Careers360 Logo
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

Edited By Komal Miglani | Updated on Sep 23, 2023 06:30 PM IST

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?'

This Story also Contains
  1. Sequences And Series Class 11 Questions And Answers PDF Free Download
  2. Sequences and Series Class 11 Solutions - Important Formulae
  3. Sequences and Series Class 11 NCERT Solutions (Exercise)
  4. Importance of solving NCERT questions for class 11 Math Chapter 8
  5. NCERT solutions for class 11 - Subject-wise
  6. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

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.

Background wave

Sequences And Series Class 11 Questions And Answers PDF Free Download

Download PDF


Sequences and Series Class 11 Solutions - Important Formulae

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., r=an+1an, for all n > 1.

The general term or nth term of GP is an=ar(n1).

The nth term of a GP from the end: an=1rn1, where l is the last term.

If a, b, and c are three consecutive terms of a GP, then b2=ac.

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 G=ab.

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: r=(ba)1n+1.

The GM of a1, a2, a3,…, an is given by: GM=(a1a2a3an)1n.

Product of n GMs is G1×G2×G3××Gn=Gn=(ab)n2.

Relationship between A.M. and G.M.

A.M.=a+b2 and G.M.=ab
Thus, we have

A.M.G.M.=a+b2ab=a+b2ab2=(ab)220

Thus, we get the relationship A.M.G.M. always.

Sequences and Series Class 11 NCERT Solutions (Exercise)

Sequences and Series class 11 questions and answers: Exercise: 8.1
Page number: 138-139
Total Questions: 14

Question:1 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

an=n(n+2)

Answer:

Given : an=n(n+2)

a1=1(1+2)=3

a2=2(2+2)=8

a3=3(3+2)=15

a4=4(4+2)=24

a5=5(5+2)=35

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:

an=nn+1

Answer:

Given : an=nn+1

a1=11+1=12

a2=22+1=23

a3=33+1=34

a4=44+1=45

a5=55+1=56

Therefore, the required number of terms 12,23,34,45,56

Question:3 Write the first five terms of each of the sequences in Exercises 1 to 6 whose terms are:

an=2n

Answer:

Given : an=2n

a1=21=2

a2=22=4

a3=23=8

a4=24=16

a5=25=32

Therefore, the required number of terms =2,4,8,16,32.

Question:4 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

an=2n36

Answer:

Given : an=2n36

a1=2×136=16

a2=2×236=16

a3=2×336=36=12

a4=2×436=56

a5=2×536=76

Therefore, the required number of terms =16,16,12,56,76

Question:5 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

an=(1)n15n+1

Answer:

Given : an=(1)n15n+1

a1=(1)1151+1=(1)0.52=25

a2=(1)2152+1=(1)1.53=125

a3=(1)3153+1=(1)2.54=625

a4=(1)4154+1=(1)3.55=3125

a5=(1)5155+1=(1)4.56=15625

Therefore, the required number of terms =25,125,625,3125,15625

Question:6 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

an=nn2+54

Answer:

Given : an=nn2+54

a1=1.12+54=64=32

a2=2.22+54=184=92

a3=3.32+54=424=212

a4=4.42+54=844=21

a5=5.52+54=1504=752

Therefore, the required number of terms =32,92,212,21,752

Question:7 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

an=4n3;a17,a24

Answer:

an=4n3

Put n=17,

a17=4×173=683=65

Put n = 24,

a24=4×243=963=93

Hence, we have a17=65 and a24=93.

Question:8 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

an=n22n;a7

Answer:

Given : an=n22n

Put n=7,

a7=7227=49128

Hence, we have a7=49128

Question:9 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

an=(1)n1n3,a9

Answer:

Given : an=(1)n1n3

Put n = 9,

a9=(1)9193=(1).(729)=729

The value of a9=729

Question:10 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

an=n(n2)n+3;a20

Answer:

Given : an=n(n2)n+3

Put n = 20,

a20=20(202)20+3=36023

Hence, value of a20=36023

Question:11 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

a1=3,an=3an1+2foralln>1

Answer:

Given : a1=3,an=3an1+2foralln>1

a2=3a21+2=3a1+2=3(3)+2=11

a3=3a31+2=3a2+2=3(11)+2=35

a4=3a41+2=3a3+2=3(35)+2=107

a5=3a51+2=3a4+2=3(107)+2=323

Hence, the five terms of the series are 3,11,35,107,323

Series =3+11+35+107+323+...............

