NCERT Solutions Class 11 Maths Chapter 9

NCERT Solutions for Class 11 Mathematics Chapter 9 – Sequences and series

Mathematics is an important aspect of every field. It is majorly used in various disciplines of life. The applications of Mathematics make day-to-day dealings of life easier. Hence, it is included as a core subject in almost every academic curriculum.

You have already studied the various patterns of numbers and alphabets in the lower classes. It is nothing but an arrangement with certain repetitions considered as the part of sequence and series. The main topics covered in the chapter are the introduction to sequence and series, Arithmetic Progression (A.P.), Geometric Progression (G.P.), the relationship between A.M and G.M and sum to ‘n’ terms of special series.

NCERT Solutions for Class 11 Mathematics Chapter 9 has detailed coverage of every topic covered in the NCERT Class 9th Mathematics textbook. Students will be able to understand all the concepts related to sequence and series once you refer to NCERT Solutions. How the chapter is presented in the NCERT Solutions for Class 11 Mathematics Chapter 9 will help students gain interest in the subject and assist them in acquiring related knowledge. NCERT solutions will definitely prove fruitful to students in their class assignments, tests and preparation.

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Key Topics Covered In Chapter 9 Class 11 Mathematics

The arrangement of numbers, objects, and alphabets in a particular pattern are called sequence, whereas placing of numbers, objects, and alphabets one after the other without repetition of the pattern is called series. The chapter sequence and series are a combination of both. To get a particular series and understand its associated sequence, one needs to know certain rules based on it. Students will get a clear understanding of this concept in this chapter.

They would be able to derive a series with the help of a few formulas given in the chapter. This will make their calculations while finding a series easier and save their time. The various patterns covered for calculating series in this chapter are AP and GP. Every bit of this chapter is included in our NCERT Solutions for Class 11 Mathematics Chapter 9, available on the  Extramarks’ website.

NCERT Solutions for Class 11 Mathematics Chapter 9 require students to use their critical thinking ability and apply a wide range of formulas they have learnt. 



In this chapter, we will learn about the sequence and series. When we put a collection of objects in an orderly manner in a certain pattern using some sets of rules, it is called a sequence and series. The sequence and series have an important application in many human activities. When the sequence follows a pattern, it’s called a progression.

We have already studied the arithmetic progression in our previous classes. In this chapter, we will learn some more concepts about the arithmetic progression, series, geometric progression, the relationship between arithmetic and geometric progression and the sum of ‘n’ terms of the series. 



In the section of this chapter, we will understand the sequence with the help of some examples listed below:

  • 2, 4, 8, 16, 32, …………, 1024
  • 3, .3, .33, .333, .3333……and so on.

The different elements in a sequence are called terms, for example (a1, a2, a3, a4,…… an)


There are two types of sequences:

  • Infinite sequence

When the sequence contains an infinite number of elements, then it is called an infinite sequence. 

  • Finite sequence 

When the sequence contains a finite number of elements, it’s called a finite sequence. 


Fibonacci sequence :


A1 = A2 = 1, 

A3 = A1 + A2

An = An-2 + An-1,

 n > 2



In this section, we will learn about the series.

There is an example given below that shows the series:

a1 +  a2 +  a3 +  a4 + …… an + …..

The example of the series is associated with the sequence given above.

The important points you should note about the series are:

  • The sequence of the series can be finite or infinite.
  • The series is offered in the compact form called sigma notation, denoted by the ‘∑’.


Arithmetic progression (A.P)

In the section of this chapter, we will learn about the arithmetic progression or arithmetic sequence. There is a series given below which follows arithmetic progression:

a1 +  a2 +  a3 +  a4 + …… an + …..

an+1 = an + d

Where a = first term, d = common difference of A.P

There are some properties that verify an A.P. They are as follows:

  • If a constant is added to each term of A.P., then the resulting sequence will be in A.P
  • If a constant is subtracted from each term of A.P., then the resulting sequence will be in A.P
  • If a constant is multiplied by each term of A.P., then the resulting sequence will be in A.P
  • If a constant/nonzero is divided fry each term of A.P., then the resulting sequence will be in A.P

The series given below follows arithmetic progression:

a, a + d, a + 2d, a + 3d…a + (n + 1)d



  • The last number of A.P,

l = a + (n – 1)d


  • The sum of n numbers in A.P, 

S= n/2 [2a + (n – 1)d]


S= n/2 [a + l]


Arithmetic mean

If we have a sequence, say a, b & c, then the arithmetic mean is given by b = (a + c) /2


Geometric progressions (G.P)

In this part of the chapter, we will learn about the Geometric progression or Geometric sequence. 

There is a series given below which follows arithmetic progression:

a1 +  a2 +  a3 +  a4 + …… an + …..

 ak+1 / ak = r (constant), for k 1

Where, a = first term, r = common ratio of G.P

The series given below follows a Geometric progression:

a, ar, ar2, ar3,…..



  • General ‘n’ term of a G.P,

a=  arn-1

  • The sum of n numbers in G.P, 

S= a(1-rn) / (1-r)  


S= a(rn-1) / (r-1)


Geometric mean

If we have a sequence, say a, b & c, then the geometric mean is given by b = √a.c


Relationship between A.M and G.M

Let us check A.M & G.M.’s relationship of two given numbers a and b.

A.M = (a + b) /2

G.M = √a.b

Then, A.M – G.M = (a + b) /2 – √a.b = (√a – √b)2 / 2 0. 

This is the relationship between A.M G.M


Sum to ‘n’ terms of special series.

There is a special series, and we will find the sum of ‘n’ terms of this series. 

  • 1 + 2 + 3 + 4 +…….+n (sum of first ‘n’ numbers)

Sn = n(n + 1) / 2 

  • 12 + 22 + 32 + 42 +…….+n2 (sum of first square ‘n’ numbers)

Sn = n(n + 1)(2n + 1) / 6

  • 13 + 23 + 33 + 43 +…….+n3(sum of first cube ‘n’ numbers)

Sn = [n(n + 1)]2 / 4

NCERT Solutions for Class 11 Mathematics Chapter 9 Exercise & Solutions

The NCERT Solutions for Class 11 Mathematics Chapter 9 is mostly practice-oriented. The more you practice, the better you will get. . Hence, multiple exercises are given in this chapter in the NCERT textbook. You must have all the accurate solutions to these exercises. As a result, we have provided a detailed solution to every question given in the chapter in the NCERT Solutions for Class 11 Mathematics Chapter 9 available on the Extramarks’ website. The solutions will give you a complete analysis of the steps you should follow when solving the problems related to sequence and series. Thus, helping you to develop clear mindsets while solving the problems. .


Click on the  link below  to view exercise-specific questions and solutions for NCERT Solutions for Class 11 Mathematics Chapter 9:

  • Chapter 9 Class 11 Mathematics: Exercise 9.1
  • Chapter 9 Class 11 Mathematics: Exercise 9.2
  • Chapter 9 Class 11 Mathematics: Exercise 9.3
  • Chapter 9 Class 11 Mathematics: Exercise 9.4
  • Chapter 9 Class 11 Mathematics: Miscellaneous Exercise  


Along with Class 11 Mathematics solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

  • NCERT Solutions Class 1, 
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NCERT Exemplar Class 11 Mathematics

NCERT Class 11th Mathematics lays a strong foundation for the aspirants preparing for competitive examinations like JEE, NEET, KVPY, WBJEE and many other engineering and medical-related examinations. To score well and be in the top percentile in these examinations, students must have a clear-cut understanding of every concept covered in Class 11 Mathematics.

To excel in it, students need to do rigorous practice. Hence, they must have access to NCERT-related questions. NCERT Exemplar Class 11th Mathematics thus has all NCERT-based questions with solutions. . Students can find each topic covered with various types of questions with varying difficulty levels that they will face in the examination.

