NCERT Solutions Class 11 Maths Chapter 9 – Download PDF

Get here NCERT Solutions Class 11 Maths Chapter 9. These NCERT Solutions for Class 11 of Maths subject includes detailed answers of all the questions in Chapter 9 – Sequences and Series provided in NCERT Book which is prescribed for class 11 in schools.

Resource: National Council of Educational Research and Training (NCERT) Solutions
Class: 11th Class
Subject: Maths
Chapter: Chapter 9 – Sequences and Series

NCERT Solutions Class 11 Maths Chapter 9 – Free Download PDF

NCERT Solutions Class 11 Maths Chapter 9 – Sequences and Series

NCERT Solutions For Class 11 Maths Chapter 9 Sequences and Series in PDF format is available here. Students can learn class 11 Maths sequences and Series, with the help of these materials, notes in both online and offline. This chapter is also an important one as some of the topics based on this chapter, you will learn in your higher classes as well. So try to understand the concepts of sequences and series with the help of NCERT solutions available here.

NCERT solutions for class 11 cover the topics which are covered in NCERT books and syllabus (Updated 2018-2019) prescribed by CBSE Board for sequence and series class 11 notes. These materials are designed and formulated by experts of BYJU’S to make sure the students get the proper knowledge of relevant topics.

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. The major topics covered here for Class 11 Chapter 9 are;

• Sequences and Series
• Arithmetic Progression(AP) and Geometric Progression(GP)
• Relation between Arithmetic Mean and Geometric Mean
• Sum of Nth terms

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.

Exercise 9.1

Q1: a_s = s (s + 3) is the s^th term of a sequence. Find the starting five terms of the sequence.

Answer:

a_s = s (s + 3)

Putting s = 1, 2, 3, 4 and 5 respectively, in a_s = 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: a_s = ss+2 is the s^th term of a sequence. Find the starting five terms of the sequence.

Answer:

a_s = 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: a_s = 5^s is the s^th term of a sequence. Find the starting five terms of the sequence.

Answer:

a_s = 5^s

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

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: a_s = 2s+23 is the s^th term of a sequence. Find the starting five terms of the sequence.

Answer:

a_s = 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: a_s = (−1)s–1+3s+1 is the s^th term of a sequence. Find the starting five terms of the sequence.

Answer:

a_s = (−1)s–1+3s+1

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

a₁ = (−1)1–1+31+1 = 1 + 9 = 10

a₂ = (−1)2–1+32+1 = – 1 + 27 = 26

a₃ = (−1)3–1+33+1 = 1 + 81 = 82

a₄ = (−1)4–1+34+1 = – 1 + 243 = 242

a₅ = (−1)5–1+35+1 = 1 + 729 = 730

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

 

Q6: a_s = (−1)s–13s+1 is the s^th term of a sequence. Find the starting five terms of the sequence.

Answer:

a_s = (−1)s–13s+1

Putting s = 1, 2, 3, 4 and 5 respectively, in (−1)s–13s+1

a₁ = (−1)1–131+1=(−1)032=19

a₂ = (−1)2–132+1=(−1)133=−127

a₃ = (−1)3–133+1=(−1)234=181

a₄ = (−1)4–134+1=(−1)335=−1243

a₅ = (−1)5–135+1=(−1)436=1729

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

 

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

Answer:

a_s = s + 3^s

Putting s = 17 and 22 respectively, in a_s = s + 3^s

a₁₇ = 17 + 3¹⁷

a₂₂ = 22 + 3²²

 

Q8: a_s = 9+ss2 is the s^th term of a sequence. Obtain the 24^th term in the required sequence.

Answer:

a_s = 9+ss2

Putting s = 24, in a_s = 9+ss2

a₂₄ = 9+24242=33576

 

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

Answer:

a_s = (−1)s×6s

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

a₁₁ = (−1)11×611=–611

a₁₂ =  (−1)12×612=612

 

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

Answer:

a_s = s+32+s

Putting s = 21 and 82, in a_s = 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, a_s = 2 a_{s – 1} for all s > 1

Answer:

a₁ = 2, a_s = 2 a_{s – 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, a_s = as–1–12 for all s > 3

Answer:

a₁ = 2, a_s = as–1–12 for all s > 3

a2–1–12=a1–12=2–12=12a3=a3–1–12=a2–12=12–12=–14a4=a4–1–12=a3–12=−14–12=–58a5=a5–1–12=a4–12=−58–12=–1316

The starting five terms of the sequence are 2,12,–14,–58and–1316

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, a_s = a_{s – 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 a_s = a_{s – 1} + a_{s – 2}, n > 1. Obtain as+1as  for n = 2, 4.

Answer:

1 = a₁ = a₂ and a_s = a_{s – 1} + a_{s – 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

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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 [1^st term]

Difference, d = 2

The standard equation of A.P,

a + (s – 1) d = 4001

1 + (s – 1) 2 = 4001

s = 2001

S_S = s2[2a+(s–1)d]Ss=20012[2.1+(2001–1)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 [1^st term]

Difference, d = 5

The standard equation of A.P,

a + (s – 1) d = 95

15 + (s – 1) 5 = 95

s = 17

S_S = s2[2a+(s–1)d]Ss=172[2.15+(17–1)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 21^st 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 [1^st term]

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

s2[2a+(s–1)d] −25=s2[2.(−6)+(n–1)12]–50=n[−12+n2–12]–50=n[−25+n2]–100=n(−25+n)n2–25n+100=0n2–5n–20n+100=0(n–5)(n–20)=0n=5or20

 

Q5. If m ^th term is 1 / n and n^th 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 n^th term is 1 / m

So, acc. to the above condition

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

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

(2) – (1), we get,

(n – 1) d – (m – 1) d = 1m–1n=n–mmn

(n – m) d = n–mmn

d = 1mn

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

a + (m – 1) 1mn = 1 / m

a = a=1n–1n+1mn=1mn Smn=mn2[2a+(mn–1)d]Smn=mn2[21mn+(mn–1)1mn]=1+12(mn–1)=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+(s–1)d]116=s2[2.(25)+(s–1)(−3)]116=s[50–3s+3]232=s(53–3s)=53s–s23s2–53n+232=03s2–24n–29n+232=0(3s–29)(n–8)=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 n^th term is 5n + 1.

