NCERT Solutions For Class 11 Maths Chapter 9

NCERT Solutions Class 11 Maths Sequences and Series

Class 11 is an important phase of a student's life because the topics which are taught in class 11 are basics of the topics which will be taught in class 12. Students studying in class 11 should try to understand the chapters in a better way so that they don’t get confused while they will be learning the tough topics in class 12. NCERT Solutions for class 11 maths chapter 9 sequence and series are provided here so that students can refer to these solutions when they are facing difficulties to solve the mathematical problems. The NCERT Solutions for class 11 maths chapter 9 deals with the topics of Sequences and Series. The various problems that are asked during this topic are finding out the different problems related to the topics of sequences and series. Figuring out the initial terms of the sequences is important to understand, as it is the most commonly asked for problem type in Sequences and Series. The NCERT Solutions for Class 11 maths deals with the various topics in mathematics.

NCERT Solutions Class 11 Maths Chapter 9 Exercises

Exercise 9.1

Q1: as = s (s + 3) is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = s (s + 3)

Putting s = 1, 2, 3, 4 and 5 respectively, in as = s (s + 3)

a1 = 1 (1 + 3) = 4

a2 = 2 (2 + 3) = 10

a3 = 3 (3 + 3) = 18

a4 = 4 (4 + 3) = 28

a5 = 5 (5 + 3) = 40

The starting five terms of the sequence are 4, 10, 18, 28 and 40

 

Q2: as = ss+2 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = ss+2

Putting s = 1, 2, 3, 4 and 5 respectively, in ss+2

a1 = 11+2 = 13

a2 = 22+2 = 24

a3 = 33+2 = 35

a4 = 44+2 = 46

a5 = 55+2 = 57

The starting five terms of the sequence are 12,23,34,45and56

 

Q3: as = 5s is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = 5s

Putting s = 1, 2, 3, 4 and 5 respectively, in as = 5s

a1 = 51 = 5

a2 = 52 = 25

a3 = 53 = 125

a4 = 54 = 625

a5 = 55 = 3125

The starting five terms of the sequence are 5, 25, 125, 625 and 3125.

 

Q4: as = 2s+23 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = 2s+23

Putting s = 1, 2, 3, 4 and 5 respectively, in 2s+23

a1 = 2.1+23=43

a2 = 2.2+23=63

a3 = 2.3+23=83

a4 = 2.4+23=103

a5 = 2.5+23=123

The starting five terms of the sequence are 43,63,83,103and123

 

Q5: as = (1)s1+3s+1 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = (1)s1+3s+1

Putting s = 1, 2, 3, 4 and 5 respectively, in (1)s1+3s+1

a1 = (1)11+31+1 = 1 + 9 = 10

a2 = (1)21+32+1 = – 1 + 27 = 26

a3 = (1)31+33+1 = 1 + 81 = 82

a4 = (1)41+34+1 = – 1 + 243 = 242

a5 = (1)51+35+1 = 1 + 729 = 730

The starting five terms of the sequence are 10, 26, 82, 242 and 730.

 

Q6: as = (1)s13s+1 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = (1)s13s+1

Putting s = 1, 2, 3, 4 and 5 respectively, in (1)s13s+1

a1 = (1)1131+1=(1)032=19

a2 = (1)2132+1=(1)133=127

a3 = (1)3133+1=(1)234=181

a4 = (1)4134+1=(1)335=1243

a5 = (1)5135+1=(1)436=1729

The starting five terms of the sequence are 19,127,181,1243,and1729

 

Q7: as = s + 3s is the sth term of a sequence. Obtain the 17th and 22nd term in the required sequence.

Answer:

as = s + 3s

Putting s = 17 and 22 respectively, in as = s + 3s

a17 = 17 + 317

a22 = 22 + 322

 

Q8: as = 9+ss2 is the sth term of a sequence. Obtain the 24th term in the required sequence.

Answer:

as = 9+ss2

Putting s = 24, in as = 9+ss2

a24 = 9+24242=33576

 

Q9: as = (1)s×6s is the sth term of a sequence. Obtain the 11th and 12th term in the required sequence.

