# NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series

Practising the NCERT Solutions Class 11 Chapter 9 Sequences and Series can help the students develop a thorough understanding of the topics explained in the Chapter. These solutions prepared by highly experienced teachers at BYJUâ€™S according to the latest CBSE Syllabus 2022-23 and are bound to help the students in addressing challenging questions with utmost confidence. Using the NCERT Solutions provided here, students can learn new methods of solving a particular problem in expeditious time to improve their performance in Class 11 Maths exam. These solutions have been designed after undertaking extensive research on each question and their problem-solving methods.

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Exercise 9.1 Solutions: 14 Questions

Exercise 9.2 Solutions: 18 Questions

Exercise 9.3 Solutions: 32 Questions

Exercise 9.4 Solutions: 10 Questions

Miscellaneous Exercise On Chapter 9 Solutions: 32 Questions

### Access Answers to NCERT Class 11 Maths Chapter 9 â€“ Sequences and Series

Exercise 9.1 Page No: 180

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

1. an = n (n + 2)Â

Solution:

Given,

nth term of a sequence an = n (n + 2)Â

On substitutingÂ nÂ = 1, 2, 3, 4, and 5, we get the first five terms

a1 = 1(1 + 2) = 3

a2 = 2(2 + 2) = 8

a3 = 3(3 + 2) = 15

a4 = 4(4 + 2) = 24

a5 = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

2.Â an = n/n+1

Solution:

Given nth term, an = n/n+1

On substitutingÂ nÂ = 1, 2, 3, 4, 5, we get

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

3. anÂ = 2n

Solution:

Given nth term, anÂ = 2n

On substitutingÂ nÂ = 1, 2, 3, 4, 5, we get

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

4. Â anÂ = (2n â€“ 3)/6

Solution:

Given nth term, anÂ = (2n â€“ 3)/6

On substitutingÂ nÂ = 1, 2, 3, 4, 5, we get

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

5. an = (-1)n-1 5n+1

Solution:

Given nth term, an = (-1)n-1 5n+1

On substitutingÂ nÂ = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, â€“125, 625, â€“3125, and 15625.

6.

Solution:

On substitutingÂ nÂ = 1, 2, 3, 4, 5, we get first 5 terms

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

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

7. an = 4n â€“ 3; a17, a24

Solution:

Given,

nthÂ term of the sequence is an = 4n â€“ 3

On substitutingÂ nÂ = 17, we get

a17 = 4(17) â€“ 3 = 68 â€“ 3 = 65

Next, on substitutingÂ nÂ = 24, we get

a24 = 4(24) â€“ 3 = 96 â€“ 3 = 93

8. anÂ =Â n2/2nÂ ;Â a7

Solution:

Given,

nthÂ term of the sequence is an = n2/2n

Now, on substitutingÂ nÂ = 7, we get

a7 = 72/27 = 49/ 128

9. an = (-1)n-1 n3; a9

Solution:

Given,

nthÂ term of the sequence is an = (-1)n-1 n3

On substitutingÂ nÂ = 9, we get

a9 = (-1)9-1 (9)3 = 1 x 729 = 729

10.

Solution:

On substitutingÂ nÂ = 20, we get

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

11. a1 = 3, an = 3an-1 + 2 for all n > 1

Solution:

Given, an = 3an-1 + 2 and a1 = 3

Then,

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

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

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

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

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 â€¦â€¦.

12. a1 = -1, an = an-1/n, n â‰¥ 2

Solution:

Given,

an = an-1/n and a1 = -1

Then,

a2 = a1/2 = -1/2

a3 = a2/3 = -1/6

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + â€¦â€¦.

13. a1 = a2 = 2, an = an-1 â€“ 1, n > 2

Solution:

Given,

a1 = a2, an = an-1 â€“ 1

Then,

a3 = a2 â€“ 1 = 2 â€“ 1 = 1

a4 = a3 â€“ 1 = 1 â€“ 1 = 0

a5 = a4 â€“ 1 = 0 â€“ 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + â€¦â€¦

14. The Fibonacci sequence is defined by

1 = a1 = a2 and an = an â€“ 1 + an â€“ 2, n > 2

Find an+1/an, for n = 1, 2, 3, 4, 5Â

Solution:

Given,

1 = a1 = a2

an = an â€“ 1 + an â€“ 2, n > 2

So,

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

Thus,

Exercise 9.2 Page No: 185

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

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, â€¦1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term,Â aÂ = 1

Common difference,Â dÂ = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n â€“ 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

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

Solution:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, â€¦ 995.

It clearly forms a sequence in A.P.

Where, the first term,Â aÂ = 105

Common difference,Â dÂ = 5

Now,

a + (n -1)d = 995

105 + (n â€“ 1)(5) = 995

105 + 5n â€“ 5 = 995

5n = 995 â€“ 105 + 5 = 895

n = 895/5

n = 179

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

3. 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 20thÂ term is â€“112.