Moreover, they will also find tricks to solve these questions in less time with greater accuracy and will become more proficient in their calculations. The subject matter experts have designed the book after analysing the past year’s papers and the competitive examinations. The challenging questions will aid in developing the strong mindsets among the students, and as a result, they will start thinking more rationally. Thus, the book builds students’ approach, making them capable of solving difficult questions with ease and becoming more confident in the process. 


Key Features of NCERT Solutions for Class 11 Mathematics Chapter 9

The more you revise, the more you retain. Hence, Extramarks NCERT Solutions for Class 11 Mathematics Chapter 9 provides a complete revision guide to every student irrespective of their level. Along with the revision guides, students will also get access to study notes, solved questions from NCERT textbook and Exemplars, and so on. 

The key features are as follows: 

  • The entire chapter has been summarised in a point-wise manner in our NCERT solutions. 
  • Our NCERT solutions are prepared by subject matter experts working conscientiously and diligently to prepare authentic, concise answers which students can trust and enjoy the process of learning.
  • All the important formulas are listed in a structured way for the students to revise quickly.
  • After completing the NCERT Solutions for Class 11 Mathematics Chapter 9, students will be able to apply all the concepts related to sequence and series in other chapters.

Q.1 Write the first five terms of the sequence whose nth terms is:
an = n(n+2)


We havean= n(n+2)Substituting n=1,2,3,4,5 respectively, we geta1= 1(1+2)      =3a2= 2(2+2)      =8a3= 3(3+2)      =15a4= 4(4+2)      =24a5= 5(5+2)      =35Therefore, the first five terms are: 3, 8, 15, 24, 35.


Write the first five terms of the sequence whose nth terms is:an=nn+1.


We havean=nn+1Substituting n=1,2,3,4,5 respectively, we geta1=11+1      =12a2=22+1      =23a3= 33+1      =34a4=44+1      =45a5=55+1      =56Therefore, the first five terms are: 12,23,34,45,56.

Q.3 Write the first five terms of the sequence whose nth terms is:
an = 2n


We have, an= 2n Substituting n=1,2,3,4,5 respectively, we geta1= 21      =2a2= 22      =4a3= 23      =8a4=24      =16a5= 25      =32Therefore, the first five terms are: 2, 4, 8, 16, 32.


Write the first five terms of the sequence whose nth terms is:an=2n36.


We havean=2n36Substituting n=1,2,3,4,5 respectively, we geta1=2(1)36      =16a2=2(2)36      =16 a3=2(3)36      =36      =12a4=2(4)36      =56a5=2(5)36      =76Therefore, the first five terms are: 16,16,12,56,76.

Q.5 Write the first five terms of the sequence whose nth terms is an = (– 1)n–1 5n+1.


We have an=(1)n15n+1Substituting n=1, 2, 3, 4, 5 respectively, we geta1=(1)1151+1      =25a2=(1)2152+1      =125a3=(1)3153+1      =625 a4=(1)4154+1      =3125a5=(1)5155+1      =15625Therefore, the first five terms are:25,125,625,3125,15625.


Write the first five terms of the sequence whose nth terms isan=nn2+54


We have an=nn2+54Substituting n = 1, 2, 3, 4, 5 respectively, we geta1=112+54      =64=32a2=222+54      =92a3=332+54      =424=212a4=442+54      =21a5=552+54      =1504      =752Therefore, the first five terms are: 32,  92,  212,  21,  752.


Find the indicated terms in each of the sequences given below whose nth terms are:1. an= 4n3; a17, a242. an=n22n;  a73. an= (1)n1n3; a94. an=n(n2)n+3; a20



Given:an=4n3Substituting n=17 and 24 respectively, we geta17=4(17)3       =683       =65a24=4(24)3       =963       =93


Given:     an=n22nSubstituting n=7, we geta7=7227       =49128Thus, the value of a7 is 49128.


Given:     an=(1)n1n3Substituting n=9, we get    a9=(1)91(9)3 =(1)8(729) =729Thus, the value of a9 is 729. 4. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A1E1@ Given:     an=n(n2)n+3Substituting n=20, we get  a20=20(202)20+3 =20(18)23 =36023Thus, the value of a20 is 36023.


Write the first five terms of each of the sequences given below andobtain the corresponding series:1. a1=3, an=3an-1+2, for all n>12. a1=1, an=an-1n, n23. a1=a2=2, an=an11, n > 2



Given:an=3an1+2 and a1=3Substituting n=2, we geta2=3a21+2      =3a1+2      =3(3)+2[Putting a1=3]      =11Substituting n=3, we geta3=3a31+2      =3a2+2      =3(11)+2[Putting a2=11]      =35Substituting n=4, we geta4=3a41+2      =3a3+2      =3(35)+2[Putting a3=35]      =107 Substituting n=5, we geta5=3a51+2      =3a4+2      =3(107)+2[Putting a4=107]      =323Thus, the first five terms are 3, 11, 35, 107 and 323.Corresponding series is:  3+11+35+107+323+...


Given:a1=1,an=an1n,n2Substituting n=2, we geta2=a212      =a12      =12[Putting a1=1]Substituting n=3, we geta3=a313      =a23      =(12)3[Putting a2=12] =16Substituting n=4, we geta4=a414      =a34 =(16)4[Putting a3=16]      =124Substituting n=5, we geta5=a515      =a45      =(124)5[Putting a4=124]      =1120Thus, the first five terms are 1, 12, 16, 124 and 1120.The corresponding series is:(1)+(12)+(16)+(124)+(1120)+...


Given:     an=an11,  a1=a2=2, n>2Substituting n=3, we geta3=a311      =a21      =21[Putting a2=2]      =1 Substituting n=4, we geta4=a411      =a31      =11[Putting a3=1]      =0Substituting n=5, we geta5=a411      =a31      =01[Putting a4=0]      =1Thus, the first five terms of series are:2, 2, 1, 0,1The series is:  2+ 2+ 1+ 0+(1)+...


The Fibonacci sequence is defined by 1=a1=a2 and an=an1+an2,  n>2.Find an+1an, for n=1,2,3,4,5.


TheFibonaccisequenceisdefinedby1=a1=a2andan=an1+an2,n>2.Substitutingn=3,wegeta3=a2+a1 =1+1 =2Forn=1,wegeta1+1a1=a2a1=11a2a1=1anda3a2=21=2 an+1an=an+an1an-1+an2Substitutingn=3,wegeta3+1a3=a3+a31a31+a32 a4a3=a3+a2a2+a1 =a2+a1+a2a2+a1a3=a2+a1 =1+1+11+1=32Substituting n=4, we geta4=a3+a2=2+1=3a4+1a4=a4+a3a3+a2 a5a4=3+22+1=53Substitutingn=5,wegeta5=a4+a3=3+2=5 a5+1a5=a5+a4a4+a3 a6a5=5+33+2=85Thus,therequiredtermsare:1,2,32,53,85.

Q.10 Find the sum of odd integers from 1 to 2001.


The odd integers from 1 to 2001 are:1,3,5,7,9,11,13,...,1999,2001a=1,d=31=2  l=a+(n1)d2001=1+(n1)2 20011=(n1)2 n1=20002 n=1000+1 =1001    Sn=n2(a+l)      S1001=10012(1+2001)       S1001=10012(2002)                  =1001(1001)                    =1002001Thus, the sum of odd integers from 1 to 2001 is 1002001.

Q.11 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.


The multiples of 5 from 100 to 1000 are:105,105,110,115,120,  125,...,995 a=100,d=105100=5 l=a+n1d995=105+n15995105=n15 n1=8905 n=178+1 =179 Sn=n2a+l S179=1792105+995 =1792×1100 =98450Thus, the sum of the natural numbers which are multiple of 5 between 100 to 1000 is 98450.

Q.12 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.