Answer:

Given,

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

n^th term = a_n = a + (n – 1) d

a_n = 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+(s–1)d]=s2[2.6+(s–1)d]=s2[12+(s–1)5]=s2[12+5s–5]=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+(s–1)d] = (ms + ns²)s2[2a+(s–1)d]=ms+ns2s2[2a+sd–d]=ms+ns2sa+s2d2–sd2=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 18^thterm.

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

18^th term = a18=182[2a+(18–1)d] 18thtermof1stA.P18thtermof2ndA.P=a18=182[2a+(18–1)d]=a1+17d1a2+17d2=179321

Hence, 179 : 321 is the required ratio of the 18^th 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 = S_m = m2[2a+(m–1)d]

n ^th term = S_n = n2[2a+(n–1)d]

As mentioned in the given condition,

m2[2a+(m–1)d]=n2[2a+(n–1)d]m[2a+(m–1)d]=n[2a+(n–1)d]2am+(m–1)md=2an+(n–1)nd2a(m–n)+d[m(m–1)–n(n–1)]=02a(m–n)+d[m2–m–n2+n]=02a(m–n)+d[(m+n)(m–n)–(m–n)]=02a(m–n)+d[(m–n)(m+n–1)]=02a+d(m+n–1)=0d=−2am+n–1Sm+n=m+n2[2a+(m+n–1)d]Sm+n=m+n2[2a+(m+n–1)−2am+n–1]Sm+n=m+n2[2a–2a]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(n–o)+qn(m–o)+ro(m–n)=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+(m–1)d]=p2a1+(m–1)d=2pm……(i)Sn=n2[2a1+(n–1)d]=q2a1+(n–1)d=2qn……(ii)So=o2[2a1+(o–1)d]=r2a1+(o–1)d=2ro……(iii)Subtract(i)−(ii),weget(m–1)d–(n–1)d=2pm–2qnd(m–1–n+1)=2pn–2qmmnd(m–n)=2(pn–qm)mnd=2(pn–qm)mn(m–n)….(4) Subtract(ii)−(iii),weget(n–1)d–(o–1)d=2qn–2rod(n–1–o+1)=2qo–2rnnod(n–o)=2(qo–rn)nod=2(qo–rn)no(n–o)….(5)

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

2(pn–qm)mn(m–n)=2(qo–rn)no(n–o)no(n–o)(pn–qm)=mn(m–n)(qo–rn)o(n–o)(pn–qm)=m(m–n)(qo–rn)(n–o)(pno–qmo)=(m–n)(mqo–mrn)Multiplyboththesidesby1mno,weget,(pm–qn)(n–o)=(qn–ro)(m–n)pm(n–o)–qn(n–o)=qn(m–n)–ro(m–n)pm(n–o)+qn(n–o+m–n)+ro(m–n)=0pm(n–o)+qn(m–o)+ro(m–n)=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 ^thand 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+(r–1)d]s2[2a+(s–1)d]=r2s22a+(r–1)d2a+(s–1)d=rs…..(i)Substitutingr=2r–1ands=2s–1in(i),weget,2a+(2r–1–1)d2a+(2s–1–1)d=2r–12s–12a+(2r–2)d2a+(2s–2)d=2r–12s–1a+(r–1)da+(s–1)d=2r–12s–1….(ii)rthtermofA.PsthtermofA.P=a+(r–1)da+(s–1)d=2r–12s–1rthtermofA.PsthtermofA.P=2r–12s–1

Hence proved

 

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

Answer:

Given,

S_r = r2[2a+(r–1)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 = S_p = p2[2a+(p–1)d] = 3 p ² + 5p

As mentioned in the given condition,

pa+d2p2–d2p=3p2+5pd2p2+(a–d2)p=3p2+5pEquating,thecoefficientsofp2onboththesides,wegetd2=3d=6Equating,thecoefficientsofponboththesides,wegeta–d2=5a–3=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 ₄  = 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+qmpm–1+qm–1 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+qmpm–1+qm–1(p+q)(pm–1+qm–1)=2(pm+qm)pm+pqm–1+qpm–1+qm=2pm+2qmpqm–1+qpm–1=pm+qmqm–1(p–q)=pm–1(p–q)qm–1=pm–1(pq)m–1=1=(pq)0m–1=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 7^th 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 ₄  = a + 4 d …….

Q ₇ = a + 7 d

Q _{n – 1} = a + (n – 1) d

As mentioned in the given condition,

a+7da+(n–1)d=591+7(30n+1)1+(n–1)(30n+1)=59n+1+7(30)n+1+30(n–1)=59n+21131m–29=599n+1899=155n–145155n–9n=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 30^th installment if he increases the installment by Rs 5 every month?

Answer:

Given,

His first installment is Rs 100

His 2^nd 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 30^th installment is Rs 245.

 

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

Answer:

Given,

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

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

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

Ss=180∘(s–2)s2[2a+(s–1)d=180∘(s–2)s[240+(s–1)5=360(s–2)240s+s(s–1)5=360(s–2)240s+5s2–5s=360s–7205s2+235s–360s+720=05s2–125s+720=0s2–25s+144=0s2–16s–9s+144=0s(s–16)–9(s–16)=0(s–16)(s–9)=0s=16ors=9

 

Exercise 9.3

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Q1. Find 20^th and n^th 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=ar20–1=52(12)19=5(2)(2)19=5(2)20 an=arn–1=52(12)n–1=5(2)(2)n–1=5(2)n

 

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

Soln:

Given,

Common ratio = r = 2

Assume = first term  = a

a8=ar8–1=ar7⇒ar7=192a(2)7=(2)6(3) ⇒a=(2)6×3(2)7=32 a12=ar12–1=(32)(2)11=(3)(2)10=3072

 

Q3. If p is the 5^th, q is the 8^th and s is the 11^th term of a G.P. Prove that q² = ps

Soln:

Let the first term be ‘a’ and the common ratio be r for the G.P.