Solution:

Given,

The first term (a) of an A.P = 2

Letâ€™s assumeÂ dÂ be the common difference of the A.P.

So, the A.P. will be 2, 2 +Â d, 2 + 2d, 2 + 3d, â€¦

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = Â¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20 â€“ 1)d = 2 + (19) (-6) = 2 â€“ 114 = -112

Therefore, the 20thÂ term of the A.P. is â€“112.

4. How many terms of the A.P. -6, -11/2, -5, â€¦.Â are needed to give the sum â€“25?

Solution:

Letâ€™s consider the sum ofÂ nÂ terms of the given A.P. as â€“25.

We known that,

Sn = n/2 [2a + (n-1)d]

whereÂ nÂ = number of terms,Â aÂ = first term, andÂ dÂ = common difference

So here,Â aÂ = â€“6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

5. In an A.P., ifÂ pthÂ term isÂ 1/q andÂ qthÂ term isÂ 1/p, prove that the sum of firstÂ pqÂ terms is Â½ (pq + 1) where p â‰  q. Â

Solution:

6. If the sum of a certain number of terms of the A.P. 25, 22, 19, â€¦ is 116. Find the last term

Solution:

Given A.P.,

25, 22, 19, â€¦

Here,

First term, a = 25 and

Common difference, d = 22 â€“ 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

Sn = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 â€“ 3n + 3]

232 = n [53 â€“ 3n]

232 = 53n â€“ 3n2

3n2 â€“ 53n + 232 = 0

3n2 â€“ 24n â€“ 29n+ 232 = 0

3n(n â€“ 8) â€“ 29(n â€“ 8) = 0

(3n â€“ 29) (n â€“ 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, 8th term is the last term of the A.P.

a8 = 25 + (8 â€“ 1)(-3)

= 25 â€“ 21

= 4

7. Find the sum toÂ nÂ terms of the A.P., whoseÂ kthÂ term is 5kÂ + 1.

Solution:

Given, theÂ kthÂ term of the A.P. is 5kÂ + 1.

kthÂ term =Â akÂ =Â aÂ + (kÂ â€“ 1)d

And,

aÂ + (kÂ â€“ 1)dÂ = 5kÂ + 1

aÂ +Â kdÂ â€“Â dÂ = 5kÂ + 1

On comparing the coefficient ofÂ k, we getÂ dÂ = 5

aÂ â€“Â dÂ = 1

aÂ â€“ 5 = 1

â‡’Â aÂ = 6

8. If the sum ofÂ nÂ terms of an A.P. is (pnÂ +Â qn2), whereÂ pÂ andÂ qÂ are constants, find the common difference.

Solution:

We know that,

Sn = n/2 [2a + (n-1)d]

From the question we have,

On comparing the coefficients ofÂ n2Â on both sides, we get

d/2 = q

Hence, dÂ = 2q

Therefore, the common difference of the A.P. is 2q.

9. The sums ofÂ nÂ terms of two arithmetic progressions are in the ratio 5nÂ + 4: 9nÂ + 6. Find the ratio of their 18thÂ terms.

Solution:

LetÂ a1,Â a2, andÂ d1,Â d2Â be the first terms and the common difference of the first and second arithmetic progression respectively.

Then, from the question we have

10. If the sum of firstÂ pÂ terms of an A.P. is equal to the sum of the firstÂ qÂ terms, then find the sum of the first (pÂ +Â q) terms.

Solution:

Letâ€™s takeÂ aÂ andÂ dÂ to be the first term and the common difference of the A.P. respectively.

Then, it given that

Therefore, the sum of (p + q) terms of the A.P. is 0.

11. Sum of the firstÂ p, qÂ andÂ rÂ terms of an A.P. areÂ a, bÂ andÂ c, respectively.

Prove thatÂ

Solution:

LetÂ a1Â andÂ dÂ be the first term and the common difference of the A.P. respectively.

Then according to the question, we have

Now, subtracting (2) from (1), we get

12. The ratio of the sums ofÂ mÂ andÂ nÂ terms of an A.P. isÂ m2:Â n2. Show that the ratio ofÂ mthÂ andÂ nthÂ term is (2mÂ â€“ 1): (2nÂ â€“ 1).

Solution:

Letâ€™s consider thatÂ aÂ andÂ bÂ to be the first term and the common difference of the A.P. respectively.

Then from the question, we have

Hence, the given result is proved.

13. If the sum ofÂ nÂ terms of an A.P. isÂ 3n2 + 5n and itsÂ mthÂ term is 164, find the value ofÂ m.

Solution:

Letâ€™s considerÂ aÂ andÂ bÂ to be the first term and the common difference of the A.P. respectively.

amÂ =Â aÂ + (mÂ â€“ 1)dÂ = 164 â€¦ (1)

We the sum of the terms is given by,

Sn = n/2 [2a + (n-1)d]

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

Solution:

Letâ€™s assume A1, A2, A3, A4, and A5Â to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.