            First term of A.P.=2Let common difference=dAccording to given condition:Sum of first 5 terms=14(Sum of next 5 terms)         S5=14(S10S5)      4S5=S10S5      5S5=S10 5[52{2×2+(51)d}]=[102{2×2+(101)d}][∵Sn=n2{2a+(n1)d}]     252(4+4d)=5(4+9d)       25(2+2d)=5(4+9d)           50+50d=20+45d                       5d=2050                         d=305         =6Now, sum of 20 terms.          T20=2+(201)×6 [∵an=a+(n1)d]         =2114         =112Thus, the 20th term is 112.


How many terms of the A.P.  6,112,5,...  are needed to givethe sum 25?


The given A.P. is 6,112,5,...Let the sum of n terms of A.P. is 25.Then, Sn=n2{2a+(n1)d} a=6 d=112(6)       =112+6=1225=n2{2×6+(n1)×12}[∵Sn=n2{2a+(n1)d}]25=n2{12+(n1)×12}100=n(24+n1)100=25n+n2n225n+100=0 (n20)(n5)=0        n=20,  5Thus, the number of terms of A.P. is 5 or 20.


In an A.P., if pth term is 1p and qth term is 1q, prove that the sum of first pq terms is 12(pq+1), where pq.


                    Let first term of A.P.=aLet common difference of A.P.=dSince, Tp=1q a+(p1)d=1q ...(i)And, Tq=1pa+(q1)d=1p ...(ii)Subtracting equation(ii) from equation(i), we get   (p1q+1)d=1q1p                (pq)d=pqpq                d=1pqSubstituting value of d in equation(i),weget      a+(p1)1pq=1q         a+ppq1pq=1q         a=1pqNow,         Spq=pq2{2×1pq+(pq1)1pq}[∵ Sn=n2{2a+(n1)d}] Spq=pq2{2×1pq+(pq1)1pq}   =pq2(2pq+11pq)   =pq2(1pq+1)   =12(1+pq)Thus, the sum of pq terms is 12(pq+1).

Q.15 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.


Given A.P. 25, 22, 19,First term of A.P.=25Common difference of A.P.=2225=3Let sum of n terms of A.P.=116i.e.,Sn=116 n22×25+n13=116n503n+3=23253n3n2=2323n253n+232=03n224n29n+232=03nn829n8=0n83n29=0n=8,293Since, n is a natural number.So,n=8Last term of A.P.=a+n1d=25+813=2521=4Thus, the last term of A.P. is 4.

Q.16 Find the sum to n terms of the A.P., whose kth term is 5k + 1.


kth term of A.P.=5k+1Tk=5k+1Substituting k=1,2,3,... respectively.        T1=5(1)+1    =6        T2=5(2)+1    =11        T3=5(3)+1    =16Common difference=T2T1    =116    =5 Sum of n terms,Sn=n2{2a+(n1)d}    =n2{2×6+(n1)5}    =n2(12+5n5)    =n2(7+5n)    =n2(5n+7)Thus, the sum of n terms is n2(5n+7).

Q.17 If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.


Given,Sn=(pn+qn2) Putting n=1,2,3respectively,weget      S1=(p×1+q×12)=p+q      S2=(p×2+q×22)=2p+4q      S3=(p×3+q×32)=3p+9q       T1=S2S1=(2p+4q)(p+q) =p+3q T2=S3S2=(3p+9q)(2p+4q)=p+5qCommon difference,d=T2T1=(p+5q)(p+3q)=2qThus, common difference is 2q.

Q.18 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.


Since,Sum of n terms of an A.P.Sum of n terms of another A.P.=5n+ 49n+ 6    n2{2a+(n1)d}n2{2A+(n1)D}=5n+ 49n+ 6        {a+(n12)d}{A+(n12)D}=5n+ 49n+ 6    ...(i)Since, T18=a+17d. So, Substituting n12=17 orn=35 in equation (i), we get {a+17d}{A+17D}=5×35+ 49×35+ 618th term of an A.P.18th term of another A.P.=175+ 4315+ 6 =179321Thus, the ratio of 18th terms of two A.P. is 179:321.

Q.19 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.


first term of A.P.=aand common difference of A.P.=d  Sp=Sq(Given)    p2{2a+(p1)d}=q2{2a+(q1)d}     2ap+p(p1)d=2aq+q(q1)d                2ap2aq=q(q1)dp(p1)d                  2a(pq)=(q2qp2+p)d                  2a(pq)={pq(p2q2)d}                  2a(pq)={(pq)(pq)(p+q)d}                  2a(pq)=(pq){1(p+q)d}                                2a={1(p+q)d} Now,          Sp+q=p+q2{2a+(p+q1)d}                       Sp+q=p+q2[{1(p+q)}d+(p+q1)d]                      Sp+q=p+q2{1(p+q)+(p+q)1}d                 Sp+q=0Thus, the sum of (p+q) terms is zero.


Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.Prove that apqr+bqrp+crpq=0.


Let first term of A.P. be A and common difference be d.Given:Sp=aa=p2{2A+(p1)d} ap=12{2A+(p1)d} Sq=bb=p2{2A+(q1)d} bq=12{2A+(q1)d}and Sr=cc=r2{2A+(r1)d} cr=12{2A+(r1)d} L.H.S.=ap(qr)+bq(rp)+cr(pq)   =12{2A+(p1)d}(qr)+12{2A+(q1)d}(rp) +12{2A+(r1)d}(pq)   =12[2A(qr+rp+pq)+d{(p1)(qr)+(q1)(rp)+(r1)(pq)}]   =12{2A(0)+d(pqprq+r+qrqpr+p+rprqp+q)}   =12(0+0)   =0=R.H.S.Hence proved.

Q.21 The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).


Let first term of an A.P.=aand commong difference of A.P.=dIt is given that: Sm:Sn=m2:n2SmSn=m2n2 m2{2a+(m1)d}n2{2a+(n1)d}=m2n2 {2a+(m1)d}{2a+(n1)d}=mn 2an+(mnn)d=2am+(mnm)d 2an2am=(mnm)d(mnn)d 2a(nm)=d(mnmmn+n) 2a(nm)=d(m+n) 2a=dNow,TmTn=a+(m1)2aa+(n1)2a =a+(m1)2aa+(n1)2a =1+2m21+2n2 =2m12n1Thus, the ratio of mth and nth term is (2m1):(2n1).

Q.22 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.


Itis given thatSn=3n2+ 5nReplace n by n1, we getSn-1=3n12+ 5n1=3n26n+3+ 5n5=3n2n2So,Tn=SnSn-1=3n2+ 5n3n2n2=6n+2Since,Tm=164So,6m+2=164m=16426=1626=27Thus, the value of m is 27.

Q.23 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.


Let 5 numbers between 8 and 26 are A1, A2, A3, A4 and A5.Then, A.P. is 8, A1, A2, A3, A4, A5, 26.∵ Last term=a+(n1)d       26=8+(71)d      26=8+6d          d=2686    =186    =3Then,A1=8+d    =8+3    =11              A2=8+2d    =8+2(3)    =14              A3=8+3d    =8+3(3)    =17              A4=8+4d    =8+4(3)    =20 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGOmaiaaiAdacqGH9aqpcaaI4aGaey4kaSIaaGOnaiaaykW7caWGKbaabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGKbGaeyypa0ZaaSaaaeaacaaIYaGaaGOnaiabgkHiTiaaiIdaaeaacaaI2aaaaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9maalaaabaGaaGymaiaaiIdaaeaacaaI2aaaaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9iaaiodaaeaacaWGubGaamiAaiaadwgacaWGUbGaaiilaiaaykW7caaMc8UaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadgeadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaI4aGaey4kaSIaamizaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9iaaiIdacqGHRaWkcaaIZaaabaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8Uaeyypa0JaaGymaiaaigdaaeaacaaMc8UaaGPaVlaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyqamaaBaaaleaacaaIYaaabeaakiabg2da9iaaiIdacqGHRaWkcaaIYaGaamizaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9iaaiIdacqGHRaWkcaaIYaWaaeWaaeaacaaIZaaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8Uaeyypa0JaaGymaiaaisdaaeaacaaMc8UaaGPaVlaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyqamaaBaaaleaacaaIZaaabeaakiabg2da9iaaiIdacqGHRaWkcaaIZaGaamizaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9iaaiIdacqGHRaWkcaaIZaWaaeWaaeaacaaIZaaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8Uaeyypa0JaaGymaiaaiEdaaeaacaaMc8UaaGPaVlaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyqamaaBaaaleaacaaI0aaabeaakiabg2da9iaaiIdacqGHRaWkcaaI0aGaamizaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9iaaiIdacqGHRaWkcaaI0aWaaeWaaeaacaaIZaaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8Uaeyypa0JaaGOmaiaaicdaaaaa@42BF@    A5=8+5d    =8+5(3)    =23Thus, the 5 terms between 8 and 26 are 11, 14, 17, 20 and 23.