Given condition

a5=ar5–1=ar4=p....(1) a8=ar8–1=ar7=q....(2) a11=ar11–1=ar10=s....(3)

Eqn (2) divided by Eqn (1), we have

ar7ar7=qp r3=qp  . . . . .  (4)

Eqn (3) divided by Eqn (2), we have

ar10ar7=sq ⇒r3=sq  . . . . .  (5)

From eqn (4) and eqn (5), we get

qp=sq ⇒q2=ps

Hence proved

 

Q4. If first term of a G.P is -3 and the 4^th term is square of the 2^nd term. Find the 7^th term.

Soln:

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

Therefore, a = -3

We know that, a_n = ar^{n – 1}

a₄ = ar³ = (-3)r³

a₂ = ar¹ = (-3)r

Given condition

(-3)r³ = [(-3) r]²

⇒−3r3=9r2⇒r=–3a7=ar7–1=a

r⁶ = (-3) (-3)⁶ = – (3)⁷ = -2187

Hence, the 7^th term is -2187.

 

Q5. Which term of

(a) 2,22–√,4,....is128?

(b) 3–√,3,33–√....is729?

(c) 13,19,127,....is119683?

Soln:

(a) Given sequence = 2,22–√,4,....

Here,

First term = a = 2

Common ratio = r = 22√2=2–√

Assume n^th term = 128

an=arn–1 ⇒(2)(2–√)n–1=128 ⇒(2)(2)n–12=(2)7 ⇒(2)n–12+1=(2)7 \frac{n – 1}{2} + 1 = 7 ⇒ \frac{n – 1}{2} = 6 ⇒ n – 1 = 12 ⇒ n = 13

Hence, 128 is the 13^th term

(b) Given sequence = \sqrt{3}, 3, 3\sqrt{3} . . . .

Here,

First term = a = \sqrt{3}

Common ratio = r = \frac{3}{\sqrt{3}} = \sqrt{3}

Assume n^th term = 729

a_{n} = a r^{n – 1} a r^{n – 1} = 729 \Rightarrow (\sqrt{3})(\sqrt{3})^{n – 1} = 729 \Rightarrow (3)^{\frac{1}{2}}(3)^{\frac{n – 1}{2}} = (3)^{6} \Rightarrow (3)^{\frac{1}{2} + \frac{n – 1}{2}} = (3)^{6}

Therefore,

\frac{1}{2} + frac{n – 1}{2} = 6 \Rightarrow \frac{1 + n – 1}{2} = 6 \Rightarrow n = 12

Hence, 729 is the 12^th term of the sequence.

(c) Given sequence = \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . . . .

Here,

First term = a = \frac{1}{3}

Common term = r = \frac{1}{9} \div \frac{1}{3} = \frac{1}{3}

Assume the nth term be \frac{1}{19683} a_{n} = a r^{n – 1} a r^{n – 1} = \frac{1}{19683} \Rightarrow \left ( \frac{1}{3} \right ) \left ( \frac{1}{3} \right )^{n – 1} = \frac{1}{19683} \Rightarrow \left ( \frac{1}{3} \right )^{n} = \left ( \frac{1}{3} \right )^{9}

n = 9

Hence, \frac{1}{19683} is the 9^th term of the sequence.

 

Q6. If \frac{2}{7}, x, -\frac{7}{2} are in G.P then find the value of x?

Soln:

Given sequence = \frac{2}{7}, x, -\frac{7}{2}

Common ratio = \frac{x}{\frac{-2}{7}} = \frac{-7x}{2}

Also, \frac{\frac{-7}{2}}{x} = \frac{-7}{2x} \frac{-7x}{2} = \frac{-7}{2x} \Rightarrow x^{2} = \frac{-2 \times 7}{-2 \times 7} = 1 \Rightarrow x = \sqrt{1} \Rightarrow x = \pm 1

Hence, the sequence is in G.P if x = \pm 1

 

Q7. The sum of the first 20 terms of the G.P 0.15, 0.015, 0.0015 . . . .

Soln:

Given G.P = 0.15, 0.015, 0.0015 . . . .

Here,

First term = a = 0.15

Common ration = r = \frac{0.015}{0.15} = 0.1 S_{n} = \frac{a(1 – r^{n})}{1 – r} = \frac{0.15[1 – (0.1)^20]}{1 – 0.1}= \frac{0.15}{0.9}[1 – (0.1)^{20}] = \frac{15}{90}[1 – (0.1)^{20}] = \frac{1}{6}[1 – (0.1)^{20}]

 

Q8. \sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . . . is a G.P series. Find sum till the n^th term.

Soln:

Given series = \sqrt{7}, \sqrt{21}, 3\sqrt{7}, . . . .

Here,

First term = a = \sqrt{7}

Common ratio = r = \frac{\sqrt{21}}{7} = \sqrt{3} S_{n} = \frac{a(1 – r^{n})}{1 – r} = \frac{\sqrt{7}[1 – (\sqrt{3})^{n}]}{1 – \sqrt{3}} = \frac{\sqrt{7}[1 – (\sqrt{3})^{n}]}{1 – \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{\sqrt{7}(\sqrt{3} + 1)[1 – (\sqrt{3})^{n}]}{1 – 3} = \frac{ – \sqrt{7}(\sqrt{3} + 1)[1 – (\sqrt{3})^{n}]}{2}

 

Q9. What is the sum of the n^th term of the G.P series

1, -a, a², -a³ . . . .  (if a ≠ -1)?

Soln:

Given series = 1, -a, a², -a³ . . .

Here,

The first term of the G.P = a₁ = 1

Common ratio of the G.P = r = -a

S_{n} = \frac{a_{1}(1 – r^{n})}{1 – r} S_{n} = \frac{1[1 – (-a)^{n}]}{1 – (-a)} = \frac{[1 – (-a)^{n}]}{1 + a}

 

Q10. What is the sum of the n^th term of the G.P series

x³, x⁵, x⁷ . . . . (if x ≠ ± 1)

Soln:

Given series = x³, x⁵, x⁷ . . . .