Here we have,

aÂ = 8,Â bÂ = 26,Â nÂ = 7

So,

26 = 8 + (7 â€“ 1)Â d

6dÂ = 26 â€“ 8 = 18

dÂ = 3

Now,

A1Â =Â aÂ +Â dÂ = 8 + 3 = 11

A2Â =Â aÂ + 2dÂ = 8 + 2 Ã— 3 = 8 + 6 = 14

A3Â =Â aÂ + 3dÂ = 8 + 3 Ã— 3 = 8 + 9 = 17

A4Â =Â aÂ + 4dÂ = 8 + 4 Ã— 3 = 8 + 12 = 20

A5Â =Â aÂ + 5dÂ = 8 + 5 Ã— 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

15. IfÂ is the A.M. betweenÂ aÂ andÂ b, then find the value ofÂ n.

Solution:

The A.M between a and b is given by, (a + b)/2

Then according to the question,

Thus, the value of n is 1.

16. Between 1 and 31,Â mÂ numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7thÂ and (mÂ â€“ 1)thÂ numbers is 5: 9. Find the value ofÂ m.

Solution:

Letâ€™s consider a1, a2, â€¦ amÂ beÂ mÂ numbers such that 1, a1, a2, â€¦ am, 31 is an A.P.

And here,

aÂ = 1,Â bÂ = 31,Â nÂ =Â mÂ + 2

So, 31 = 1 + (mÂ + 2 â€“ 1) (d)

30 = (mÂ + 1)Â d

d = 30/ (m + 1) â€¦â€¦. (1)

Now,

a1Â =Â aÂ +Â d

a2Â =Â aÂ + 2d

a3Â =Â aÂ + 3dÂ â€¦

Hence, a7Â =Â aÂ + 7d

amâ€“1Â =Â aÂ + (mÂ â€“ 1)Â d

According to the question, we have

Therefore, the value of m is 14.

17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30thÂ instalment?

Solution:

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105 and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And the, A.P. is 100, 105, 110, â€¦

Where, first term,Â aÂ = 100

Common difference,Â dÂ = 5

So, the 30th term in this A.P. will be

A30Â  =Â aÂ + (30 â€“ 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30thÂ instalment will be Rs 245.

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

Solution:

Itâ€™s understood from the question that, the angles of the polygon will form an A.P. with common differenceÂ dÂ = 5Â° and first termÂ a = 120Â°.

And, we know that the sum of all angles of a polygon withÂ nÂ sides is 180Â° (nÂ â€“ 2).

Thus, we can say

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.

Exercise 9.3 Page No: 192

1. Find the 20thÂ andÂ nthterms of the G.P. 5/2, 5/4, 5/8, â€¦â€¦â€¦

Solution:

Given G.P. isÂ 5/2, 5/4, 5/8, â€¦â€¦â€¦

Here,Â aÂ = First term =Â 5/2

rÂ = Common ratio =Â (5/4)/(5/2) = Â½

Thus, the 20th term and nth term

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

Solution:

Given,

The common ratio of the G.P.,Â rÂ = 2

And, letÂ aÂ be the first term of the G.P.

Now,

a8Â =Â arÂ 8â€“1Â =Â ar7

ar7Â = 192

a(2)7Â = 192

a(2)7Â = (2)6Â (3)

3. The 5th, 8thÂ and 11thÂ terms of a G.P. areÂ p,Â qÂ andÂ s, respectively. Show thatÂ q2Â =Â ps.

Solution:

Letâ€™s takeÂ aÂ to be the first term andÂ rÂ to be the common ratio of the G.P.

Then according to the question, we have

a5Â =Â aÂ r5â€“1Â =Â aÂ r4Â =Â pÂ â€¦ (i)

a8Â =Â aÂ r8â€“1Â =Â aÂ r7Â =Â qÂ â€¦ (ii)

a11Â = aÂ r11â€“1Â =Â aÂ r10Â =Â sÂ â€¦ (iii)

Dividing equation (ii) by (i), we get

4. The 4thÂ term of a G.P. is square of its second term, and the first term is â€“3. Determine its 7thÂ term.

Solution:

Letâ€™s considerÂ aÂ to be the first term andÂ rÂ to be the common ratio of the G.P.

Given, aÂ = â€“3

And we know that,

anÂ =Â arnâ€“1

So, a4Â =Â ar3Â = (â€“3)Â r3

a2Â =Â a r1Â = (â€“3)Â r

Then from the question, we have

(â€“3)Â r3Â = [(â€“3)Â r]2

â‡’ â€“3r3Â = 9Â r2

â‡’Â rÂ = â€“3

a7Â =Â aÂ rÂ 7â€“1Â =Â aÂ r6Â = (â€“3) (â€“3)6Â = â€“ (3)7Â = â€“2187

Therefore, the seventh term of the G.P. is â€“2187.