If an+bnan1+bn1  is the A.M. between a and b, then find the value of n.


A.M. of a and b=a+b2an+bnan1+bn1=a+b2 a+ban1+bn1=2an+bnan+abn1+ban1+bn=2an+2bn                    abn1bn=anban1                 bn1ab=an1ab                  bn1=an1                  abn1=1=ab0                  n=1

Q.25 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.


Let m numbers between 1 and 31 are A1, A2, A3, A4,...Am-1,Am.Then, A.P. is 1, A1, A2, A3, A4,...Am-1,Am,  31.∵ Last term=a+(n1)d       31=1+(m+21)d       30=(m+1)d          d=30m+1Then,         A1=1+d    =1+30m+1    =m+31m+1                   A7=1+7d     =1+7(30m+1)    =m+211m+1            Am1=1+(m1)d    =1+(m1)(30m+1)    =m+1+30(m1)m+1    =31m29m+1According to question:A7Am1=(m+211m+1)(31m29m+1)      59=m+21131m29   155m145=9m+1899     155m9m=1899+145               146m=2044     m=2044146 =14Thus, the value of m is 14.

Q.26 The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.


The difference between any two consecutiveangles=5°The smallest angle of a polygon=120°Let number of sides in given polygon be n.Then,Sum of angles of a polygon=(n2)180°The series formed by angles of polygon is:120°,125°,130°,135°,140°, ... Sum of n angles of polygon of n sides=n2{2×120°+(n1)×5°}=n2{240°+5°n5°}=n2{235°+5°n}Therefore,n2{235°+5°n}=(n2)180°5°n2+235°n=360°n720°5°n2125°n+720°=0        n225°n+144°=0          (n9)(n16)=0                n=9,16Number of sides in polygon is 9 because number of angles in a polygon is equal to the number of sides. Sum of 16 angles will be equal to sum of 9 angles if some angles are 0° or negative, which is impossible. Therefore number of sides in the polygon is 9.


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


The given G.P. is: 5 2 , 5 4 , 5 8 , MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWGubGaamiAaiaadwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaae4raiaab6cacaqGqbGaaeOlaiaabccacaqGPbGaae4CaiaabQdacaWLjaGaaCzcaiaaxMaaieqacaWFGaWaaSaaaeaacaaI1aaabaGaaGOmaaaacaGGSaWaaSaaaeaacaaI1aaabaGaaGinaaaacaGGSaWaaSaaaeaacaaI1aaabaGaaGioaaaacaGGSaGaaiOlaiaac6cacaGGUaaaaa@534B@ a=52 r=(54)(52)=12T20=ar201 [∵Tn=arn1] =52(12)19 =5220andTn=arn1 =52(12)n1 =52n

Q.28 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.


Common ratio of given G.P.=28th term of A.P.=192ar81=192    a.27=192          a=19227=1.512th term of A.P.=ar121 =1.5(2)11=1.5×2048=3072

Q.29 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.


Let first term of G.P. be a and common ratio be r.  5th term of G.P.=p  T5=p    p=ar4  8th term of G.P.=q  T8=q    q=ar711th term of G.P.=sT11=s    s=ar10L.H.S.=q2       =ar72       =a2r14R.H.S.=ps       =ar4ar10       =a2r14So,     L.H.S.=R.H.S.Hence proved.

Q.30 The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.


        First term of G.P.=3Let common ratio of G.P.=rAccording to given condition, T4=(T2)2ar3=(ar)2  (3)r3=(3)2r2             r=3 Then,T7=ar6       =(3)(3)6       =(3)7       =2187Thus, 7th term of G.P. is 2187.


Which term of the following sequences:(a). 2, 22, 4, ... is 128?(b). 3, 33, 3, ... is 729?(c).  13, 19, 127,...  is  119683?


(a)Let tn=128, a=2,    r=222=2              tn=arn1         128=2(2)n1      27=(2)1+n12         7=1+n126×2=n1       n=13Thus, 128 is 13th term of G.P.(b)  Let tn=729, a=3,    r=33=3             tn=arn1         729=3(3)n1        36=(3)12+n12         6=12+n126×2=1+n1        n=12Thus, 729 is 12th term of G.P.(c)Let tn=729, a=13,   r=(19)(13)=13 tn=arn1    119683=13(13)n1       (13)9=(13)n         9=n        n=9Thus, 119683 is 9th term of G.P.


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


27,x,72 are in G.P., thenCommon ratio=x27   =7x2Also, common ratio=72x   =72xThen, 7x2=72x    x2=1     x=±1


Find the sum to indicated number of terms in each of the geometricprogressions given below.i. 0.15, 0.015, 0.0015,… 20 terms.ii. 7,  21,  37,  ...  n terms.iii. 1, a, a2, a3, ... n terms (if a1).iv. x3, x5, x7, ... n terms (if x± 1).


i.0.15, 0.015, 0.0015, 20 termsSince,Sn=a(1rn)1r                 a=0.15,        r=0.0150.15=0.1Sn=0.15{1(0.1)20}10.1       =0.15{1(0.1)20}0.9       =1590{1(0.1)20}       =16{1(0.1)20} ii.7,21,  37,...ntermsSince,Sn=a(rn1)r1 a=7,         r=217=3Sn=7{(3)n1}31×3+13+1        =7{(3)n1}31×(3+1) iii.1,a,a2,a3,...ntermsFirst term of G.P.=1Common ratio of G.P.=a1         =aSince,Sn=a(1rn)1r Sn=1{1(a)n}1+a iv.x3,x5,x7,... n termsFirst term of G.P.=x3 Common ratio of G.P.=x5x3         =x2Since,  Sn=a(rn1)r1   Sn=x3{1(x2)n}1x2 Sn=x3(1x2n)1x2


Evaluate:  k=111(2+3k).


We have,k=111(2+3k)=(2+31)+(2+32)+(2+33)+...+(2+311)=2×11+(31+32+33+...+311)=22+3(3111)31=22+32(3111)


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


Let three terms in A.P. are ar,a,ar.Then according to given conditions: ar+a+ar=3910a(1r+1+r)=3910 ...(i)andar×a×ar=1       a3=1         a=1Substituting value of a in equation(i),we get        1(1r+1+r)=3910    (1+r+r2r)=3910     10(r2+r+1)=39r10r2+10r+10=39r10r229r+10=010r225r4r+10=0 5r(2r5)2(2r5)=0   (2r5)(2r5)=0 r=52,52 Thus, first term of G.P. is 1 and common difference is 52.Three terms are: 152,1,1×5225,1,52Or  125,1,1×2552,1,25  

Q.36 How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?


G.P. is 3, 32, 33, a=3, r=323=3 and Sn=120Sn=a(rn1)(r1) 120=3(3n131)120=32(3n1)120×23=(3n1)81=3n      34=3n    n=4Thus, 4 terms of G.P. are needed to give the sum 120.

Q.37 The sum of 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.