Question:12 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

a1=1,an=an1n,n2

Answer:

Given : a1=1,an=an1n,n2

a2=a212=a12=12

a3=a313=a23=16

a4=a414=a34=124

a5=a515=a45=1120

Hence, five terms of series are 1,12,16,124,1120

Series

=1+12+16+124+1120.........................

Question:13 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: a1=a2=2,an=an11,n>2

Answer:

Given : a1=a2=2,an=an11,n>2

a3=a311=a21=21=1

a4=a411=a31=11=0

a5=a511=a41=01=1

Hence, the five terms of the series are 2,2,1,0,1

Series =2+2+1+0+(1)+..................

Question:14 The Fibonacci sequence is defined by 1=a1=a2andan=an1+an2,n>2

Find an+1an , for n = 1, 2, 3, 4, 5

Answer:

Given: The Fibonacci sequence is defined by 1=a1=a2andan=an1+an2,n>2

a3=a31+a32=a2+a1=1+1=2

a4=a41+a42=a3+a2=2+1=3

a5=a51+a52=a4+a3=3+2=5

a6=a61+a62=a5+a4=5+3=8

Forn=1,an+1an=a1+1a1=a2a1=11=1

Forn=2,an+1an=a2+1a2=a3a2=21=2

Forn=3,an+1an=a3+1a3=a4a3=32

Forn=4,an+1an=a4+1a4=a5a4=53

Forn=5,an+1an=a5+1a5=a6a5=85

Sequences and Series class 11 questions and answers: Exercise: 8.2
Page number: 145-147
Total Questions: 32

Question:1 Find the 20th and nth terms of the G.P. 52,54,58,....

Answer:

G.P :

52,54,58,....

first term = a

a=52

common ratio = r

r=5452=12

an=a.rn1

a20=52.(12)201

a20=52.(1219)

a20=5220

an=a.rn1

an=52.(12)n1

an=52.12n1

an=52n the nth term of G.P

Question:2 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Answer:

First term = a

common ratio = r = 2

8th term is 192

an=a.rn1

a8=a.(2)81

192=a.(2)7

a=26.327

a=32

an=a.rn1

a12=32.(2)121

a12=32.(2)11

a12=3.(2)10

a12=3072 is the 12th term of a G.P.

Question:3 The 5th,8th andnd11th terms of a G.P. are p, q, and s, respectively. Show that q2=ps

Answer:

To prove : q2=ps

Let first term and common ratio = r

a5=a.r4=p..................(1)

a8=a.r7=q..................(2)

a11=a.r10=s..................(3)

Dividing equation 2 by 1, we have

a.r7a.r4=qp

r3=qp

Dividing equation 3 by 2, we have

a.r10a.r7=sq

r3=sq

Equating values of r3 , we have

qp=sq

q2=ps

Hence proved.

Question:4 The 4th term of a G.P. is the square of its second term, and the first term is -3. Determine its 7th term.

Answer:

First term =a= -3

4th term of a G.P. is the square of its second term

a4=(a2)2

a.r41=(a.r21)2

a.r3=a2.r2

r=a=3

a7=a.r71

a7=(3).(3)6

a7=(3)7=2187

Thus, the seventh term is -2187.

Question:5 Which term of the following sequences:

(a) 2,22,4.,....is128?

(b) Which term of the following sequences: 3,3,33,...is729?

(c) Which term of the following sequences: 13,19,127,....is119683?

Answer:

(a) Given : GP=2,22,4.,............

a=2andr=222=2

nth term is given as 128.

an=a.rn1

128=2.(2)n1

64=(2)n1

26=(2)n1

212=(2)n1

n1=12

n=12+1=13

The 13th term is 128.

(b) Given : GP=3,3,33,........

a=3andr=33=3

n The term is given as 729.

an=a.rn1

729=3.(3)n1

729=(3)n

(3)12=(3)n

n=12

The 12th term is 729.

(c) Given : GP=13,19,127,............

a=13andr=1913=13

nth term is given as 119683

an=a.rn1

119683=13.(13)n1

119683=13n

139=13n

n=9

Thus, n = 9.

Question:6 For what values of x, the numbers 27,x,72 are in G.P.?

Answer:

GP=27,x,72

Common ratio = r.

r=x27=72x

x2=1

x=±1

Thus, for x=±1, the ennumbers will be in GP.

Question:7 Find the sum of the indicated number of terms in each of the geometric progressions in 0.15, 0.015, 0.0015, ... 20 terms.