5. Which term of the following sequences:

(a) 2, 2âˆš2, 4,â€¦ is 128 ? (b) âˆš3, 3, 3âˆš3,â€¦ is 729 ?

(c) 1/3, 1/9, 1/27, â€¦ is 1/19683 ?

Solution:

(a) The given sequence, 2, 2âˆš2, 4,â€¦

We have,

a = 2 and r = 2âˆš2/2 = âˆš2

Taking the nth term of this sequence as 128, we have

Therefore, the 13th term of the given sequence is 128.

(ii) Given sequence, âˆš3, 3, 3âˆš3,â€¦

We have,

a = âˆš3 and r = 3/âˆš3 = âˆš3

Taking the nth term of this sequence to be 729, we have

Therefore, the 12th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, â€¦

a = 1/3 and r = (1/9)/(1/3) = 1/3

Taking the nth term of this sequence to be 1/19683, we have

Therefore, the 9th term of the given sequence is 1/19683.

6. For what values ofÂ x,Â the numbersÂ -2/7, x, -7/2 are in G.P?

Solution:

The given numbers are -2/7, x, -7/2.

Common ratioÂ = x/(-2/7) = -7x/2

Also, common ratio =Â (-7/2)/x = -7/2x

Therefore, forÂ xÂ = Â± 1, the given numbers will be in G.P.

7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 â€¦

Solution:

Given G.P., 0.15, 0.015, 0.00015, â€¦

Here,Â aÂ = 0.15 and r = 0.015/0.15 = 0.1Â

8. Find the sum toÂ nÂ terms in the geometric progression âˆš7, âˆš21, 3âˆš7, â€¦.

Solution:

The given G.P is âˆš7, âˆš21, 3âˆš7, â€¦.

Here,

a = âˆš7 and

9. Find the sum toÂ nÂ terms in the geometric progression 1, -a, a2, -a3 â€¦. (if a â‰  -1)

Solution:

The given G.P. isÂ 1, -a, a2, -a3 â€¦.

Here, the first term =Â a1Â = 1

And the common ratio =Â rÂ = â€“Â a

We know that,

10. Find the sum toÂ nÂ terms in the geometric progression x3, x5, x7, â€¦ (if x â‰  Â±1 )

Solution:

Given G.P. isÂ x3, x5, x7, â€¦

Here, we haveÂ aÂ =Â x3Â andÂ rÂ =Â x5/x3 = x2

11. Evaluate:

Solution:

12. The sum of first three terms of a G.P. isÂ 39/10 and their product is 1. Find the common ratio and the terms.

Solution:

LetÂ a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 â€¦â€¦ (1)

(a/r) (a) (ar) = 1 â€¦â€¦.. (2)

From (2), we have

a3 = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

1/r + 1 + r = 39/10

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

10 + 10r + 10r2 = 39r

10r2 â€“ 29r + 10 = 0

10r2 â€“ 25r â€“ 4r + 10 = 0

5r(2r â€“ 5) â€“ 2(2r â€“ 5) = 0

(5r â€“ 2) (2r â€“ 5) = 0

Thus,

r = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

13. How many terms of G.P. 3, 32, 33, â€¦ are needed to give the sum 120?

Solution:

Given G.P. is 3, 32, 33, â€¦

Letâ€™s consider thatÂ nÂ terms of this G.P. be required to obtain the sum of 120.

We know that,

Here,Â aÂ = 3 andÂ rÂ = 3

Equating the exponents we get, nÂ = 4

Therefore, four terms of the given G.P. are required to obtain the sum as 120.

14. 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 toÂ nÂ terms of the G.P.

Solution:

Letâ€™s assume the G.P. to beÂ a,Â ar,Â ar2,Â ar3, â€¦

Then according to the question, we have

aÂ +Â arÂ +Â ar2Â = 16 andÂ ar3Â +Â ar4Â +Â ar5Â = 128

aÂ (1 +Â rÂ +Â r2) = 16 â€¦ (1) and,

ar3(1 +Â rÂ +Â r2) = 128 â€¦ (2)

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

r3 = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as

15. Given a G.P. withÂ aÂ = 729 and 7thÂ term 64, determine S7.

Solution:

Given,

aÂ = 729 and a7Â = 64

LetÂ rÂ be the common ratio of the G.P.

Then we know that,Â anÂ =Â a rnâ€“1

a7Â =Â ar7â€“1Â = (729)r6

â‡’ 64 = 729Â r6

r6 = 64/729

r6 = (2/3)6

r = 2/3

And, we know that

16. Find a G.P. for which sum of the first two terms is â€“4 and the fifth term is 4 times the third term.

Solution:

ConsiderÂ aÂ to be the first term andÂ rÂ to be the common ratio of the G.P.