Let first term of G.P. is a and common ratio is r. Then,six terms of G.P. are: a, ar, ar2, ar3, ar4, ar5, ar6.According to first condition: a+ar+ar2=16a(1+r+r2)=16       ...(i)     ar3+ar4+ar5=128 ar3(1+r+r2)=128   ...(ii)Dividing equation(ii) by equation(i), we get     ar3(1+r+r2)a(1+r+r2)=12816 r3=8   r=2Putting r=2 in equation(i), we get        a(1+2+22)=16 a=167 and r=2Sum of n terms is:   Sn=a(rn1)(r1) =167(2n1)(21)=167(2n1)

Q.38 Given a G.P. with a = 729 and 7th term 64, determine S7.


First term of G.P.=(a)          =7297th term of G.P.=T7          =64Then,  ar6=64  729r6=64     r6=64729     r6=2636       r=23 Sn=a(1rn)(1r)S7=729(12n3n)(123)   S7=729×3(12737)  S7=729×3(11282187)  S7=2187(21871282187)  S7=2059Thus, sum of 7 terms of G.P. is 2059.

Q.39 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.


Let G.P. be: a, ar, ar2, ar3, ar4, ar5,...Then according to first condition,    a+ar=4ora(1+r)=4or    a=41+r ...(i)Then according to second condition, T5=4T3 ar4=4(ar2)   r2=4    r=±2Putting r=2 in equation (i), we get     a=41+2       =43Putting r=2 in equation (i), we get    a=412       =41       =4Therefore, G.P. when a=43 and r=243, 43(2), 43(2)2, 43(2)3, 43(2)4, 43(2)5,...or       43,83,163,323,643,1283,...Therefore, G.P. when a=4 and r=24, 4(2), 4(2)2, 4(2)3, 4(2)4, 4(2)5,...or  4,8,16,32,...Therefore, the required G.P. is43,83,163,323,643,1283,...or4,8,16,32,...

Q.40 If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.


Let first term of G.P. be ‘a’ and common ratio be ‘r’.   T4=xar3=x T10=yar9=y T16=zar15=zNow,yx=ar9ar3       =r6and   zy=ar15ar9       =r6 yx=zyx,y,z are in G.P.

Q.41 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .


LetSn=8+88+888+8888+     =8(1+11+111+1111+...)     =89(9+99+999+9999+...)     =89{(101)+(1021)+(1031)+(1041)+...}     =89{10+102+103+104+...1111...n times}      =89{10(10n1)101n} [∵Sn=a(rn1)r1]     =89{10(10n1)9n}     =8081(10n1)89n


Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2, 12.


Sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 12=2×128+4×32+8×8+16×2+32×12=256+128+64+32+16=2561125112=2563213212=256×2×3132=496

Q.43 Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn – 1 and A, AR, AR2, … ARn–1 form a G.P, and find the common ratio.


The products of the corresponding terms of the sequencesa, ar, ar2, arn 1 and A, AR, AR2, ARn1=aA, aARr, aAR2r2, ,aARn 1rn 1 Common ratio=aARraA=RrCommon ratio=aAR2r2aARr=RrSince, common ratio is same i.e., rR.So, aA, aARr, aAR2r2, ,aARn 1rn 1 is in G.P.

Q.44 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.


Let four numbers of G.P. are a, ar, ar2, ar3.According to first condition,   T3T1=9ar2a=9ar21=9 ...iAccording to second condition, T2T4=18arar3=18   ar1r2=18   arr21=18 ...iiDividing equationii by equation i, we get       arr21ar21=189      r=2Putting r=2, in equationi, we geta221=9      a=93          =3Thus, the four terms of G.P. are:3, 32, 322, 323i.e.,    3,6,12,24


If the pth, qth and rth terms of a G.P. are a and b, respectively.Prove that  aqr brp cpq=1.


Let first term of G.P. is A and common ratio is R.Given that:Tp=aARp1=aTq=bARq1=bTr=cARr1=cL.H.S.=aqrbrpcpq       =(ARp1)qr(ARq1)rp(ARr1)pq       =Aqr+rp+pqR(p1)(qr)+(q1)(rp)(r1)(pq)       =A0Rpqprq+r+qrqpr+p+rprqp+q       =A0R0       =1×1       =1=R.H.S.

Q.46 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.


First term of G.P.=a    nth term of G.P.=bLet common ratio=rThe G.P. isa, ar, ar2, ar3, ar4,..., band b=arn-1Product of n terms=a×ar×ar2×ar3×ar4×...×b   =a×ar×ar2×ar3×ar4×...×arn-1    =an×r1+2+3+4+...+(n1)        P=an×r(n1)(n1+1)2        P=an×rn(n1)2Squarring both sides, we getP2=a2nrn(n1) =(a2rn1)n =(a×arn1)n P2=(ab)n [∵b=arn1]Hence Proved.


Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from  (n+1)th to (2n)th term is 1rn.


LetfirsttermofG.P.isaandcommonratioisr.Then,SumoffirstntermsSn=arn-1r-1      Sumof2ntermsS2n=ar2n-1r-1Sumoftermsfrom n+1thto2nthterm=S2n-Sn =ar2n-1r-1arn-1r-1=ar-1r2n-1-rn+1=arnr-1rn-1The required ratio=SnS2n-Sn=arn-1r-1arnr-1rn-1=1rnThus,  the ratio of the sum of first n terms of a G.P. to the sum of terms from  n+1th to 2nth term is 1rn.

Q.48 If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.


Since, a, b, c, d are in G.P.So, ba=cb=dc=r(let)     b=ar, c=br=ar2 and d=cr=ar4L.H.S.=(a2+b2+c2)(b2+c2+d2)    =(a2+a2r2+a2r4)(a2r2+a2r4+a2r6)   =a2(1+r2+r4)×a2r2(1+r2+r4)   =a4r2(1+r2+r4)2R.H.S.=(ab + bc + cd)2   =(a×ar + ar×ar2 + ar2×ar4)2   =a4r2(1+r2+r4)2L.H.S.=R.H.S.Therefore,(a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2Hence proved.

Q.49 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.


Let two numbers between 3 and 81 are x and y in such a waythat 3, x, y, 81 are in G.P.So, first term of G.P.=3Common difference=r   4thterm of G.P.=ar3        3r3=81           r3=813           r=273     =3 x=T2     =ar     =3×3     =9andy=T3     =ar2     =3×32     =27Thus, two numbers between 3 and 81 are 9 and 27.


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


Geometric mean between a and b=aban+1+bn+1an+bn=abab12an+bn=an+1+bn+1an+12b12+a12bn+12=an+1+bn+1an+12b12an+1=bn+1a12bn+12an+12b12a12=bn+12b12a12 an+12=bn+12 abn+12=1abn+12=ba0n+12=0n=12


The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+22):(322).


Let two numbers be a and b.G.M. of a and b=abThen, according to given condition,     a+b=6ab ...(i)∵         (ab)2=(a+b)24ab         (ab)2=(6ab)24ab      =36ab4ab      =32ab     ab=32ab     ab=42ab ...(ii) Solving equation(i) and equation(ii), we geta=(3+22)ab and b=(322)abThen,         ratio of the numbers=(3+22)ab(322)ab      =(3+22):(322)


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


Let two numbers are a and b.Then, A.M.=a+b2 A=a+b2     a+b=2A ...i∵     G.M.=ab           G=ab         G2=ab ...ii∵     ab=a+b24ab      =2A24G2     ab=2A2G2 ...iiiSolving equation i and equationiii, we geta=A+A2G2 and b=AA2G2Therefore, the two numbers are:A±A+GAG.

Q.53 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 2nd hour, 4th hour and nth hour?


Originally present bacteria in a culture=30         Number of bacteria after one hour=60Since, number of bacteria doubles every hour.So, common ratio(r) in G.P.=2        First term(a) in G.P.=30Number of bacteria after 2nd hour   =ar2   =30×22   =120Number of bacteria after 4th hour   =ar4   =30×24   =480 Number of bacteria after nth hour   =arn   =30×2n   =30(2n)

Q.54 What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?