Here,

First term = a = x³

Common ratio = r = x²

S_{n} = \frac{a(1 – r^{n})}{1 – r} = \frac{x^{3}[1 – (x^{2})^{n}]}{1 – x^{2}} = \frac{x^{3}(1 – x^{2n})}{1 – x^{2}}

 

Q11. Evaluate \sum_{k = 1}^{11}(2 + 3^{k})

Soln:

\sum_{k = 1}^{11}(2 + 3^{k}) = \sum_{k = 1}^{11}(2) + \sum_{k = 1}^{11} 3^{k} = 2(11) + \sum_{k = 1}^{11} 3^{k} = 22 + \sum_{k = 1}^{11} 3^{k} . . . . . (1) \sum_{k = 1}^{11} 3^{k} = 3^{1} + 3^{2} + 3^{3} + . . . . + 3^{11}

The above sequence is in a G.P 3, 3², 3³, . . .

S_{n} = \frac{a(r^{n} – 1)}{ r – 1} \Rightarrow S_{n} = \frac{3[(3)^{11} – 1]}{3 – 1} \Rightarrow S_{n} = \frac{3}{2}(3^{11} – 1) \sum_{k = 1}^{11} 3^{k} = \frac{3}{2}(3^{11} – 1) ….(2)

Substituting eqn (2) in eqn (1), we get

\sum_{k = 1}^{11} (2 + 3^{k}) = 22 + \frac{3}{2}(3^{11} – 1)

 

Q12. Sum of 1^st 3 terms in a G.P is \frac{39}{10} and the product of the terms is 1. What is the the terms and common ratio of the sequence?

Soln:

Assume \frac{a}{r}, a, ar be the 1^st 3 terms in the G.P

\frac{a}{r} + a + ar = \frac{39}{10} . . . . (1)\left (\frac{a}{r} \right ) \left (a \right ) \left ( ar \right ) = 1 . . . . (2)

From eqn (2), we get a³ = 1

a = 1

Substituting (a = 1) in eqn (1), we get

\frac{1}{r} + 1 + r = \frac{39}{10} \Rightarrow 1 + r + r^{2} = \frac{39}{10}r \Rightarrow 10 + 10 r + 10 r^{2} -39 r = 0 \Rightarrow 10 r^{2} – 29 r + 10 = 0 \Rightarrow 10 r^{2} – 25 r – 4 r + 10 = 0 ]\Rightarrow 5r (2r – 5) -2(2r – 5) = 0 ]\Rightarrow (5r – 2) (2r – 5) = 0 ]\Rightarrow r = \frac{2}{5} or \frac{5}{2}

Hence,

The terms are \frac{5}{2}, 1 \; and \; \frac{2}{5}

 

Q13. For the G.P 3, (3)², (3)³ …. sum of how many terms is 120?

Soln:

Given G.P = 3, (3)², (3)³ ….

Let n number of terms are required for obtaing the sum of 120

S_{n} = \frac{a(1 – r)^{n}}{1 – r}

Here,

First term = a = 3

Common ratio = r = 3

S_{n} = 120 = \frac{3(3^{n} – 1)}{3 – 1} \Rightarrow 120 = \frac{3(3^{n} – 1)}{2} \Rightarrow \frac{120 \times 2}{3} = 3^{n} – 1 \Rightarrow 3^{n} – 1 = 80 \Rightarrow 3^{n} = 81 \Rightarrow 3^{n} = 3^{4}

n = 4

Hence, 4 terms are required to get the sum of 120

 

Q14. In a G.P sum of the first 3 terms of are 16 where as sum for the next 3 terms are 128.

Find common ratio, first term and sum of n terms for the G.P

Soln:

Assume G.P = a, ar, a(r)², a(r)³, . . .

Given condition =

a + a(r)² + ar + = 16 and a(r)⁵ + a(r)⁴ + a(r)³ = 128

a (1 + (r)² + r) = 16 …… (1)

a(r)³(1+ (r)² + r) = 128 …….. (2)

eqn (2) divided by eqn (1), we get

\frac{ar^{3}(1 + r + r^{2})}{a(1 + r + r^{2})} = \frac{128}{16}

r³ = 8

r = 2

substituting r = 2 in eqn(1), we get

a(1 + 2 + 4) = 16

a(7) = 16

a = \frac{16}{7} S_{n} = \frac{16}{7} \frac{(2^{n} – 1)}{2 – 1}  = \frac{16}{7}(2^{n} – 1)

 

Q15. Given first term ‘a’ = 729 and the 7^th term = 64, then determine S₇.

Soln:

Given a = 729 , a₇ = 64

Assume common ratio = r

We know that,

a_n = a r^{n – 1}

a₇ = a r^{7 – 1} = (729) (r)⁶

64 = 729 (r)⁶

r^{6} = \frac{64}{729} r^{6} = \left ( \frac{2}{3}\right )^{6} r = \frac{2}{3}

Also, we know that

S_{n} = \frac{a(1 – r^{n})}{1 – r} S_{7} = \frac{729 \left [ 1 – \left ( \frac{2}{3} \right )^{7} \right ] }{1 – \frac{2}{3}}= 3 \times 729 \left [ 1 – \left ( \frac{2}{3} \right )^{7} \right ] = (3)^{7} \left [ \frac{(3)^{7} – (2)^{7}}{(3)^{7}} \right ] = (3)^{7} – (2)^{7}

= 2187 – 128

= 2059

 

Q16. The sum of first 2 terms of a G.P is -4 and 5^th term is 4 times that of third term. Find the G.P

Soln:

Assume first term = a

Common ratio = r

S_{2} = -4 = \frac{a(1 – r^{2})}{1 – r} ….. (1)a_{5} = 4 \times a_{3} \Rightarrow ar^{4} = 4ar^{2} \Rightarrow r^{2} = 4 r = \pm 2

From (1), we get

-4 = \frac{a\left [ 1 – (2)^{2} \right ]}{1 – 2} \; for \; r = 2 -4 = \frac{a(1 – 4)}{-1}

-4 = a(3)

a = \frac{-4}{3}

Also, -4 = \frac{a\left [ 1 – (-2)^{2} \right ]}{1 – (-2)} \; for \; r =  -2 -4 = \frac{a(1 – 4)}{1 + 2} -4 = \frac{a(- 3)}{3}

a = 4

Hence, required G.P is \frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3} … \; or \; 4, -8, 16, -32….