Given, S2 = -4

Then, from the question we have

And,

a5 = 4 x a3

ar4 = 4ar2

r2 = 4

r = Â± 2

Using the value of r in (1), we have

Therefore, the required G.P is

-4/3, -8/3, -16/3, â€¦. Or 4, -8, 16, -32, â€¦â€¦

17. If the 4th, 10thÂ and 16thÂ terms of a G.P. areÂ x, yÂ andÂ z, respectively. Prove thatÂ x,Â y,Â zÂ are in G.P.

Solution:

LetÂ aÂ be the first term andÂ rÂ be the common ratio of the G.P.

According to the given condition,

a4Â =Â aÂ r3Â =Â xÂ â€¦ (1)

a10Â =Â aÂ r9Â =Â yÂ â€¦ (2)

a16Â =Â a r15Â =Â zÂ â€¦ (3)

On dividing (2) by (1), we get

18. Find the sum toÂ nÂ terms of the sequence, 8, 88, 888, 8888â€¦

Solution:

Given sequence: 8, 88, 888, 8888â€¦

This sequence is not a G.P.

But, it can be changed to G.P. by writing the terms as

SnÂ = 8 + 88 + 888 + 8888 + â€¦â€¦â€¦â€¦â€¦.. toÂ nÂ terms

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

Solution:

The required sum =Â 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x Â½

= 64[4 + 2 + 1 + Â½ + 1/22]

Now, itâ€™s seen that

4, 2, 1,Â Â½, 1/22 is a G.P.

With first term,Â aÂ = 4

Common ratio,Â rÂ =1/2

We know,

Therefore, the required sum =Â 64(31/4) = (16)(31) = 496

20. Show that the products of the corresponding terms of the sequencesÂ a, ar, ar2, â€¦arn-1 and A, AR, AR2, â€¦ ARn-1Â form a G.P, and find the common ratio.

Solution:

To be proved: The sequence,Â aA,Â arAR,Â ar2AR2, â€¦arnâ€“1ARnâ€“1, forms a G.P.

Now, we have

Therefore, the above sequence forms a G.P. and the common ratio isÂ rR.

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

Solution:

ConsiderÂ aÂ to be theÂ first term andÂ rÂ to be the common ratio of the G.P.

Then,

a1Â =Â a,Â a2Â =Â ar,Â a3Â =Â ar2,Â a4Â =Â ar3

From the question, we have

a3Â =Â a1Â + 9

ar2Â =Â aÂ + 9 â€¦ (i)

a2Â =Â a4Â + 18

arÂ =Â ar3Â + 18 â€¦ (ii)

So, from (1) and (2), we get

a(r2Â â€“ 1) = 9 â€¦ (iii)

arÂ (1â€“Â r2) = 18 â€¦ (iv)

Now, dividing (4) by (3), we get

-r = 2

r = -2

On substituting the value ofÂ rÂ in (i), we get

4aÂ =Â aÂ + 9

3aÂ = 9

âˆ´Â aÂ = 3

Therefore, the first four numbers of the G.P. are 3, 3(â€“ 2), 3(â€“2)2, and 3(â€“2)3

i.e., 3Â¸â€“6, 12, and â€“24.

22. If the pth, qth and rthÂ terms of a G.P. areÂ a, bÂ andÂ c, respectively. Prove that aq-r br-p cp-q = 1

Solution:

Letâ€™s take AÂ to be the first term andÂ RÂ to be the common ratio of the G.P.

Then according to the question, we have

ARpâ€“1Â =Â a

ARqâ€“1Â =Â b

ARrâ€“1Â =Â c

Then,

aqâ€“rÂ brâ€“pÂ cpâ€“q

=Â Aqâ€“rÂ Ã—Â R(pâ€“1) (qâ€“r)Â Ã— Arâ€“pÂ Ã—Â R(qâ€“1) (râ€“p)Â Ã—Â Apâ€“qÂ Ã—Â R(rÂ â€“1)(pâ€“q)

=Â AqÂ â€“Â rÂ +Â rÂ â€“Â pÂ +Â pÂ â€“Â qÂ Ã—Â RÂ (prÂ â€“Â prÂ â€“Â qÂ +Â r) + (rqÂ â€“Â rÂ +Â pÂ â€“Â pq) + (prÂ â€“Â pÂ â€“Â qrÂ +Â q)

=Â A0Â Ã—Â R0

= 1

Hence proved.

23. If the first and theÂ nthÂ term of a G.P. areÂ aÂ adÂ b, respectively, and ifÂ PÂ is the product ofÂ nÂ terms, prove thatÂ P2Â = (ab)n.