Principal amount deposited in the bank=₹500Rate of interest in the bank=10%p.a.Number of years for amount deposited=10 yearsAmount after one year in the bank=500(1+10100)=500(1110)Amount after two years in the bank=500(1+10100)2=500(1110)2Amount after three years in the bank=500(1+10100)3=500(1110)3So, G.P. formed by using amount obtained from bank is 500(1110),500(1110)2,500(1110)3,...First term of G.P.=500(1110)Common ratio of G.P.=1110Number of years=10Since, Tn=arn1T10=500(1110)×(1110)101        =500(1.1)10Thus, ₹ 500 amounts to ₹500(1.1)10 in 10 years after its deposit in a bank.

Q.55 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.


Let roots of a quadratic equation be a and b.A.M. of a and b=8   a+b2=8    a+b=16 ...(i)  G.M. of a and b=5     ab=5         ab=25 ...(ii)Quadratic equation, whose roots are a and b,         (xa)(xb)=0x2(a+b)x+ab=0          x216x+25=0 [From equation (i) and (ii)]Thus, the required quadratic equation is  x216x+25=0.

Q.56 Find the sum to n terms of each of the series in Exercises 1 to 7.

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +…
2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
3. 3 × 12 + 5 × 22 + 7 × 32 + …

1 1×2 + 1 2×3 + 1 3×4 + MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaadaWcaaqaaGqabiaa=fdaaeaacaWFXaGaa831aiaa=jdaaaGaa83kamaalaaabaGaa8xmaaqaaiaa=jdacaWFxdGaa83maaaacaWFRaWaaSaaaeaacaWFXaaabaGaa83maiaa=DnacaWF0aaaaiaa=TcacaWFUaGaa8Nlaiaa=5caaaa@47AA@

5. 52 + 62 + 72 + … + 202
6. 3 × 8 + 6 × 11 + 9 × 14 + …
7. 12 + (12 + 22) + (12 + 22 + 32) + …


1. Let  Sn=1×2+2×3+3×4+4×5 +...+n×n+1         an=n×n+1        Sn=n×n+1    =n2+n    =nn+12n+16+nn+12    =nn+122n+13+1    =nn+122n+43    =nn+12×2n+23        Sn=13nn+1n+22. Let  Sn=1×2×3+2×3×4+3×4×5+...+n×n+1×n+2         an=n×n+1×n+2        Sn=n×n+1×n+2    =n3+3n2+2n    =n3+3n2+2n    =nn+122+3.nn+12n+16+2.nn+12    =n2n+124+nn+12n+12+2nn+12    =nn+12nn+12+2n+1+2    =nn+12n2+n+4n+2+42    =nn+12n2+5n+62    =nn+1n+2n+34        Sn=14nn+1n+2n+33. an=2n+1×n2        Sn=2n3+n2    =2n3+n2    =2nn+122+nn+12n+16    =2.n2n+124+nn+12n+16    =n2n+122+nn+12n+16    =nn+12nn+1+2n+13    =nn+123nn+1+2n+13    =nn+123n2+3n+2n+13    =nn+123n2+5n+13        Sn=16nn+13n2+5n+14. Let  Sn=11×2+12×3+13×4+...+1nn+1    =1112+1213+1314+...+1n1n+1    =111n+1    =n+11n+1    =nn+15. 52+62+72+... + 202       =12+22+32+42+52+62+72+... + 20212+22+32+42       =20242       =2020+12×20+1644+12×4+16       =20×21×4164×5×96       =172201806       =170406       =28406. Let Sn=3×8+6×11+9×14+...        Tn=3,6,9,...nthterm8,11,14,...nth term   =3+n138+n13   =3n3n+5   =9n2+15n       Sn=Tn    =9n2+15n    =9n2+15n    =9×nn+12n+16+15×nn+12    =3×nn+12n+12+15×nn+12    =3nn+122n+1+5    =3nn+122n+6    =3nn+1n+37. Let       Sn=12+12+22+12+22+32+ ...  +12+22+32+...+n2Here,  Tn=12+22+32+...+n2=n2=nn+12n+16=16n2n2+3n+1=162n3+3n2+n    Sn=Tn=162n3+3n2+n=162n3+3n2+n=162nn+122+3nn+12n+16+nn+12=16n2n+122+nn+12n+12+nn+12=16×nn+12nn+1+2n+1+1=16×nn+12n2+n+2n+2=112nn+1n2+3n+2=112nn+1n+1n+2    Sn=nn+12n+212

Q.57 Find the sum to n terms of the series in whose nth terms is given by:

1. n (n + 1) (n + 4).
2. n2 + 2n
3. (2n – 1)2



Let  Tn=n(n+1)(n+4)     =n(n2+5n+4)    =n3+5n2+4n        Sn=Tn    =(n3+5n2+4n)    =n3+5n2+4n    ={n(n+1)2}2+5n(n+1)(2n+1)6+4n(n+1)2    =n(n+1)2{n(n+1)2+5.(2n+1)3+4}    =n(n+1)2{3n(n+1)+10(2n+1)+246}    =n(n+1)2(3n2+3n+20n+10+246)Sn=n(n+1)12(3n2+23n+34)


Let   Tn=n2+2nSn=Tn       =(n2+2n)       =n2+2n       =n(n+1)(2n+1)6+(2+22+23+24+...+2n) =n(n+1)(2n+1)6+2(2n1)21    [∵Sn=a(rn1)r1]       =n(n+1)(2n+1)6+2(2n1) Sn=16n(n+1)(2n+1)+2(2n1)


Let   tn=(2n1)2     =4n24n+1         Sn=(4n24n+1)      =4n24n+1      =4n(n+1)(2n+1)64n(n+1)2+n      =2n(n+1)(2n+1)32n(n+1)+n      =2n(2n2+3n+1)32n22n+n      =2(2n3+3n2+n)32n2n      =4n3+6n2+2n6n23n3      =4n3n3       =n3(4n21)Sn=n3(2n1)(2n+1)


Find the sum to infinity in each of the following Geometric Progression.1. 1,13,19,...2. 6, 1.2, 0.24,...3. 5,207,8049,...4.  34,316,-364,...5. Prove that:312×314×318...=36. Let x = 1+a+a2+... and y = 1+b+b2+..., where a<1 and b>1. Prove that 1+ab+a2b2+...=xyx+y-1


1. Given G.P. is:1,13,19,...a=1, r=131=13 S=a1r =1113 =123 =32=1.52. The given G.P. is:6, 1.2, 0.24, a=6, r=1.26=0.2S=a1r =610.2 =60.8 =608 =7.5Thus, the sum of infinite terms is 7.5.3. The given G.P. is:5,207,8049,...        a=5,         r=2075=47S=a1r =5147 =537 =353 =11.6Thus, the sum of infinite terms is 11.6.4. The given G.P. is:34,316,364,...a=34,  r=31634 =14S=a1r=34114=34×45=355. L.H.S.=312×314×318...=312+14+18+...=312112∵S=a1r=31212=31=3=R.H.S.Hence, it was to be proved.6. x =1+a+a2+...    =11aS=a1rand y =1+b+b2+...    =11bL.H.S.=1+ab+a2b2+...     =11abR.H.S.=xyx+y1     =11a11b11a+11b1     =11b+1a1a1b     =12ba1ba+ab     =12ba1+b+aab     =11abSo,L.H.S.=R.H.S.

Q.59 Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.


Let first term of A.P. be a and common difference be d.Then,Tm+n+Tm-n={a+(m+n1)d}+{a+(mn1)d}   =2a+(m+n1+mn1)d   =2a+2(m1)d   =2{a+(m1)d}  =2TmThus,Tm+n+Tm-n=2TmHence proved.

Q.60 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.


Let three numbers in A.P. are ad, a, a+d.Then, according to first condition: (ad)+a+(a+d)=24  3a=24     a=243 =8According to second condition:        (ad)a(a+d)=440     (8d)8(8+d)=440       (8d)(8+d)=4408    64d2=55    d2=6455          =9       d=±3The three numbers in A.P. are:83,8,8+3 or 8+3,8,835,  8,  11 or 11, 8, 5Thus, the three numbers are 5, 8, 11.