 

Q17. If p, q, and s are fourth, tenth and sixteenth terms of G.P prove that p, q, s are in G.P

Soln:

Let

First term = a

Common ratio = r

According to given condition,

a₄ = ar³ = p ………. (1)

a₁₀ = ar⁹ = q ………. (2)

a₁₆ = ar¹⁵ = x ………. (3)

eqn (2) divided by eqn (1)

\frac{q}{p} = \frac{ar^{9}}{ar^{3}} \Rightarrow \frac{q}{p} = r^{6}

eqn (3) divided by eqn (2)

\frac{s}{q} = \frac{ar^{15}}{ar^{9}} \Rightarrow \frac{s}{q} = r^{6} \frac{q}{p} = \frac{s}{q}

 

Q18. 8, 88, 888, 8888, …… are in sequence find sum of the n^th terms

Soln:

Given sequence 8, 88, 888, 8888, ……

This sequence isn’t a G.P

It may be changed to a G.P if the terms are written as S_n = 8 + 88 + 888 + 8888 + …..  to n terms

= \frac{8}{9}\left [ 9 + 99 + 999 + 9999 + ……….\; to \; n \; terms \right ] = \frac{8}{9}\left [ (10 – 1) + (10^{2} – 1) + (10^{3} – 1) + (10^{4} – 1) + ……….\; to \; n \; terms \right ] = \frac{8}{9}\left [ (10 + 10^{2} + …. n \; terms) – (1 + 1 + 1 ……….\; to \; n \; terms) \right ] = \frac{8}{9}\left [ \frac{10(10^{n} – 1)}{10 – 1} – n \right ] = \frac{8}{9}\left [ \frac{10(10^{n} – 1)}{9} – n \right ] = \frac{80}{81}\left ( 10^{n} – 1 \right ) – \frac{8}{9}n

 

Q19. If 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ are in sequences find the sum of the products

Soln:

Required sum = 2 \times 128 + 4 \times 32 + 8\times 8 + 16 \times 2 + 32 \times \frac{1}{2} = 64 \left [ 4 + 2 + 1 + \frac{1}{2} + \frac{1}{2^{2}} \right ]

Here, 4, 2, 1, \frac{1}{2}, \frac{1}{2^{2}} is a G.P

Now first term, a = 4

And common ratio, r = \frac{1}{2}

We know that,

S_{n} = \frac{a(1 – r^{n})}{1 – r} S_{5} = \frac{4\left [ 1 – \left ( \frac{1}{2} \right )^{5}\right ]}{1 – \frac{1}{2}} = \frac{4\left [ 1 – \frac{1}{32} \right ]}{ \frac{1}{2}} = 8 \left ( \frac{32 – 1}{32} \right ) = \frac{31}{4}

Required sum = 64\left ( \frac{31}{4} \right ) = (16)(31) = 496

 

Q20. Prove that product of corresponding terms from a, ar, ar²,_… ar^n-1 and A, AR, AR²,_… AR^n-1 are in sequences find common ratio

Soln:

To show = aA, ar(AR), ar² (AR²), … ar^{n – 1} (AR^{n – 1}) is a G.P series

\frac{Second \; term}{First \; term} = \frac{arAR}{aA} = rR \frac{Third \; term}{Second \; term} = \frac{ar^{2}AR^{2}}{arAR} = rR

Hence, the sequences are in G.P and rR is common ratio.

 

Q21. If in a G.P of 4 numbers 3^rd term is greater to 1^st term by 9, and the 2^nd term is greater to 4^th by 18.

Soln:

Let

First term = a

Common ratio = r

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

From given conditions,

a₃ = a₁ + 9 => ar² = a + 9 ………. (1)

a₂ = a₄ + 18 => ar = ar³ + 18 ………. (2)

From eqn (1) and eqn (2), we get

a(r² – 1) = 9 ……… (3)

ar(1 – r²) = 18 ……..(4)

Dividing eqn (4) by eqn (3), we get

\frac{ar(1 – r^{2})}{a( 1 – r^{2})} = \frac{18}{9}

=>  – r = 2

=> r = -2

Substituting r = -2 in eqn (1), we get

4a = a + 9

=> 3a = 9

=> a = 3

Hence, first four terms of G.P are 3, 3(-2), 3(-2)², and 3(-2)³

i.e. 3, -6, 12, -24

 

Q22. If in a G.P a, b and c are the p^th, q^th and r^th terms. Prove that

a^{q – r}.b^{r – p}.c^{p – q} = 1

Soln:

Let

First term = a

Common ratio = r

From the given condition

AR^{p – 1} = a

AR^{q – 1} = b

AR^{r – 1} = c

a^{q – r}.b^{r – p}.c^{p – q}

= A^{q – r} \times R^{(p – 1)(q – r)} \times A^{r – p} \times R^{(q – 1)(r – p)} \times A^{p – q} \times R^{(r – 1)(p – q)} = A^{q – r + r – p + p – q} \times R^{(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr – q)}

= A⁰ x R⁰

= 1

Hence proved

 

Q23. If in a G.P ‘a’ and ‘b’ are the first and n^th term and product of n^th terms is P.