Solution:

Given, the first term of the G.P isÂ aÂ and the last term isÂ b.

Thus,

The G.P. isÂ a,Â ar,Â ar2,Â ar3, â€¦Â arnâ€“1, whereÂ rÂ is the common ratio.

Then,

bÂ =Â arnâ€“1Â  â€¦ (1)

PÂ = Product ofÂ nÂ terms

= (a) (ar) (ar2) â€¦ (arnâ€“1)

= (aÂ Ã—Â aÂ Ã—â€¦a) (rÂ Ã—Â r2Â Ã— â€¦rnâ€“1)

=Â anÂ rÂ 1 + 2 +â€¦(nâ€“1)Â  â€¦ (2)

Here, 1, 2, â€¦(nÂ â€“ 1) is an A.P.

And, the product of n terms P is given by,

24. Show that the ratio of the sum of firstÂ nÂ terms of a G.P. to the sum of terms fromÂ .

Solution:

LetÂ aÂ be the first term andÂ rÂ be the common ratio of the G.P.

Since there areÂ nÂ terms from (nÂ +1)thÂ to (2n)thÂ term,

Sum of terms from(nÂ + 1)thÂ to (2n)thÂ term

aÂ nÂ +1Â =Â arÂ n + 1Â â€“ 1Â =Â arn

Thus, required ratio =

Thus, the ratio of the sum of firstÂ nÂ terms of a G.P. to the sum of terms from (nÂ + 1)thÂ to (2n)thÂ term isÂ .

25. IfÂ a, b, cÂ andÂ dÂ are in G.P. show thatÂ (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

Solution:

Given, a,Â b,Â c,Â dÂ are in G.P.

So, we have

b2Â =Â acÂ  â€¦ (2)

c2Â =Â bdÂ  â€¦ (3)

Taking the R.H.S. we have

R.H.S.

= (abÂ +Â bcÂ +Â cd)2

= (abÂ +Â adÂ +Â cd)2Â  [Using (1)]

= [abÂ +Â dÂ (aÂ +Â c)]2

=Â a2b2Â + 2abdÂ (aÂ +Â c) +Â d2Â (aÂ +Â c)2

=Â a2b2Â +2a2bdÂ + 2acbdÂ +Â d2(a2Â + 2acÂ +Â c2)

=Â a2b2Â + 2a2c2Â + 2b2c2Â +Â d2a2Â + 2d2b2Â +Â d2c2Â  [Using (1) and (2)]

=Â a2b2Â +Â a2c2Â +Â a2c2Â +Â b2c2Â +Â b2c2Â +Â d2a2Â +Â d2b2Â +Â d2b2Â +Â d2c2

=Â a2b2Â +Â a2c2Â +Â a2d2Â +Â b2Â Ã—Â b2Â +Â b2c2Â +Â b2d2Â +Â c2b2Â +Â c2Â Ã—Â c2Â +Â c2d2

[Using (2) and (3) and rearranging terms]

=Â a2(b2Â +Â c2Â +Â d2) +Â b2Â (b2Â +Â c2Â +Â d2) +Â c2Â (b2+Â c2Â +Â d2)

= (a2Â +Â b2Â +Â c2) (b2Â +Â c2Â +Â d2)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

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

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

Solution:

Letâ€™s assumeÂ G1Â andÂ G2Â to be two numbers between 3 and 81 such that the series 3,Â G1,Â G2, 81 forms a G.P.

And letÂ aÂ be the first term andÂ rÂ be the common ratio of the G.P.

Now, we have the 1st term as 3 and the 4th term as 81.

81 = (3)Â (r)3

r3Â = 27

âˆ´Â rÂ = 3 (Taking real roots only)

ForÂ rÂ = 3,

G1Â =Â arÂ = (3) (3) = 9

G2Â =Â ar2Â = (3) (3)2Â = 27

Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

27. Find the value ofÂ nÂ so thatÂ may be the geometric mean betweenÂ aÂ andÂ b.

Solution:

We know that,

The G. M. ofÂ aÂ andÂ bÂ isÂ given by âˆšab.

Then from the question, we have

By squaring both sides, we get

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio.

Solution:

Consider the two numbers beÂ aÂ andÂ b.

Then, G.M. =Â âˆšab.

From the question, we have

29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the

numbers are .

Solution:

Given thatÂ AÂ andÂ GÂ are A.M. and G.M. between two positive numbers.

And, let these two positive numbers beÂ aÂ andÂ b.

30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2ndÂ hour, 4thÂ hour andÂ nthÂ hour?

Solution:

Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here we have,Â aÂ = 30 andÂ rÂ = 2

So,Â a3Â =Â ar2Â = (30) (2)2Â = 120

Thus, the number of bacteria at the end of 2ndÂ hour will be 120.