Answer:

Geometric progressions are 0.15, 0.015, 0.0015, ... .....

a = 0.15 , r = 0.1 , n = 20

Sn=a(1rn)1r

S20=0.15(1(0.1)20)10.1

S20=0.15(1(0.1)20)0.9

S20=0.150.9(1(0.1)20)

S20=1590(1(0.1)20)

S20=16(10.120)

Question:8 Find the sum to the indicated number of terms in each of the geometric progressions in 7,21,37,....nterms

Answer:

GP=7,21,37,...............

a=7andr=217=3

Sn=a(1rn)1r

Sn=7(13n)13

Sn=7(13n)13×1+31+3

Sn=7(13n)13(1+3)

Sn=7(13n)2(1+3)

Sn=7(1+3)2(3n1)

Question:9 Find the sum to the indicated number of terms in each of the geometric progressions in a,a,a3,...nterms(ifa1)

Answer:

The sum to the indicated number of terms in each of the geometric progressions is:

GP=1,a,a2,a3,.............

a=1andr=a

Sn=a(1rn)1r

Sn=1(1(a)n)1(a)

Sn=1(1(a)n)1+a

Sn=1(a)n1+a

Question:10 Find the sum to the indicated number of terms in each of the geometric progressions in x3,x5,x7...nterms(ifx±1)

Answer:

GP=x3,x5,x7.....................

a=x3andr=x5x3=x2

Sn=a(1rn)1r

Sn=x3(1(x2)n)1x2

Sn=x3(1(x2)n)1x2

Question:11 Evaluate k=111(2+3k)

Answer:

Given :

k=111(2+3k)

k=111(2+3k)=k=1112+k=1113k

=22+k=1113k...............(1)

k=1113k=31+32+33+....................311

These terms form GP with a = 3 and r = 3.

Sn=a(1rn)1r

Sn=3(1311)13

Sn=3(1311)2

Sn=3(3111)2 =k=1113k

k=111(2+3k) =22+3(3111)2

Question:12 The sum of the first three terms of a G.P. is 3910 and their product is 1. Find the common ratio and the terms.

Answer:

Given: The sum of the first three terms of a G.P. is 3910 and their product is 1.

Let three terms be ar,a,ar .

Sn=a(1rn)1r

S3=a(1r3)1r=3910

ar+a+ar=3910.........1

The product of 3 terms is 1.

ar×a×ar=1

a3=1

a=1

Put the value of an in equation 1,

1r+1+r=3910

10(1+r+r2)=39(r)

10r229r+10=0

10r225r4r+10=0

5r(2r5)2(2r5)=0

(2r5)(5r2)=0

r=52,r=25

The three terms of AP are 52,1,25 .

Question:13 How many terms of G.P. $3, 3 ^ 2 3 ^ 3 … are needed to give the sum 120?

Answer:

G.P.= 3,32,33 , …............

Sum = 120

These terms are GP with a = 3 and r = 3.

Sn=a(1rn)1r

120=3(13n)13

120×23=(13n)

80=(13n)

3n=1+80=81

3n=81

3n=34

n=4

Hence, we have a value of n as 4 to get a sum of 120.

Question:14 The sum of the first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio, and the sum to n terms of the G.P.

Answer:

Let GP be a,ar,ar2,ar3,ar4,ar5,ar6................................

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

a+ar+ar2=16

a(1+r+r2)=16...............................(1)

Given: the sum of the next three terms is 128.

ar3+ar4+ar5=128

ar3(1+r+r2)=128...............................(2)

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

ar3(1+r+r2)a(1+r+r2)=12816

r3=8

r3=23

r=2

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

a(1+2+22)=16

a(7)=16

a=167

Sn=a(1rn)1r

Sn=167(12n)12

Sn=167(2n1)

Question:15 Given a G.P. with a = 729 and 7th term 64, determine s7

Answer:

Given a G.P. with a = 729 and 7th term 64.

an=a.rn1

64=729.r71

r6=64729

r6=(23)6

r=23

Sn=a(1rn)1r

S7=729(1(23)7)123

S7=729(1(23)7)13

S7=3×729(372737)

S7=(3727)

S7=2187128

S7=2059

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

a5=4.a3

a.r51=4.a.r31

a.r4=4.a.r2

r2=4

r=±2

If r = 2, then

Sn=a(1rn)1r

a(122)12=4

a(14)1=4

a(3)=4

a=43

If r = - 2, then

Sn=a(1rn)1r

a(1(2)2)1(2)=4

a(14)3=4

a(3)=12

a=123=4

Thus, required GP is 43,83,163,......... or 4,8,16,32,...........