Show P² = (ab)ⁿ

Soln:

Given:

First term = a

Last term = b

Therefore,

G.P = a, ar, ar², ar³ . . . ar^{n – 1}, where r is the common ratio.

b = ar^{n – 1}…….. (1)

P = product of n terms

= (a)(ar)(ar²)…(ar^{n – 1})

= (a x a x …. a)(r x r² x … r^{n – 1})

= an r 1 + 2 + … (n – 1) …… (2)

Here, 1, 2, …..(n – 1) is an A.P

1 + 2 + …. + (n – 1)

= \frac{n – 1}{2}\left [ 2 + (n – 1 – 1) \times 1 \right ] = \frac{n – 1}{2}[2 + n – 2] = \frac{n(n – 1)}{2} P = a^{n}r^{\frac{n(n -1)}{2}} P = a^{2n}r^{n(n -1)} = [a^{2}r^{n – 1}]^{n}

= (ab)^{n}

Hence proved

 

Q24. Prove that the ratio of sum of first n terms to sum of terms from (n + 1)^th to (2n)^th terms is 1/ rⁿ .

Soln:

Let

First term = a

Common ratio = r

Sum of the first n terms = \frac{a(1- r^{n})}{(1 – r)}

Since there are ‘n’ number of terms from (n + 1)^th to (2n)^th term,

Sum of the terms

S_{n} = \frac{a_{n + 1(1 – r^{n})}}{1 – r}

Required ratio = \frac{a (1 – r^{n})}{(1 – r)} \times \frac{(1 – r)}{ar^{n}(1 – r^{n})} = \frac{1}{r^{n}}

Thus, the ratio is 1/rⁿ

 

Q25. Prove that:

(a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

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

Soln:

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

Therefore

bc = ad …………….. (1)

b² = ac ……………… (2)

c² = bd ……………… (3)

It is to be proven that,

(a² + b² + c²) (b² + c² + d²) = (ab + bc – cd)²

R.H.S

= (ab + bc + cd)²

= (ab + ad + cd)²           [From (1)]                                                      

= [ab + d(a + c)]²

= a² (b²) + 2abd(a + c) + (d²)(a + c)²

= (a²) (b²) + 2a²bd + 2acbd + (d²)(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²      [From (1) and (2)]

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

= a²b² + a²c² + a²d² + b² x b² + b²c² + b²d² + c²b² + c² x c² + c²d²
[From eqn (1) and (2)]

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

= (a² + b² + c²) (b² + c² + d²) = L.H.S

L.H.S = R.H.S

 

Q26. Between 3 and 81 insert two numbers such that the sequence is a G.P series.

 Soln:

Let the numbers be g₁ and g₂

So from the given condition 3, g_1, g₂, 81 is in G.P

Assume first term = a

Common ratio = r

Therefore 81 = (3)(r)³

=> r³ = 27

r = 3 (taking real roots only)

For r = 3,

Q26.

G₁ = ar = (3)(3) = 9

G₂ = ar² = (3)(3)² = 27

Hence, the numbers are 9 and 27

 

Q27. If \frac{a^{2 + 1} + b^{n + 1}}{a^{n} + b^{n}} is the geometric mean for a and b. What is the value of n?

Soln:

Mean of a and b is \sqrt{ab}

From the given condition \frac{a^{2 + 1} + b^{n + 1}}{a^{n} + b^{n}} = \sqrt{ab}

Squaring both the sides, we get

\frac{(a^{2 + 1} + b^{n + 1})^{2}}{(a^{n} + b^{n})^{2}} = ab

=> a^{2n +1} + 2a^{n + 1} b^{n + 1} + b^{2n + 1} = (ab)(a²ⁿ + 2aⁿbⁿ + b²ⁿ)

=> a^{2n +1} + 2a^{n + 1} b^{n + 1} + b^{2n + 1} = a^{2n +1}b + 2a^{n + 1} b^{n + 1} + ab^{2n + 1}

=> a^{2n +1} + b^{2n + 1} = a^{2n +1}b + ab^{2n + 1}

=> a^{2n +1} – a^{2n +1}b = ab^{2n + 1} – b^{2n + 2}

=> a^{2n +1} (a – b) = b^{2n + 1} (a – b)

=> \left ( \frac{a}{b} \right )^{2n + 1} = 1 = \left ( \frac{a}{b} \right )^{0}

=> 2n + 1 = 0

=> n = \frac{-1}{2}

 

Q28. Prove that the ratio of two numbers is  (3 + 2\sqrt{2})(3 – 2\sqrt{2}) if the sum of the two numbers is 6 times their geometric mean.

Soln:

Let the two numbers be a and b.

G.M = \sqrt{ab}

From given conditions,

a + b = 6\sqrt{ab} ……. (1)

=> (a + b)² = 36(ab)

Also,

(a – b)² = (a + b)² – 4ab = 36ab – 4ab = 32 ab

=> a – b = \sqrt{32}\sqrt{ab}

= 4\sqrt{2}\sqrt{ab} ……. (2)

Eqn(1) + Eqn (2)

2a = (6 + 4 \sqrt{2})\sqrt{ab}

=> a = (3 + 2 \sqrt{2})\sqrt{ab}

Putting the value of a in eqn (1),we get

b = 6\sqrt{ab} – (3 + 2\sqrt{2})\sqrt{ab}

=> b = (3 – 2\sqrt{2})\sqrt{ab} \frac{a}{b} = \frac{(3 + 2\sqrt{2})\sqrt{ab}}{(3 – 2\sqrt{2})\sqrt{ab}} = \frac{3 + 2\sqrt{2}}{3 – 2\sqrt{2}}

Hence, required ratio = \left ( 3 + 2\sqrt{2} \right ) : \left ( 3 – 2\sqrt{2} \right )

 

Q29. For two positive numbers if G and A are the G.M and A.M show that the numbers are A \pm \sqrt{(A + G)(A – G)}

Soln:

Given G and A are A.M and G.M

Assume the numbers be a and b.