And, a5Â =Â ar4Â = (30) (2)4Â = 480

The number of bacteria at the end of 4thÂ hour will be 480.

anÂ +1Â =Â arnÂ = (30) 2n

Therefore, the number of bacteria at the end ofÂ nthÂ hour will be 30(2)n.

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

Solution:

Given,

The amount deposited in the bank is Rs 500.

At the end of first year, amount = Rs 500(1 + 1/10)Â = Rs 500 (1.1)

At the end of 2ndÂ year, amount = Rs 500 (1.1) (1.1)

At the end of 3rdÂ year, amount = Rs 500 (1.1) (1.1) (1.1) and so onâ€¦.

Therefore,

The amount at the end of 10 years = Rs 500 (1.1) (1.1) â€¦ (10 times)

= Rs 500(1.1)10

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

Solution:

Letâ€™s consider the roots of the quadratic equation to beÂ aÂ andÂ b.

Then, we have

We know that,

A quadratic equation can be formed as,

x2 â€“Â xÂ (Sum of roots) + (Product of roots) = 0

x2Â â€“Â xÂ (aÂ +Â b) + (ab) = 0

x2Â â€“ 16xÂ + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation isÂ x2Â â€“ 16xÂ + 25 = 0

Exercise 9.4 Page No: 196

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

1. 1 Ã— 2 + 2 Ã— 3 + 3 Ã— 4 + 4 Ã— 5 + â€¦

Solution:

Given series is 1 Ã— 2 + 2 Ã— 3 + 3 Ã— 4 + 4 Ã— 5 + â€¦

Itâ€™s seen that,

nthÂ term,Â anÂ =Â nÂ (Â nÂ + 1)

Then, the sum of n terms of the series can be expressed as

2. 1 Ã— 2 Ã— 3 + 2 Ã— 3 Ã— 4 + 3 Ã— 4 Ã— 5 + â€¦

Solution:

Given series is 1 Ã— 2 Ã— 3 + 2 Ã— 3 Ã— 4 + 3 Ã— 4 Ã— 5 + â€¦

Itâ€™s seen that,

nthÂ term,Â anÂ =Â nÂ (Â nÂ + 1) (Â nÂ + 2)

= (n2Â +Â n) (nÂ + 2)

=Â n3Â + 3n2Â + 2n

Then, the sum of n terms of the series can be expressed as

3. 3 Ã— 12Â + 5 Ã— 22Â + 7 Ã— 32Â + â€¦

Solution:

Given series is 3 Ã—12Â + 5 Ã— 22Â + 7 Ã— 32Â + â€¦

Itâ€™s seen that,

nthÂ term,Â anÂ = ( 2nÂ + 1)Â n2Â = 2n3Â +Â n2

Then, the sum of n terms of the series can be expressed as

4. Find the sum toÂ nÂ terms of the seriesÂ

Solution:

5. Find the sum toÂ nÂ terms of the seriesÂ 52Â + 62Â + 72Â + â€¦ + 202

Solution:

Given series is 52Â + 62Â + 72Â + â€¦ + 202

Itâ€™s seen that,

nthÂ term,Â anÂ = (Â nÂ + 4)2Â =Â n2Â + 8nÂ + 16

Then, the sum of n terms of the series can be expressed as

6. Find the sum toÂ nÂ terms of the series 3 Ã— 8 + 6 Ã— 11 + 9 Ã— 14 +â€¦

Solution:

Given series is 3 Ã— 8 + 6 Ã— 11 + 9 Ã— 14 + â€¦

Itâ€™s found out that,

anÂ = (nthÂ term of 3, 6, 9 â€¦) Ã— (nthÂ term of 8, 11, 14, â€¦)

= (3n) (3nÂ + 5)

= 9n2Â + 15n

Then, the sum of n terms of the series can be expressed as

7. Find the sum toÂ nÂ terms of the series 12Â + (12Â + 22) + (12Â + 22Â + 32) + â€¦

Solution:

Given series is 12Â + (12Â + 22) + (12Â + 22Â + 32Â ) + â€¦

Finding the nth term, we have

anÂ = (12Â + 22Â + 32Â +â€¦â€¦.+Â n2)

Now, the sum of n terms of the series can be expressed as

8. Find the sum toÂ nÂ terms of the series whoseÂ nthÂ term is given byÂ nÂ (nÂ + 1) (nÂ + 4).

Solution:

Given,

anÂ =Â nÂ (nÂ + 1) (nÂ + 4) =Â n(n2Â + 5nÂ + 4) =Â n3Â + 5n2Â + 4n

Now, the sum of n terms of the series can be expressed as

9. Find the sum toÂ nÂ terms of the series whoseÂ nthÂ terms is given byÂ n2Â + 2n

Solution:

Given,

nth term of the series as:

anÂ = n2Â + 2n

Then, the sum of n terms of the series can be expressed as

10. Find the sum toÂ nÂ terms of the series whoseÂ nthÂ terms is given by (2nÂ â€“ 1)2

Solution:

Given,

nth term of the series as:

anÂ =Â (2nÂ â€“ 1)2Â = 4n2Â â€“ 4nÂ + 1

Then, the sum of n terms of the series can be expressed as

Miscellaneous Exercise Page No: 199

1. Show that the sum of (mÂ +Â n)thÂ and (mÂ â€“Â n)thÂ terms of an A.P. is equal to twice theÂ mthÂ term.

Solution:

Letâ€™s takeÂ aÂ andÂ dÂ to be the first term and the common difference of the A.P. respectively.