Answer:

as = (1)s×6s

Putting s = 11 and 12, in as = (1)s×6s

a11 = (1)11×611=611

a12 (1)12×612=612

 

Q10. as = s+32+s is the sth term of a sequence. Obtain the 21st and 82nd term in the required sequence.

Answer:

as = s+32+s

Putting s = 21 and 82, in as = s+32+s

a21=21+32+21=2423 a82=82+32+82=8584

 

Q11. Find out the starting five terms and also the following series of the required sequence given below:

a1 = 2, as = 2 as – 1 for all s > 1

Answer:

a1 = 2, as = 2 as – 1 for all s > 1

a2 = 2 a1 = 2 x 2 = 4

a3 = 2 a2 = 2 x 4 = 8

a4 = 2 a3 = 2 x 8 = 16

a5 = 2 a4 = 2 x 16 = 32

The starting five terms of the sequence are 2, 4, 8, 16 and 32

The following series is 2 + 4 + 8 + 16 + 32 + …….

 

Q12. Find out the starting five terms and also the following series of the required sequence given below:

a1 = 2, as = as112 for all s > 3

Answer:

a1 = 2, as = as112 for all s > 3

a2112=a112=212=12a3=a3112=a212=1212=14a4=a4112=a312=1412=58a5=a5112=a412=5812=1316

The starting five terms of the sequence are 2,12,14,58and1316

The following series is 2+12+(14)+(58)+(1316)+

 

Q13.  Find out the starting five terms and also the following series of the required sequence given below:

Answer

a1 = a2 = 4, as = as – 1 + 1 for all s > 4

a3 = a 3 – 1 + 1 = a2 +1 = 4 + 1 = 5

a4 = a 4 – 1 + 1 = a3 + 1 = 5 + 1 = 6

a5 = a 5 – 1 + 1 = a4 + 1 = 6 + 1 = 7

The starting five terms of the sequence are 4, 4, 5, 6 and 7

The following series is 4 + 4 + 5 + 6 + 7 + ….

 

Q14. The sequence of Fibonacci is described as 1 = a1 = a2 and as = as – 1 + as – 2, n > 1. Obtain as+1as  for n = 2, 4.

Answer:

1 = a1 = a2 and as = as – 1 + as – 2, n > 1.

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

For s = 2 and 4 respectively,

a2+1a2=a3a2=21=2 a4+1a4=a5a4=53

 

Exercise 9.2

Q1. Obtain the sum of all integers which are odd between 1 and 4001 including 1 and 4001.

Answer:

The integers which are odd between 1 and 2001 are 1, 3, 5, 7, 9, ……………. , 3997, 3999, 4001.

The sequence is in A.P form.

a = 1 [1st term]

Difference, d = 2

The standard equation of A.P,

a + (s – 1) d = 4001

1 + (s – 1) 2 = 4001

s = 2001

SS = s2[2a+(s1)d]Ss=20012[2.1+(20011)2]=20012[2+4000]=20012×4002=2001×2001=4004001

Hence, the sum of all integers which are odd between 1 and 4001 is 4004001.

 

Q2. Obtain the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5.

Answer:

All natural numbers which are lying between 10 and 100, and are multiples of 5, are 15, 20, 25,  … , 95.

The sequence is in A.P form.

a = 15 [1st term]

Difference, d = 5

The standard equation of A.P,

a + (s – 1) d = 95

15 + (s – 1) 5 = 95

s = 17

SS = s2[2a+(s1)d]Ss=172[2.15+(171)5]=172[30+80]=172×110=17×55=935

Hence, the sum of all natural numbers which are lying between 10 and 100, and are multiples of 5 is 935.

 

Q3. Prove that the 21st term in the sequence is – 118 provided that the sequence is in A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms and 2 is the first term.

Answer:

Given:

2 is the first term, a = 2

The sequence is in A.P form,

In A. P = 2, 2 + d, 2 + 2d, 2 + 3d ……

In A.P form, the sum of the starting 5 terms is one – fourth of the next 5 terms,

So, according to the given condition,

10 + 10d = (1 / 4) (10 + 35d)

40 + 40d = 10 + 35d

d = – 6

a21 = a + (21 – 1) d = 2 + 20 (- 6) = – 118

Hence proved

 

Q4. Obtain the number of terms in A.P which are needed to get – 25 from the sum of – 6, (- 11 / 2), – 5, ….