Question:17 If the 4th,10th,16th terms of a G.P. are x, y and z, respectively. Prove that x,y, and z are in G.P.

Answer:

Let x,y, and z be in G.P.

Let first term and common ratio = r

a4=a.r3=x..................(1)

a10=a.r9=y..................(2)

a16=a.r15=z..................(3)

Dividing equation 2 by 1, we have

a.r9a.r3=yx

r4=yx

Dividing equation 3 by 2, we have

a.r15a.r9=zy

r4=zy

Equating values of r4 , we have

yx=zy

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

Sn=8+88+888+8888+............. to n terms

Sn=89[9+99+999+9999+................]

Sn=89[(101)+(1021)+(1031)+(1041)+................]

Sn=89[(10+102+103+........)(1+1+1.....................)]

Sn=89[10(10n1)101(n)]

Sn=89[10(10n1)9(n)]

Sn=8081(10n1)8n9

Question:19 Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,1/2

Answer:

Requiredsum=2×128+4×32+8×8+16×2+32×12

Requiredsum=64[4+2+1+12+122]

Here, 4,2,1,12,122 is a GP.

first term = a = 4

common ratio =r

r=12

Sn=a(1rn)1r

S5=4(1(12)5)112

S5=4(1(12)5)12

S5=8(1(132))

S5=8(3132)

S5=314

Requiredsum=64[314]

Requiredsum=16×31=496

Question:20 Show that the products of the corresponding terms of the sequences a,ar,ar,...arn1andA,AR,AR2....ARn1 form a G.P, and find the common ratio.

Answer:

To prove : aA,arAR,ar2AR2,................... is a GP.

secondtermfirstterm=arARaA=rR

thirdtermsecondterm=ar2AR2arAR=rR

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

Question:21 Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Answer:

Let the first term be a and the common ratio be r.

a1=a,a2=ar,a3=ar2,a4=ar3

Given: the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

a3=a1+9

ar2=a+9

a(r21)=9.................1

a2=a4+18

ar=ar3+18

ar(1r2)=18......................2

By dividing equation 2 by 1 we get

ar(1r2)a(1r2)=189

r=2

Putting the value of r, we get

4a=a+9

4aa=9

3a=9

a=3

Thus, four terms of GP are 3,6,12,24.

Question:22 If the pthqth,rth terms of a G.P. are a, b and c, respectively. Prove that aqrbrpCpq=1

Answer:

To prove : aqrbrpCpq=1

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

According to the given information, we have

ap=A.Rp1=a

aq=A.Rq1=b

ar=A.Rr1=c

L.H.S : aqrbrpCpq

Aqr.R(qr)(p1).Arp.R(rp)(q1).Apq.R(pq)(r1)

Aqr+rp+pq.R(qprpq+r)+(rqpq+pr)+(prpqr+q)

A0.R0=1 =RHS

Thus, LHS = RHS.

Hence proved.

Question:23 If the first and the nth terms of a G.P. are a and b, respectively, and if P is the product of n terms, prove that $P^2 = ( ab)^n.

Answer:

Given First term =a and n th term = b.

Common ratio = r.

To prove : P2=(ab)n

Then , GP=a,ar,ar2,ar3,ar4,..........................

an=a.rn1=b..................................1

P = product of n terms

P=(a).(ar).(ar2).(ar3)..............(arn1)

P=(a.a.a...............a)((1).(r).(r2).(r3)..............(rn1))

P=(an)(r1+2+.........(n1))........................................2

Here, 1+2+.........(n1) is aananP.

sum=n2[2a+(n1)d]

n12[2(1)+(n11)1]

n12[2+n2]

n12[n]

n(n1)2

Put in equation (2),

P=(an)(rn(n1)2)

P2=(a2n)(rn(n1))

P2=(a.a.r(n1))n

P2=(a.b)n

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 (n+1)thto(2n)th term is 1rn

Answer:

Let the first term =a and common ratio = r.

sumofnterms=a(1rn)1r

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

The sum of terms from (n+1) to 2n.

Sn=a(n+1)(1rn)1r

a(n+1)=a.rn+11=arn

Thus, the required ratio = a(1rn)1r×1rarn(1rn)

=1rn

Thus, the common ratio of the sum of the first n terms of a G.P. to the sum of terms from (n+1)thto(2n)th term is 1rn.

Question:25 If a, b, c, and d are in G.P. show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2.

