2/5
Hence the required terms are 2/5, 1 and 5/2 or 5/2,1 and 2/5.
Q13.How many terms of G.P. 3, 3^{²}, 3^{³}, … are needed to give the sum 120?
Ans. The given series is 3, 3^{²}, 3^{³}, …
Where, a = 2, r= 3²/3 = 3
Let the sum of n terms is = 120
Since r > 1, therefore the sum of n terms is given as
Comparing the exponents of both sides
n + 1 = 5
n = 5-1 = 4
Therefore 4 terms of the given G.P give the sum 120.
Q14.The sum of first three terms of a G.P is 16 and the sum of next three terms is 128. Determine the first term, the common ratio, and the sum to n terms of the G.P.
Ans.Let the first term of the G.P is a and the common ratio is r
Therefore G.P is a, ar,ar²,ar³……..
According to first condition of question
Sum of first three terms = 16
a+ ar+ ar² =16
a(1 + r +r²) = 16…….(i)
According to second condition of question
The sum of next three terms(4^{th}, 5^{th}, 6^{th}terms)
ar^{3}+ar^{4}+ar^{5}= 128
ar³( 1 + r + r²) = 128……(ii)
Dividing equation (i) by equation (ii)
r³ = 8
r = 2
Putting the value of r =2 in equation (i)
a(1 + r +r²) = 16
a(1 + 2 + 2²) = 16
a(7) = 16
a = 16/7
Therefore the first term of the sequence is 16/7, common ratio is 2 and sum of n terms is 16/7(2^{n}-1).
Q15.Given a G.P. with a = 729 and 7^{th} term 64, determine S_{7}
Ans.We are given a G.P in which a = 729, 7^{th }term =a_{7}=64
a_{n}=ar^{n-1}
a_{7}=ar^{7-1}
64= (729)r^{6}
Putting the value n = 7,a=729 and r= 2/3
S_{7}=2187 – 128 = 2059
Q16. Find a GP for which the sum of the first two terms is -4 and fifth term is 4 times the third term.
Ans. Let a , r be the first term and common ratio of the GP respectively
First term of GP = a and second term is = ar
according to first condition of the question
a + ar = -4
a(1 + r) = -4……(i)
According to second condition of the question
fifth term = 4× the third term
ar^{5-1} = 4× ar^{3-1})
ar^{4} = 4ar^{2}
r² = 4
r = ±2
If r = 2, then putting the value of r=2 in equation (i)
a(1 + 2) = -4
a = -4/3
So, the GP is -4/3, (-4/3) × 2, (-4/3)×2²…..= -4/3, -8/3, -16/3……
If r = -2, then putting the value of r= -2 in equation (i)
a(1 -2) = -4
-a = -4
a = 4
So, the GP is 4, 4×-2, -4×(-2)²…..,= 4, -8, -16……..
Hence the required GP is either -4/3, -8/3, -16/3……or 4, -8, -16……..
You can study rests of the solutions through the link
You can compensate us
Paytm number 9891436286
The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions for class 9 maths
NCERT Solutions for class 9 science
NCERT Solutions for class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT Solutions for Class 10 Science
NCERT Solutions for class 11 maths
Chapter 1-Sets | Chapter 9-Sequences and Series |
Chapter 2- Relations and functions | Chapter 10- Straight Lines |
Chapter 3- Trigonometry | Chapter 11-Conic Sections |
Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
NCERT Solutions for Class 11 Physics
chapter 3-Motion in a Straight Line
NCERT Solutions for Class 11 Chemistry
Chapter 1-Some basic concepts of chemistry
NCERT Solutions for Class 11 Biology
NCERT solutions for class 12 maths
Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
Pretty nice post. I just stumbled upon your weblog and wanted to say that Ive truly enjoyed browsing your blog posts. In any case I’ll be subscribing to your rss feed and I hope you write again very soon!