Q.61 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3(S2 – S1).


∵S1=Sn=n22a+n1dS2=S2n=2n22a+2n1dS3=S3n=3n22a+3n1dR.H.S.=3S2S1=32n22a+2n1dn22a+n1d=3n222a+2n1d2a+n1d=3n24a2a+4n2n+1d=3n22a+3n1d=S3=L.H.S.Thus,S3=3S2S1 is proved.

Q.62 Find the sum of all numbers between 200 and 400 which are divisible by 7.


The numbers between 200 to 400 divisible by 7 are:203,210,217,224,231,...,399.Here, a=203, d=210203=7,  l=399∵         l=a+(n1)d        399=203+(n1)71967=n1 28+1=n     n=29  Sn=n2(a+l)S29=292(203+399)         =292×602         =8729Thus, the sum of multiples of 7 between 200 and 400 is 8729.

Q.63 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.


The numbers from 1 to 100 divisible by 2:        2,4,6,8,...,100Here, a=2, d=42=2,  l=100∵          l=a+(n1)d         100=2+(n1)2    982=n1       n=49+1 =50∵   Sn=n2(a+l) S50=502(2+100) =25×102 =2550The numbers from 1 to 100 divisible by 5:5,10,15,20,...,100Here, a=5, d=105=5,  l=100∵         l=a+(n1)d        100=5+(n1)5    955=n1      n=19+1 =20∵  Sn=n2(a+l)S20=202(5+100) =10×105 =1050The numbers from 1 to 100 divisible by 2 and 5:10,20,30,...,100Here, a=10, d=2010=10,  l=100∵         l=a+(n1)d        100=10+(n1)10   9010=n1      n=10∵  Sn=n2(a+l)S10=102(10+100)        =5×110        =550Sum of numbers divisible by 2 or 5=Sum of multiples of 2+Sum of multiples of 5Sum of multiples of 10=2550+1050550=3050

Q.64 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.


Two digit numbers which when divided by 4, yields 1 as remainder are:13,17,21,...,97 a=13,d=1713=4,l=97∵l=a+(n1)d    97=13+(n1)d9713=(n1)4       844=n1 n=21+1     =22 ∵     Sn=n2(a+l)   S22=222(13+97)     =222×110     =1210Thus, the sum of 22 terms is 1210.


If f is a function satisfying fx+y=fxfy for all x, yN such that f1=3 and x=1nfx=120, find the value of n.


Since,f(x+y)=f(x)f(y), f(1)=3 and x=1nf(x)=120.So,  f(2)=f(1+1) =f(1)f(1) =3×3 =32          f(3)=f(2+1) =f(2)f(1) =32×3 =33          f(4)=f(2+2) =f(2)f(2) =32×32 =34∵x=1nf(x)=120f(1)+f(2)+f(3)+f(4)+...+f(n)=120 3+32+33+34+...+3n=120       3(3n1)31=120         3n1=120×23               3n=80+1               3n=34n=4Thus, the value of n is 4.

Q.66 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.


First term of G.P.=5Common difference of G.P.=2Let number of terms in G.P.=n           Sum of n terms in G.P.=3155×(2n1)21=315 2n1=3155 2n=63+1 2n=26 n=6∵ l=arn1   =5×2(61)   =5×32   =160Thus, the last term of G.P. is 160 and number of terms in G.P. is 6.

Q.67 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.


First term of G.P.=1Let common ratio=r        T3+T5=90 ar2+ar4=90  1.r2+1.r4=90        r4+r2=90r4+r290=0  (r2+10)(r29)=0r2=9,10 r=±3 [Neglecting r2=10]Thus, the common ratio of G.P. is ±3.

Q.68 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.


Let three terms in G.P. are ar,a,ar.Then, according to first condition, we havear+a+ar=56 ...(i)On subtracting 1, 7, 21 frome each number respectively,(ar1),(a7),(ar21) are in A.P., then2(a7)=(ar1)+(ar21)2a14=ar+ar22ar+ar2a=8 ...(ii)From equation(i) and equation(ii), we get         56a2a=8 568=3aa=483=16Putting a=14 in equation (i),we get16r+16+16r=56 1r+r=561616             1r+r=4016           2+2r2=5r  2r25r+2=02r24rr+2=02r(r2)1(r2)=0       (r2)(2r1)=0r=2,12So, three numbers when a=16 and r=2,12,16×2,16,162 or 162,16,16×2i.e., 32,16,8 or 8,16,32Thus, the required three numbers are 8,16,32.

Q.69 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.


Let G.P. containing even number of terms is:a, ar, ar2, ar3,..., arn, arn+1,.,ar2n1,ar2nAccording to given condition,S2n=5 ×(a+ar2+ar4+...+ar2n1) (a+ar+ar2+ar3+...+arn+arn+1+...+ar2n1+ar2n)=5×(a+ar2+ar4+...+ar2n1) (ar+ar3+...+arn+...+ar2n)=4×(a+ar2+ar4+...+ar2n1) ar(r2n1)r1=4×a{(r2)n1}r1r(r2n1)=4(r2n1)        r=4Thus, the common ratio of G.P. is 4.

Q.70 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.


Let common difference of an A.P.=dLet number of terms in an A.P.=nThe first term of A.P.=11Sum of the first four terms of an A.P.=56 42{2×11+(41)d}=56[Sn=n2{2a+(n1)d}]3d=2822 d=63 =2Sum of the last four terms of an A.P.=112{a+(n4)d}+{a+(n3)d}+{a+(n2)d}+{a+(n1)d}=112{11+(n4)2}+{11+(n3)2}+{11+(n2)2}+{11+(n1)2}=11244+(n4+n3+n2+n1)2=1124n10=112442 n=34+104   =11Thus, the number of terms in an A.P. is 11.


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


We are given:a+bxabx=b+cxbcx=c+dxcdx(x0) a+bxabx=b+cxbcx(a+bx)(bcx)=(b+cx)(abx)abacx+b2xbcx2=bab2x+caxcbx22b2x=2acx     b2=ac      ba=cb ...(i)Andb+cxbcx=c+dxcdx(b+cx)(cdx)=(c+dx)(bcx)bcbdx+c2xcdx2=cbc2x+dbxdcx22c2x=2bdx     c2=bd     cb=dc ...(ii)From equation(i) and equation(ii), we getba=cb=dca,b,c,d are in G.P.

Q.72 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.


Let n terms of G.P. area, ar, ar2, ar3,..., arn3,arn2,arn1 Then according to the given conditions:S=a+ar+ar2+ar3+...+ arn3+arn2+arn1 =a(rn1)(r1)P=a×ar×ar2×ar3×...×arn3×arn2×arn1 =an×r1+2+3+...+(n3)+(n2)+(n1) =an×r(n1)(1+n1)2 =an×rn(n1)2Squarring both sides, we getP2=a2n×rn(n1) R=1a+1ar+1ar2+1ar3+...+1arn1 =1a(11rn)(11r) =rarn(rn1r1) Rn=rn(1n)an(rn1r1)n L.H.S.=P2Rn =a2n×rn(n1)×rn(1n)an(rn1r1)n =an×rn2n+nn2×(rn1r1)n =an×(rn1r1)n ={a(rn1)r1}n =Sn=R.H.S.

Q.73 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0.


Let first term of A.P. is A and common difference is D.Then, according to given condition:       Tp=aA+(p1)D=a ...(i)       Tq=bA+(q1)D=b...(ii)       Tr=cA+(r1)D=c...(iii) L.H.S.=(qr)a+(rp)b+(pq)c =(qr){A+(p1)D}+(rp){A+(q1)D} +(pq){A+(r1)D} =A(qr+rp+pq)+D{(qr)(p1)+(rp)(q1)+(pq)(r1)} =A(0)+D(qpqrp+r+rqrpq+p+prpqr+q) =0+0 =0=R.H.S.This was to be proved.