Answer:

Suppose,

In, A.P the sum of s terms = – 25

a = – 6 [1st term]

Difference, d = (- 11 / 2) + 6 = (1 / 2)

s2[2a+(s1)d] 25=s2[2.(6)+(n1)12]50=n[12+n212]50=n[25+n2]100=n(25+n)n225n+100=0n25n20n+100=0(n5)(n20)=0n=5or20

 

Q5. If m th term is 1 / n and nth term is 1 / m provided that the sequence is in A.P form, prove that the sum of the first mn terms is (1 / 2) (mn + 1), where m  n.

Answer:

Given,

m th term is 1 / n and nth term is 1 / m

So, acc. to the above condition

m th term in A.P form = am = a + (m – 1) d = 1 / m    ……. (1)

n th term in A.P form = an = a + (n – 1) d = 1 / n      ……. (2)

(2) – (1), we get,

(n – 1) d – (m – 1) d = 1m1n=nmmn

(n – m) d = nmmn

d = 1mn

Substituting the value of d in equation (1), we get,

a + (m – 1) 1mn = 1 / m

a = a=1n1n+1mn=1mn

Smn=mn2[2a+(mn1)d]Smn=mn2[21mn+(mn1)1mn]=1+12(mn1)=12(mn+1)

Hence proved

 

Q6. Obtain the last term if the addition of some numbers in A.P. 25, 22, 19, is 116.

Answer:

Addition of s terms in A.P be 116

Here, first term a = 25, d = – 3

Ss=s2[2a+(s1)d]116=s2[2.(25)+(s1)(3)]116=s[503s+3]232=s(533s)=53ss23s253n+232=03s224n29n+232=0(3s29)(n8)=0n=8orn=(293)

n = 8 is considered

a8 (last term) = a + (s – 1) d = 25 + 7 (- 3) = 25 – 21 = 4

Hence, the last term is 4

 

Q7. In A.P, obtain the total to s terms whose nth term is 5n + 1.

Answer:

Given,

In A.P, the total to s terms whose nth term is 5n + 1.

nth term = an = a + (n – 1) d

an = a + (n – 1) d = 5n + 1

a + nd – d = 5n + 1

By comparing the coefficient of n, we get

d = 5 and a – d = 1

a – 5 = 1

a = 6

Ss=s2[2a+(s1)d]=s2[2.6+(s1)d]=s2[12+(s1)5]=s2[12+5s5]=s2[5s+7].

 

Q8. Obtain the common difference of a sequence in A.P, when the total of s terms is (ms + ns2), where m and n are constants.

Answer:

Given,

As mentioned in the given condition,

Ss=s2[2a+(s1)d] = (ms + ns2)

s2[2a+(s1)d]=ms+ns2s2[2a+sdd]=ms+ns2sa+s2d2sd2=ms+ns2

Considering the coefficients of the of s2 , we get,

d2=n

d = 2n

Hence, in A.P the difference d = 2n

 

Q9. The ratio of the total of s terms of two arithmetic progressions are 5s + 4 : 9 + 6 . Obtain the ratio of 18th term.

Answer:

Suppose the first terms are a1 and a2 respectively, and the common differences be d1 and d2 of the first two consecutive arithmetic progressions respectively.

As mentioned in the given condition,

Totaladditionofstermsof1stA.PTotaladditionofstermsof2ndA.P=5s+49s+6s2[2a1+(s1)d1]s2[2a2+(s1)d2]=5s+49s+62a1+(s1)d12a2+(s1)d2=5s+49s+6Puttings=35in(1),weget2a1+34d12a2+34d2=5(35)+49(35)+6a1+17d1a2+17d2=179321

18th term = a18=182[2a+(181)d]

18thtermof1stA.P18thtermof2ndA.P=a18=182[2a+(181)d]=a1+17d1a2+17d2=179321

Hence, 179 : 321 is the required ratio of the 18th term.

