**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.

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**Class 11 maths NCERT Solutions of chapter 9-Sequences and Series**

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**Exercise 9.2- Sequences and Series**

**Exercise 9.4-Sequence and Series**

**NCERT solutions of 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**

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**Class 11 NCERT solutions of Physics and Chemistry**

**Chapter 1 – Some concepts of the chemistry**

**Exercise 9.3 -Sequences and Series**

**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……..

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**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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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 |

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Chapter 7- Integrals | |

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