If a(1b+1c), b(1c+1a), c(1a+1b) are in A.P., prove that a, b and c are in A.P.


Since,a1b+1c, b1c+1a,c1a+1bareinA.P. ab+ac,bc+ba, ca+cbareinA.P. aa+ab+ac,bb+bc+ba, cc+ca+cbareinA.P. a1a+1b+1c,b1b+1c+1a, c1c+1a+1bareinA.P.a,b,careinA.P.

Q.75 If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.


a,b,c,d are in G.Pba=cb=dc=r(let)b=ar,c=br and d=crb=ar,c=ar2 and d=ar3Then,(bn+cn)(an+bn)=(anrn+anr2n)(an+anrn) =anrn(1+rn)an(1+rn) =rn(cn+dn)(bn+cn)=(anr2n+anr3n)(anrn+anr2n) =anr2n(1+rn)anrn(1+rn) =rnSo,   (bn+cn)(an+bn)=(cn+dn)(bn+cn)Therefore, (an+bn),(bn+cn),(cn+dn) are G.P.

Q.76 If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.


a and b are the roots of x23x+p=0, then sum of zeroes=31             a+b=3 ...(i)Product of zeroes =p1 ab=p1ab=p...(ii)c and d are the roots of x212x+q=0, thensum of zeroes=121 c+d=12 (iii)Product of zeroes =q1 cd=q1 cd=q...(iv)Since, a,b,c,d are in G.P.So, ba=cb=dc=r(let) b=ar, c=br, d=cr b=ar, c=ar2, d=ar3From equation(i), we get a+ar=3a(1+r)=3 ...(v)From equation(iii), we get ar2+ar3=12ar2(1+r)=12 ...(vi)Dividing equation (vi) by equation(v), we getar2(1+r)a(1+r)=123r2=4    r=±2Putting r=2, in equation(v),we  get a(1+2)=3   a=33=1Putting r=2, in equation(v),we  get a(12)=3   a=31=3Case I: If a=1 and r=2, then a=1,b=1×2, c=1×22, d=1×23a=1,b=2,c=4,d=8 L.H.S.=q+pqp         =cd+abcdab         =4×8+1×24×81×2         =32+2322         =3430         =1715=R.H.S.Case II: If a=3 and r=2, then a=3,b=3×2, c=3×(2)2, d=3×(2)3a=3,b=6,c=12,d=24L.H.S.=q+pqp         =cd+abcdab         =12×24+(3)×612×24(3)×6         =28818288+18         =306270 =1715=R.H.S.Thus, (q+p):(qp)=17:15.   Hence proved.


The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n.Show that a : b = m + m2 n2: m m2 n2.


We are given,A.M. of a and bG.M. of a and b=mn      (a+b2)ab=mna+b2ab=mnApply componendo and dividendo, we get     a+b+2aba+b2ab=m+nmn        (a+b)2(ab)2=m+nmn    a+bab=m+nmnApply again componendo and dividendo, we get  a+b+aba+ba+b=m+n+mnm+nmn    2a2b=m+n+mnm+nmn         (ab)2=(m+n+mn)2(m+nmn)2[Squarring both sides]            ab=m+n+mn+2m+nmnm+n+mn2m+nmn            ab=2m+2m2n22m2m2n2            ab=m+m2n2mm2n2Thus, a:b=(m+m2n2):(mm2n2).


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



Q.79 Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + …
(ii) .6 +. 66 +. 666+…


(i) LetSn=5+55+555+           =5(1+11+111+1111+...)                 =59(9+99+999+9999+...)          =59{(101)+(1021)+(1031)+(1041)+...}          =59{10+102+103+104+...1111...n times}          =59{10(10n1)101n} [∵Sn=a(rn1)r1]          =59{10(10n1)9n}          =5081(10n1)5n9(ii) LetSn=0.6+0.66+0.666+          =6(0.1+0.11+111+1111+...)          =69(0.9+0.99+0.999+0.9999+...)          =23{(10.1)+(10.01)+(10.001)+(10.0001)+...}          =23{1+1+1+1+...n times110110211031104...}           =23{n110(1110n)1110}[∵Sn=a(rn1)r1]          =23n23×(1110n)9          =23n227×(1110n)

Q.80 Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.


2×4 +4×6+6×8 + ... +n termsTn={2+(n1)2}{4+(n1)2} =(2n)(2n+2) =4n(n+1)Putting n=20, we getT20=4(20)(20+1) =80×21 =1680Thus, the 20th term of series is 1680.

Q.81 Find the sum of the first n terms of the series: 3+ 7 + 13 + 21 + 31 +…


Let Sn=3+7+13+21+31+...+an1+anand Sn=    37132131...an1an¯  0=3+4+6+8+10+...n1termsanan=3+n128+n22      =3+n14+n2      =3+n1n+2an=3+n2+n2      =n2+n+1         Sn=n2+n+1     =nn+12n+16+nn+12+n     =n6n+12n+1+3n+1+6     =n62n2+3n+1+3n+3+6     =n62n2+6n+10Sn=n3n2+3n+5


If S1, S2, S3are the sum of first n natural numbers,their squares and their cubes,respectively, show that9S22=S31+8S1.


We are given thatS1=n    =n(n+1)2S2=n2    =n(n+1)(2n+1)6S3=n3    ={n(n+2)2}2R.H.S.=S3(1+8S1)    ={n(n+2)2}2{1+8×n(n+1)2}    =n2(n+2)24{1+4n(n+1)}    =n2(n+2)24(1+4n2+4n)    =n2(n+2)24(4n2+4n+1)    =n2(n+2)2(2n+1)24    ={3×n(n+2)(2n+1)6}2     =9{n(n+2)(2n+1)6}2 =9S22=L.H.S.This was to be proved.


Find the sum of the following series up to n terms:131+13+231+3+13+23+331+3+5+...


131+13+231+3+13+23+331+3+5+...Tn=13+23+33+...+n31+3+5+... =n3n2{2×1+(n1)2} ={n(n+1)2}2n2(2+2n2) =n2(n+1)24n2 =14(n2+2n+1) Sn=14(n2+2n+1) =14[n(n+1)(2n+1)6+2n(n+1)2+n] =14{n(n+1)(2n+1)6+n+1+1} =n4(2n2+3n+1+6n+126) =n4(2n2+9n+136) =n24(2n2+9n+13)


Show that1×22+2×32+...+n×(n+1)212×2+22×3+...+n2×(n+1)=3n+53n+1


L.H.S.=1×22+2×32+...+n×(n+1)212×2+22×3+...+n2×(n+1)  ={n×(n+1)2}{n2×(n+1)}  =(n3+2n2+n)(n3+n2)   =n3+2n2+nn3+n2  ={n(n+1)2}2+2{n(n+1)(2n+1)6}+n(n+1)2{n(n+1)2}2+n(n+1)(2n+1)6  =n(n+1)2{n(n+1)2+2×2n+13+1}n(n+1)2{n(n+1)2+2n+13}  =(3n2+3n+8n+4+66)(3n2+3n+4n+26)=3n2+11n+103n2+7n+2=(n+2)(3n+5)(n+2)(3n+1) =(3n+5)(3n+1)=R.H.S.This was to be proved.

Q.85 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.


Let 150 workers finish a job in n days.         Number of workers started the job=150Number of workers left the job per day=4The series of workers remain on the job every day150,146,144,140,...Then,Number of workers finished job in a day=Total Number of reducing workers finished job in (n+8)days        150n=150+146+144     +...(n+8) terms        150n=n+82{2×150+(n+81)(4)}        300n=(n+8){3004(n+7)}        300n=(n+8)(3004n28)         300n=4(n+8)(68n)           75n=(n260n544)      n260n544+75n=0             n2+15n544=0       (n+32)(n17)=0             n=17,32(Negative)Thus, the number of days required to complete work by 150 workers is 17 days.Therefore, number of days required to finishthe job by reducing number of workers=18+7 =25 days

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