AM = A = \frac{a + b}{2} ……… (1) GM = G = \sqrt{ab} ……… (2)

From Eqn(1) and Eqn (2), we get

a + b = 2A ……. (3)

ab = G² ……. (4)

Putting the value of a and b from (3) and (4)

(a – b)² = (a + b)² – 4ab, we get

(a – b)² = 4A² – 4G² = 4(A² – G²)

(a – b) = 2\sqrt{(A + G)(A – G)} …… (5)

From eqn (3) and (5), we get

2a  = 2A + 2\sqrt{(A + G)(A – G)}

=> a = A + \sqrt{(A + G)(A – G)}

Putting the value of a in eqn(3), we get

b  = 2A – A – \sqrt{(A + G)(A – G)} = A – \sqrt{(A + G)(A – G)}

Hence, the numbers are A \pm \sqrt{(A + G)(A – G)}

 

Q30. Bacteria over certain culture multiply twice in one hour. If in initial stage there were only 30 bacteria, what will be the amount of bacteria in the 2^nd, 4^th and n^th hour?

Soln:

Given that amount of bacteria multiplies two times in one hour

Therefore,

Bacteria amount will form G.P

Here, first term = (a) = 30

Common ratio = (r) = 2

Therefore ar² = a₃ = 30 x (2)² = 120

Therefore amount of bacteria after 2^nd hour is 120

ar⁴ = a₅ = 30 x (2)⁴ = 480

Therefore amount of bacteria after 4^th hour is 480

 arⁿ x a_{n + 1} = 30 x (2)ⁿ

Hence, amount of bacteria after n^th hour is 30 x (2)ⁿ

 

Exercise 9.4

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Q1. Find sum to n^th term for the series 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . .

Soln:

Given series = 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . . . . n^th term, a_n = n (n + 1)

S_{n} = \sum_{k = 1}^{n} a_{1} = \sum_{k = 1}^{n} k ( k + 1) = \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k \frac{n (n + 1) (2n + 1)}{6} + \frac{n (n + 1)}{2} \frac{n (n + 1)}{2} \left ( \frac{ 2n + 1}{3} + 1 \right ) \frac{n (n + 1)}{2} \left ( \frac{ 2n + 4}{3} \right ) \frac{n (n + 1) (n + 2) }{3}

 

Q2. Find sum to n^th term of the series 1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . . . . .

Soln:

Given series = 1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . . . n^{th} \; term, \; a_{n} = n (n + 1) (n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

S_{0} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} k^{3} + 3 \sum_{k = 1}^{n} k^{2} + 2 \sum^{n}_{k = 1} k = \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{3n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2} = \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{n(n + 1)(2n + 1)}{2} + n(n + 1) = \frac{n(n + 1)}{2} \left [ \frac{n(n + 1)}{2} + 2n + 1 + 2 \right ] = \frac{n(n + 1)}{2} \left [ \frac{n^{2} + n + 4n + 6}{2} \right ] = \frac{n(n + 1)}{4} \left ( n^{2} + 5n + 6 \right ) = \frac{n(n + 1)}{4} \left ( n^{2} + 2n + 3n + 6 \right ) = \frac{n(n + 1)\left [ n (n + 2) + 3 (n + 2) \right ]}{4} = \frac{n(n + 1) (n + 2) (n + 3) }{4}

 

Q3. Find sum to n^th term of the series 3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + . . .

Soln:

Given series = 3 \times 1^{2} + 5 \times 2^{2} + 7 \times 3^{2} + . . . n^{th} term,

a_n = (2n + 1) n² = 2n³ + n²

S_{n} = \sum^{n}_{k = 1} a_{k} = \sum^{n}_{k = 1} = (2k^{3} + k^{2}) = 2\sum^{n}_{k = 1} k^{3} + \sum^{n}_{k = 1} k^{2} = 2 \left [ \frac{n(n + 1)}{2} \right ]^{2} + \frac{n(n + 1)(2n + 1)}{6} = \frac{n^{2} (n + 1)^{2} }{2} + \frac{n(n + 1)(2n + 1)}{6} = \frac{n (n + 1) }{2} \left [ n(n + 1) + \frac{2n + 1}{3} \right ] = \frac{n (n + 1) }{2} \left [ \frac{3n^{2} + 3n + 2n + 1}{3} \right ] = \frac{n (n + 1) \left ( 3n^{2} + 5n + 1 \right ) }{6}

 

Q4. Find sum to n^th term of the series \frac{1}{ 1 \times 2 } + \frac{1}{ 2 \times 3 } + \frac{1}{ 3 \times 4 } + . . . .

Soln:

Given series = \frac{1}{ 1 \times 2 } + \frac{1}{ 2 \times 3 } + \frac{1}{ 3 \times 4 } + . . . . n^{th} \; term, a_{n} = \frac{1}{n(n + 1)} = \frac{1}{n} – \frac{1}{n + 1} a_{1} = \frac{1}{1} – \frac{1}{2} a_{2} = \frac{1}{2} – \frac{1}{3}a_{3} = \frac{1}{3} – \frac{1}{4} . . . a_{n} = \frac{1}{n} – \frac{1}{n + 1}

Column wise adding the above terms, we get

a_{1} +a_{2} + a_{3} + … + a_{n} = \left [ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n} \right ] – \left [ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{n + 1} +\right ] S_{n} = 1 – \frac{1}{n + 1} = \frac{n + 1 – 1}{n + 1} = \frac{n}{n + 1}

 

Q5. Find sum to n^th term of the series 5^{2} + 6^{2} + 7^{2} + . . . + 20^{2} \; n^{th} \; term, \; a_{n} = (n + 4)^{2} = n^{2} + 8 n + 16

Soln:

S_{n} = \sum^{n}_{k = 1} a_{k} = \sum^{n}_{k = 1}(k^{2} + 8k + 16) S_{n} = \sum^{n}_{k = 1} a_{k} = 8\sum^{n}_{k = 1}k + \sum^{n}_{k = 1}16 = \frac{n(n + 1)(2n + 1)}{6} + \frac{8n (n + 1)}{2} + 16 n

16^th term is (16 + 4)² = 20²

S_{16} = \frac{16(16 + 1)(2 \times 16 + 1)}{6} + \frac{8 \times 16 \times (16 + 1)}{2} + 16 \times 16 = \frac{(16)(17)(33)}{6} + \frac{(8) \times 16 \times (16 + 1)}{2} + 16 \times 16 = \frac{(16)(17)(33)}{6} + \frac{(8) (16) (17)}{2} + 256

= 1496 + 1088 + 256

= 2840

5² + 6² + 7² + …. + 20² = 2840

 

Q6. Find sum to n^th term of the series 3 \times 8 + 6 \times 11 + 9 \times 14 + …

Soln:

Given series = 3 \times 8 + 6 \times 11 + 9 \times 14 + … a_{n}

= (n^th term of 3, 6, 9, …) x (n^th term of 8, 11, 14 .. .)