Answer:

If a, b, c, and d are in G.P.

bc=ad....................(1)

b2=ac....................(2)

c2=bd....................(3)

To prove : (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2.

RHS : (ab+bc+cd)2.

(ab+ad+cd)2.

(ab+d(a+c))2.

a2b2+d2(a+c)2+2(ab)(d(a+c))

a2b2+d2(a2+c2+2ac)+2a2bd+2bcd

Using equation (1) and (2),

a2b2+2a2c2+2b2c2+d2a2+2d2b2+d2c2

a2b2+a2c2+a2c2+b2c2+b2c2+d2a2+d2b2+d2b2+d2c2

a2b2+a2c2+a2d2+b2.b2+b2c2+b2d2+c2b2+c2.c2+d2c2

a2(b2+c2+d2)+b2(b2+c2+d2)+c2(b2+c2+d2)

(b2+c2+d2)(a2+b2+c2) = LHS

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.

a4=a.r41

81=3.r3

27=r3

r=3

For r=3 ,

A=ar=(3)(3)=9

B=ar2=(3)(3)2=27

The required numbers are 9,27.

Question:27 Find the value of n so that an+1+bn+1an+bn may be the geometric mean between a and b.

Answer:

M of a and b is ab.

Given :

an+1+bn+1an+bn=ab

Squaring both sides

(an+1+bn+1an+bn)2=ab

(an+1+bn+1)2=(an+bn)2ab

(a2n+2+b2n+2+2.an+1.bn+1)=(a2n+b2n+2.an.bn)ab

(a2n+2+b2n+2+2.an+1.bn+1)=(a2n+1.b+a.b2n+1+2.an+1.bn+1)

(a2n+2+b2n+2)=(a2n+1.b+a.b2n+1)

a2n+1(ab)=b2n+1(ab)

a2n+1=b2n+1

(ab)2n+1=1

(ab)2n+1=1=(ab)0

2n+1=0

2n=1

n=12

Question:28 The sum of two numbers is 6 times their geometric mean, showing numbers are in the ratio (3+22):(322)

Answer:

Let there be two numbers a and b

Geometric mean =ab

According to the given condition,

a+b=6ab

(a+b)2=36(ab) .............................................................(1)

Also, (ab)2=(a+b)24ab=36ab4ab=32ab

(ab)=32ab

(ab)=42ab .......................................................(2)

From (1) and (2), we get

2a=(6+42)ab

a=(3+22)ab

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

b=6ab(3+22)ab

b=(322)ab

ab=(3+22)ab(322)ab

ab=(3+22)(322)

Thus, the ratio is (3+22):(322).

Question:29 If A and G are A.M. and G.M., respectively between two positive numbers, prove that the numbers are A±(A+G)(AG)

Answer:

If A and Garee A.M. and G.M., respectively between two positive numbers,
Two numberarebe a and b.

AM=A=a+b2

a+b=2A ...................................................................1

GM=G=ab

ab=G2 ...........................................................................2

We know (ab)2=(a+b)24ab

Put values from equation 1 and 2,

(ab)2=4A24G2

(ab)2=4(A2G2)

(ab)2=4(A+G)(AG)

(ab)=4(A+G)(AG) ..................................................................3

From 1 and 3, we have

2a=2A+2(A+G)(AG)

a=A+(A+G)(AG)

Put the value of an in equation 1, and we get

b=2AA(A+G)(AG)

b=A(A+G)(AG)

Thus, numbers are A±(A+G)(AG).

Question:30 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour?

Answer:

The number of bacteria in a certain culture doubles every hour. It forms GP.

Given a = 30 and r = 2.

a3=a.r31=30(2)2=120

a5=a.r51=30(2)4=480

an+1=a.rn+11=30(2)n

Thus, bacteria present at the end of the 2nd hour, 4th hour, r, and nth hour are 120,480 and 30(2)n respectively.

Question:31 What will R50 amount to in 10 years after its deposit in a bank that pays an annual interest rate of 10% compounded annually?

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

=500(1+110)=500(1.1)

At the end of the second year, the amount =500(1.1)(1.1)

At the end of the third year, the amount =500(1.1)(1.1)(1.1)

At the end of 10 years, the amount =500(1.1)(1.1)(1.1)........(10times)

=500(1.1)10

Thus, at the end of 10 years, amount =Rs.500(1.1)10.

Question:32 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation

Answer:

Let the roots of the quadratic equation be a and b.