 

Q10. In an A.P, the total of starting m terms is equal to the total of starting n terms. Obtain the total of starting (m + n) terms.

Answer:

m th term = Sm = m2[2a+(m1)d]

n th term = Sn = n2[2a+(n1)d]

As mentioned in the given condition,

m2[2a+(m1)d]=n2[2a+(n1)d]m[2a+(m1)d]=n[2a+(n1)d]2am+(m1)md=2an+(n1)nd2a(mn)+d[m(m1)n(n1)]=02a(mn)+d[m2mn2+n]=02a(mn)+d[(m+n)(mn)(mn)]=02a(mn)+d[(mn)(m+n1)]=02a+d(m+n1)=0d=2am+n1Sm+n=m+n2[2a+(m+n1)d]Sm+n=m+n2[2a+(m+n1)2am+n1]Sm+n=m+n2[2a2a]Sm+n=0

Hence, the total of starting (m + n) terms is 0.

 

Q11. . In an A.P, the total of starting m, n and o terms are p, q and r. Show that the pm(no)+qn(mo)+ro(mn)=0.

Answer:

Given,

In an A.P, the total of starting m, n and o terms are p, q and r.

As mentioned in the given condition,

Sm=m2[2a1+(m1)d]=p2a1+(m1)d=2pm(i)Sn=n2[2a1+(n1)d]=q2a1+(n1)d=2qn(ii)So=o2[2a1+(o1)d]=r2a1+(o1)d=2ro(iii)Subtract(i)(ii),weget(m1)d(n1)d=2pm2qnd(m1n+1)=2pn2qmmnd(mn)=2(pnqm)mnd=2(pnqm)mn(mn).(4) Subtract(ii)(iii),weget(n1)d(o1)d=2qn2rod(n1o+1)=2qo2rnnod(no)=2(qorn)nod=2(qorn)no(no).(5)

Now, equating the values of d in equation (4) and (5)

2(pnqm)mn(mn)=2(qorn)no(no)no(no)(pnqm)=mn(mn)(qorn)o(no)(pnqm)=m(mn)(qorn)(no)(pnoqmo)=(mn)(mqomrn)Multiplyboththesidesby1mno,weget,(pmqn)(no)=(qnro)(mn)pm(no)qn(no)=qn(mn)ro(mn)pm(no)+qn(no+mn)+ro(mn)=0pm(no)+qn(mo)+ro(mn)=0

Hence proved

 

Q12. The ratio of the total of r and s terms of two arithmetic progressions is r2 : s2. Show that the ratio of r th and s th term is (2r – 1) : (2s – 1).

Answer:

Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.

As mentioned in the given condition,

SumofrtermsSumofsterms=r2s2r2[2a+(r1)d]s2[2a+(s1)d]=r2s22a+(r1)d2a+(s1)d=rs..(i)Substitutingr=2r1ands=2s1in(i),weget,2a+(2r11)d2a+(2s11)d=2r12s12a+(2r2)d2a+(2s2)d=2r12s1a+(r1)da+(s1)d=2r12s1.(ii)rthtermofA.PsthtermofA.P=a+(r1)da+(s1)d=2r12s1rthtermofA.PsthtermofA.P=2r12s1

Hence proved

 

Q13. In an A.P, the total of n terms is 3p 2 + 5p and its rth term is 164. Obtain the value of r.

Answer:

Given,

Sr = r2[2a+(r1)d] = 164 …… (i)

Suppose the first term is a, and the common difference be d of the arithmetic progressions respectively.

As mentioned in the given condition,

p th term = Sp = p2[2a+(p1)d] = 3 p 2 + 5p

As mentioned in the given condition,

pa+d2p2d2p=3p2+5pd2p2+(ad2)p=3p2+5pEquating,thecoefficientsofp2onboththesides,wegetd2=3d=6Equating,thecoefficientsofponboththesides,wegetad2=5a3=5a=8

From equation (i), we get

8 + (r – 1) 6 = 164

(r – 1) 6 = 164 – 8

(r – 1) 6 = 156

(r – 1) = 26

r = 7

 

Q14. Place five numbers between 8 and 26 in a way such that the sequence results in an A.P form.