= (3n) (3n + 5)

= 9n² + 15n

S_{n} = \sum^{n}_{k = 1}a_{k} = \sum^{n}_{k = 1} (9k^{2} + 15k) = \sum^{n}_{k = 1}k^{2} + 15 \sum^{n}_{k = 1} k 9 \times \frac{n(n + 1)(2n + 1)}{6} + 15 \times \frac{n (n + 1)}{ 2} = \frac{3n (n + 1)(2n + 1)}{2} + \frac{ 15n (n + 1)}{ 2} = \frac{3n (n + 1)}{2}(2n + 1 + 5) = \frac{3n (n + 1)}{2}(2n + 6) = 3n (n + 1)(n + 3)

 

Q7. Find sum to n^th term of the series 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . .

Soln:

Given series = 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . a^{n} = \left (1^{2} + 2^{2} + 3^{2} + . . . . + n^{2} \right ) = \frac{n(n + 1)(2n + 1)}{6} = \frac{n(2n^{2} + 3n + 1)}{6} = \frac{2n^{3} + 3n^{2} + n)}{6} = \frac{1}{3}n^{3} + \frac{1}{2}n^{2} + \frac{1}{6}n S_{n} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} \left ( \frac{1}{3} k^{3} + \frac{1}{2} k^{2} + \frac{1}{6} k \right ) = \frac{1}{3} \sum_{k = 1}^{n} k^{3} + \frac{1}{2} \sum_{k = 1}^{n} k^{2} + \frac{1}{6} \sum_{k = 1}^{n} k = \frac{1}{3} \frac{n^{2}(n + 1)^{2}}{(2)^{2}} + \frac{1}{2} \times \frac{n(n + 1)(2n + 1)}{6} + \frac{1}{6} \times \frac{n(n + 1)}{2} = \frac{n(n + 1)}{6} \left [ \frac{n(n + 1)}{2} + \frac{2n + 1}{2} + \frac{1}{2} \right ] = \frac{n(n + 1)}{6} \left [ \frac{n^{2} + n + 2n + 1 + 1 }{2} \right ] = \frac{n(n + 1)}{6} \left [ \frac{n^{2} + n + 2n + 2 }{2} \right ] = \frac{n(n + 1)}{6} \left [ \frac{n( n + 1) + 2(n + 1) }{2} \right ] = \frac{n(n + 1)}{6} \left [ \frac{( n + 1) (n + 2) }{2} \right ] = \frac{n(n + 1)^{2} (n + 2) }{12}

 

Q8. Find sum to n^th term of the series that has n^th term given by n(n + 1)(n + 4)

Soln:

a_{n} = n(n + 1)(n + 4) = n(n^{2} + 5n + 4) = n^{3} + 5n^{2} + 4n S_{n} = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} k^{3} + 5 \sum_{k = 1}^{n} k^{2} + 4 \sum_{k = 1}^{n} k = \frac{n^{2} (n + 1)^{2}}{4} + \frac{5n (n + 1)(2n + 1)}{6} + \frac{4n( n + 1)}{2} = \frac{n(n + 1)}{2} \left [ \frac{ n(n + 1)}{2} + \frac{5(2n + 1)}{6} + 4 \right ] = \frac{n(n + 1)}{2} \left [ \frac{ 3n^{2} + 3n + 20n + 10 + 24 }{6} \right ] = \frac{n(n + 1)}{2} \left [ \frac{ 3n^{2} + 23n  + 34 }{6} \right ] \frac{n(n + 1) \left ( 3n^{2} + 23n + 34 \right ) }{12}

 

Q9. Find sum to n^th term of the series that has n^th term given by n² + 2ⁿ

Soln:
a_n = n² + 2ⁿ

S_{n} = \sum_{k = 1}^{n} k^{2} + 2^{k} = \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} 2^{k} ……………(1)

Take \sum_{k = 1}^{n} 2^{k} = 2^{1} + 2^{2} + 2^{2} + . . . .

The series 2, 2², 2³, …. Is a G.P with 2 as first term as well as common ratio.

\sum_{k = 1}^{n} 2^{k} = \frac{(2)\left [ (2)^{n} – 1 \right ]}{2 – 1} = 2 (2^{n} – 1) ……………… (2)

From (1) and (2), we get

S_{n} = \sum_{k = 1}^{n} k^{2} + 2(2^{n} – 1) = \frac{n (n + 1)(2n + 1)}{6} + 2 (2^{n} – 1)

 

Q10. Find sum to n^th term of the series that has n^th term given by (2n – 1)²

Soln:

a_n = (2n – 1)² = 4n² – 4n + 1

S_{n} = 4 \sum_{k = 1}^{n} a_{k} – 4 \sum_{k = 1}^{n} (4k^{2} – 4k + 1) = 4\sum_{k = 1}^{n} k^{2} – 4 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1 = \frac{4n (n + 1) (2n + 1)}{6} – \frac{4n (n + 1)}{2} + n = \frac{2n (n + 1) (2n + 1)}{3} – 2n (n + 1) + n = n \left [ \frac{2 (2n^{2} + 3n + 1)} {3} – 2 (n + 1) + 1 \right ] = n \left [ \frac{4n^{2} + 6n + 2 – 6n – 6 + 3} {3} \right ] = n \left [ \frac{4n^{2} – 1} {3} \right ] = \frac{n (2n + 1) (2n – 1)}{3}

 

Thus, the above problems give an idea of NCERT Solutions maths chapter 9 sequence and series for class 11. Students can also practice some of the sample papers and previous year question papers to get an idea of the types of sequence and series questions asked in the final examination of class 11.