According to the given conditions,

AM=a+b2=8

(a+b)=16

GM=ab=5

ab=25

We know that x2x(sumofroots)+(productofroots)=0

x2x(16)+(25)=0

x216x+25=0

Thus, the quadratic equation = x216x+25=0

Sequences and Series class 11 NCERT solutions: Miscellaneous Exercise
Page number: 147-148
Total Questions: 18

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

x=1nf(x)=120 , find the value of n.

Answer:

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

f(1)=3

Taking x=y=1 , we have

f(1+1)=f(2)=f(1)f(1)=33=9

f(1+1+1)=f(1+2)=f(1)f(2)=39=27

f(1+1+1+1)=f(1+3)=f(1)f(3)=327=81

f(1),f(2),f(3),f(4)..................... is 3,9,27,81,.............................. forms a GP with first term=3 and common ratio = 3.

x=1nf(x)=120=Sn

Sn=a(1rn)1r

120=3(13n)13

40=(13n)2

80=(13n)

801=(3n)

81=(3n)

3n=81

Therefore, n=4

Thus, the value of n is 4.

Question:2 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Answer:

Let the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2

Sn=a(1rn)1r

315=5(12n)12

63=(12n)1

63=(12n)

631=(2n)

64=(2n)

2n=64

Therefore, n=6

Thus, the value of n is 6.

Last term of GP=6th term =a.r^{n-1}=5.2^5=5\cdot32=160

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.

a=1

a3=a.r2=r2 a5=a.r4=r4

r2+r4=90

r4+r290=0

r2=1±1+3602

r2=1±3612

r2=10or9

r=±3

Thus, the common ratio of GP is ±3.

Question:4 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 2and 1 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Answer:

Let three terms of GP be a,ar,ar2.

Then, we have a+ar+ar2=56

a(1+r+r2)=56 ...............................................1

a1,ar7,ar221 from an AP.

ar7(a1)=ar221(ar7)

ar7a+1=ar221ar+7

ar6a=ar214ar

ar22ar+a=8

ar2arar+a=8
a(r22r+1)=8

a(r21)2=8 ....................................................................2

From equation 1 and 2, we get

7(r22r+1)=1+r+r2

7r214r+71rr2=0

6r215r+6=0

2r25r+2=0

2r24rr+2=0

2r(r2)1(r2)=0

(r2)(2r1)=0

r=2,r=12

If r = 2, GP = 8,16,32

If r=0.2, GP= 32,16,8.

Thus, the numbers required are 8,16,32.

Question:5 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Answer:

Let GP be A1,A2,A3,...................A2n

Number of terms = 2n

According to the given condition,

(A_{1},A_{2},A_{3},...................A_{2n})=5(A_{1},A_{3},...................A_{2n-1})

\Rightarrow (A_{1},A_{2},A_{3},...................A_{2n})-5(A_{1},A_{3},...................A_{2n-1})=0

\Rightarrow (A_{2},A_{4},A_{6},...................A_{2n})=4(A_{1},A_{3},...................A_{2n-1})

Lettheme be GP as a,ar,ar2,..................

ar(rn1)r1=4.a(rn1)r1

ar=4a

r=4

Thus, the common ratio is 4.

Question:6 If a+bxabx=b+cxbcx=c+dxcdx(x0) then show that a, b, c and d are in G.P.

Answer:

Given :

a+bxabx=b+cxbcx=c+dxcdx(x0)

Takin,

a+bxabx=b+cxbcx

(a+bx)(bcx)=(b+cx)(abx)

ab+b2xbcx2acx)=bab2x+acxbcx2

2b2x=2acx

b2=ac

ba=cb..................1

Taking,

b+cxbcx=c+dxcdx

(b+cx)(cdx)=(c+dx)(bcx)

bcbdx+c2xcdx2=bcc2x+bdxcdx2

2bdx=2c2x

bd=c2

dc=cb..............................2

From equation 1 and 2, we have

dc=cb=ba

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 P2Rn=Sn

Answer:

Let there be a GP =a,ar,ar2,ar3,....................

According to the given information,

S=a(rn1)r1

P=an×r(1+2+...................n1)

P=an×rn(n1)2

R=1a+1ar+1ar2+..................1arn1

R=rn1+rn2+rn3+..............r+1a.rn1

R=1a.rn1×1(rn1)r1

R=(rn1)a.rn1.(r1)

To prove : P2Rn=Sn

LHS : P2Rn

=a2n.rn(n1)(rn1)nan.rn(n1).(r1)n

=an(rn1)n(r1)n

=(a(rn1)(r1))n

=Sn=RHS

Hence proved.