Answer:

Suppose Q1, Q2, Q 3, Q 4 and Q 5 be the five required numbers.

First term, a = 8,

Last term, p = 26

s = 7,

p = a + (s – 1) d

26 = 8 + (7 – 1) d

6 d = 18

d = 3

Q1 = a + d = 8 + 3 = 11

Q2 = a + 2 d = 8 + 2.3 = 14

Q 3 = a + 3 d = 8 + 3.3 = 17

Q= a + 4 d = 8 + 3.4 = 20

Q 5 = a + 5 d = 8 + 3. 5 = 23

Hence, Q1, Q2, Q 3, Q 4 and Q 5 is equal to 11, 14, 17, 20 and 23 are the required numbers in an A.P

 

Q15. Suppose in an A.M pm+qmpm1+qm1 lies between p and q terms, then obtain the value of m.

Answer:

A.M of p and q = p+q2

As mentioned in the given condition,

p+q2=pm+qmpm1+qm1(p+q)(pm1+qm1)=2(pm+qm)pm+pqm1+qpm1+qm=2pm+2qmpqm1+qpm1=pm+qmqm1(pq)=pm1(pq)qm1=pm1(pq)m1=1=(pq)0m1=0m=1

 

Q16. Insert n numbers between 1 and 31 in a way such that the sequence results in an A.P. The ratio is 5 : 9 of the 7th and the (n – 1)th term. Find the value of n.

Answer:

Suppose Q1, Q2, Q 3, Q 4.…..  Q m, be the required n numbers.

First term, a = 1,

Last term, p = 31

s = n + 2,

p = a + (s – 1) d

31 = 1 + (n + 2 – 1) d

30 = (n + 1) d

d = 30(n+1)

Q1 = a + d

Q2 = a + 2 d

Q 3 = a + 3 d

Q= a + 4 d …….

Q 7 = a + 7 d

Q n – 1 = a + (n – 1) d

As mentioned in the given condition,

a+7da+(n1)d=591+7(30n+1)1+(n1)(30n+1)=59n+1+7(30)n+1+30(n1)=59n+21131m29=599n+1899=155n145155n9n=1899+145146n=2044n=14

Hence, the value of n = 14

 

Q17. A boy will be repaying his loans and his first installment is Rs 100. What is the amount should he pay in the 30th installment if he increases the installment by Rs 5 every month?

Answer:

Given,

His first installment is Rs 100

His 2nd installment is Rs 105 and third installment is Rs 110 and so on

The sequence of money paid by the boy is in an A.P form every month is

100, 105, 110, 115 and so on …….

a = 100 [First term]

Common difference, d = 5

a 30 = a + (30 – 1) d

= 100 + 29 (5)

= 245

Hence, the amount should be paid by him in the 30th installment is Rs 245.

 

Q18.  In a polygon, 120o is the smallest angle and 5o is the difference between consecutive angles on the interior side. Obtain the number of sides the polygon has.

Answer:

Given,

120o is the smallest angle i.e., first term, a = 120

And 5o is the difference, i.e., d = 5

Total of all angles of a polygon having s sides = 180o (s – 2)

Ss=180(s2)s2[2a+(s1)d=180(s2)s[240+(s1)5=360(s2)240s+s(s1)5=360(s2)240s+5s25s=360s7205s2+235s360s+720=05s2125s+720=0s225s+144=0s216s9s+144=0s(s16)9(s16)=0(s16)(s9)=0s=16ors=9

 

Exercise 9.3

Q1. Find 20th and nth term for the G.P 52,54,58,....

Soln:

Given G.P = 52,54,58,....

Here, first term = a = 52

Common ratio = r = 5452=12

a20=ar201=52(12)19=5(2)(2)19=5(2)20 an=arn1=52(12)n1=5(2)(2)n1=5(2)n

 

Q2. Find 12th term for the G.P that has 8th term 192 and common ratio of 2.

Soln:

Given,

Common ratio = r = 2

Assume = first term  = a

a8=ar81=ar7ar7=192a(2)7=(2)6(3) a=(2)6×3(2)7=32 a