Question:8 If a, b, c, d are in G.P, prove that (an+bn),(bn+cn),(cn+dn) are in G.P.

Answer:

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

To prove: (an+bn),(bn+cn),(cn+dn) are in G.P.

Then we can write,

b2=ac...............................1

c2=bd...............................2

ad=bc...............................3

Let (an+bn),(bn+cn),(cn+dn) be in GP.

(bn+cn)2=(an+bn)(cn+dn)

LHS: (bn+cn)2

(bn+cn)2=b2n+c2n+2bncn

(bn+cn)2=(b2)n+(c2)n+2bncn

=(ac)n+(bd)n+2bncn

=ancn+bncn+andn+bndn

=cn(an+bn)+dn(an+bn)

=(an+bn)(cn+dn)=RHS

Hence proved.

Thus, (an+bn),(bn+cn),(cn+dn) are in GP

Question:9 If a and b are the roots of x23x+p=0 and c, d are roots of x212x+q=0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

Answer:

Given: a and b are the roots of x23x+p=0

Then, a+b=3andab=p......................1

Also, c, d are roots of x212x+q=0

c+d=12andcd=q......................2

Given: a, b, c, d form a G.P

Let, a=x,b=xr,c=xr2,d=xr3

From 1 and 2, we get

x+xr=3 and xr2+xr3=12

x(1+r)=3 xr2(1+r)=12

On dividing them,

xr2(1+r)x(1+r)=123

r2=4

r=±2

When , r = 2,

x=31+2=1

When, r = -2,

x=312=3

Case (1) when r=2 and x=1,

ab=x2r=2andcd=x2r5=32

q+pqp=32+2322=3430=1715

i.e. (q + p) : (q – p) = 17:15.

Case (2) when r = -2 and x = -3,

ab=x2r=18andcd=x2r5=288

q+pqp=28818288+18=305270=1715

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 a:b=(m+m2n2):(mm2n2)

Answer:

Let two numbers be a and b.

AM=a+b2andGM=ab

According to the given condition,

a+b2ab=mn

(a+b)24ab=m2n2

(a+b)2=4ab.m2n2

(a+b)=2ab.mn ...................................................................1

(ab)2=(a+b)24ab

We get,

(ab)2=(4abm2n2)4ab

(ab)2=(4abm24abn2n2)

(ab)2=(4ab(m2n2)n2)

(ab)=(2ab(m2n2)n) .....................................................2

From 1 and 2, we get

2a=(2abn)(m+(m2n2))

a=(abn)(m+(m2n2))

Putting the value of an ann equation 1, we have

b=(2.abn)m(abn)(m+(m2n2))

b=(abn)m(abn)((m2n2))

=(abn)(m(m2n2))

a:b=ab=(abn)(m+(m2n2))(abn)(m(m2n2))

=(m+(m2n2))(m(m2n2))

a:b=(m+(m2n2)):(m(m2n2))

Question:11 Find the sum of the following series up to n terms:

(i) 5+55+555+....

(ii) .6 +. 66 +. 666+…

Answer:

(i) 5+55+555+.... is not a GP.

It can be changed in GP by writing terms such s

Sn=5+55+555+.... to n terms

Sn=59[9+99+999+9999+................]

Sn=59[(101)+(1021)+(1031)+(1041)+................]

Sn=59[(10+102+103+........)(1+1+1.....................)]

Sn=59[10(10n1)101(n)]

Sn=59[10(10n1)9(n)]

Sn=5081[(10n1)]5n9

Thus, the sum is

Sn=5081[(10n1)]5n9

(ii) The sum of 0.6 +0. 66 + 0. 666+….................

It can be written as

Sn=0.6+0.66+0.666+.......................... to n terms

Sn=6[0.1+0.11+0.111+0.1111+................]

Sn=69[0.9+0.99+0.999+0.9999+................]

Sn=69[(1110)+(11102)+(11103)+(11104)+................]

Sn=23[(1+1+1.....................nterms)110(1+110+1102+...................nterms)]

Sn=23[n110(110n1)1101]

Sn=2n3230[10(110n)9]

Sn=2n3227(110n)

Question:12 Find the 20th term of the series 2×4+4×6+×6×8+....+n terms.

Answer:

The series = 2×4+4×6+×6×8+....+n

nthterm=an=2n(2n+2)=4n2+4n

a20=2(20)[2(20)+2]

=40[40+2]

=40[42]

=1680

Thus, the 20th term of the series is 1680

Question:13 A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

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

=12%of6000+12%of5500+.............12%of500

=12%of(6000+5500+.............+500)

=12%of(500+1000+.............+6000)

Here, 500,1000,.............5500,6000 is a AP with first term =a=500 and common difference =d = 500

We know that an=a+(n1)d

6000=500+(n1)500

5500=(n1)500

11=(n1)

n=11+1=12

The sum of AP:

S12=122[2(500)+(121)500]

S12=6[1000+5500]

=6[6500]

=39000

Thus, interest to be paid :

=12%of(500+1000+.............+6000)

=12%of(39000)

=Rs.4680

Thus, cost of tractor = Rs. 12000+ Rs. 4680 = Rs. 16680

Question:14 Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in an annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

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

=10%of18000+10%of17000+.............10%of1000

=10%of(18000+17000+.............+1000)

=10%of(1000+2000+.............+18000)

Here, 1000,2000,.............17000,18000 is a AP with first term =a=1000 and common difference =d = 1000

We know that an=a+(n1)d

18000=1000+(n1)1000

17000=(n1)1000

17=(n1)

n=17+1=18

The sum of AP:

S18=182[2(1000)+(181)1000]

=9[2000+17000]

=9[19000]

=171000

Thus, interest to be paid :

=10%of(1000+2000+.............+18000)

=10%of(171000)

=Rs.17100

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

Question:15 A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail it to four different persons with the instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when the 8th set of letters is mailed.

Answer:

The number of letters mailed from a GP: 4,42,43,.............48

First term = a=4

Common ratio=r=4

Number of terms = 8

We know that the sum of GP is

Sn=a(rn1)r1

=4(481)41

=4(655361)3

=4(65535)3

=87380

Costs to mail one letter are 50 paise.

Cost of mailing 87380 letters

=Rs.87380×50100

=Rs.43690

Thus, the amount spent when the 8th set of the letter is mailed is Rs. 43690.

Question:16 A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 1the 5th year since he deposited the amount and also calculate the total amount after 20 years.

Answer:

Given: A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

=5100×10000=Rs.500

Interest in fifteen year 10000+ 14 times Rs. 500

Amount in 15 th year =Rs.10000+14×500

=Rs.10000+7000

=Rs.17000

Amount in 20 th year =Rs.10000+20×500

=Rs.10000+10000

=Rs.20000

Question:17 A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

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. 45 of the original cost.

Value at the end of 5 years

=15625×45×45×45×45×45

=5120

Thus, the value of the machine at the end of 5 years is Rs. 5120

Question:18 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day, 4 more workers dropped out on the third day, and so on.

Answer:

Let x be the number of days in which 150 workers finish the work.

According to the given information, we have

150x=150+146+142+............(x+8)terms

Series 150x=150+146+142+............(x+8)terms is a AP

First term=a=150

Common difference= -4

Number of terms = x+8

150x=x+82[2(150)+(x+81)(4)]

300x=x+8[3004x28]

300x=300x4x228x+240032x+224

x2+15x544=0

x2+32x17x544=0

x(x+32)17(x+32)=0

(x+32)(x17)=0

x=32,17

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.

NEET/JEE Offline Coaching
Get up to 90% Scholarship on your NEET/JEE preparation from India’s Leading Coaching Institutes like Aakash, ALLEN, Sri Chaitanya & Others.
Apply Now

Importance of solving NCERT questions for class 11 Math Chapter 8

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.

NCERT solutions for class 11 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 11:

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and NCERT syllabus for class 11:

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Class 11 Maths Chapter 8?

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.

2. How to find the sum of an arithmetic progression (AP) in Class 11 Maths?

The sum of the first n terms of an arithmetic progression (AP) can be calculated using the formula Sn=n2(2a+(n1)d), where a is the first term, d is the common difference, and n is the number of terms. This formula helps in finding the total sum of terms when you know the starting term and the common difference between consecutive terms.

3. What is the formula for the nth term of a geometric progression (GP)?

The nth term of a geometric progression (GP) is given by the formula Tn=arn1, where a is the first term and r is the common ratio. This formula allows you to find any term in the sequence when you know the first term and the common ratio.

4. What is the difference between arithmetic and geometric sequences?

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.

5. How to find the sum of an infinite geometric series?

The sum of an infinite geometric series, provided the common ratio r satisfies |r|<1, is calculated using the formula S=a1r, where a is the first term and r is the common ratio. This formula helps in finding the sum of an infinite series when the terms approach zero and the series converges, which is a key concept in both mathematics and real-world applications.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top