We know that, theÂ kthÂ term of an A. P. is given by

akÂ =Â aÂ + (kÂ â€“1)Â d

So, amÂ +Â nÂ =Â aÂ + (mÂ +Â nÂ â€“1)Â d

And, amÂ â€“Â nÂ =Â aÂ + (mÂ â€“Â nÂ â€“1)Â d

amÂ =Â aÂ + (mÂ â€“1)Â d

Thus,

amÂ +Â nÂ +Â amÂ â€“Â nÂ  =Â aÂ + (mÂ +Â nÂ â€“1)Â dÂ +Â aÂ + (mÂ â€“Â nÂ â€“1)Â d

= 2aÂ + (mÂ +Â nÂ â€“1 +Â mÂ â€“Â nÂ â€“1)Â d

= 2aÂ + (2mÂ â€“ 2)Â d

= 2aÂ + 2 (mÂ â€“ 1)Â d

=2 [aÂ + (mÂ â€“ 1)Â d]

= 2am

Therefore, the sum of (mÂ +Â n)thÂ and (mÂ â€“Â n)thÂ terms of an A.P. is equal to twice theÂ mthÂ term

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

Solution:

Letâ€™s consider the three numbers in A.P. asÂ aÂ â€“Â d,Â a, andÂ aÂ +Â d.

Then, from the question we have

(aÂ â€“Â d) + (a) + (aÂ +Â d) = 24 â€¦ (i)

3aÂ = 24

âˆ´Â aÂ = 8

And,

(aÂ â€“Â d)Â aÂ (aÂ +Â d) = 440 â€¦ (ii)

(8 â€“Â d) (8) (8 +Â d) = 440

(8 â€“Â d) (8 +Â d) = 55

64 â€“Â d2Â = 55

d2Â = 64 â€“ 55 = 9

âˆ´ dÂ = Â± 3

Thus,

WhenÂ dÂ = 3, the numbers are 5, 8, and 11 and

WhenÂ dÂ = â€“3, the numbers are 11, 8, and 5.

Therefore, the three numbers are 5, 8, and 11.

3. Let the sum ofÂ n, 2n, 3nÂ terms of an A.P. be S1, S2Â and S3,Â respectively, show that S3Â = 3 (S2â€“ S1)

Solution:

Letâ€™s takeÂ aÂ andÂ dÂ to be the first term and the common difference of the A.P. respectively.

So, we have

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

Solution:

First letâ€™s find the numbers between 200 and 400 which are divisible by 7.

The numbers are:

203, 210, 217, â€¦ 399

Here, the first term,Â aÂ = 203

Last term,Â lÂ = 399 and

Common difference,Â dÂ = 7

Letâ€™s consider the number of terms of the A.P. to beÂ n.

Hence, anÂ = 399 =Â aÂ + (nÂ â€“1)Â d

399 = 203 + (nÂ â€“1) 7

7 (nÂ â€“1) = 196

nÂ â€“1 = 28

nÂ = 29

Then, the sum of 29 terms of the A.P is given by:

Therefore, the required sum is 8729.

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

Solution:

First letâ€™s find the integers from 1 to 100, which are divisible by 2.

And, they are 2, 4, 6â€¦ 100.

Clearly, this forms an A.P. with the first term and common difference both equal to 2.

So, we have

100 = 2 + (nÂ â€“1) 2

nÂ = 50

Hence, the sum is

Now, the integers from 1 to 100, which are divisible by 5, are 5, 10â€¦ 100.

This also forms an A.P. with the first term and common difference both equal to 5.

So, we have

100 = 5 + (nÂ â€“1) 5

5nÂ = 100

nÂ = 20

Hence, the sum is

Lastly, the integers which are divisible by both 2 and 5, are 10, 20, â€¦ 100.

And this also forms an A.P. with the first term and common difference both equal to 10.

So, we have

100 = 10 + (nÂ â€“1) (10)

100 = 10n

nÂ = 10

Thus, the required sum = 2550 + 1050 â€“ 550 = 3050

Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

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

Solution:

We have to first find the two-digit numbers, which when divided by 4, yield 1 as remainder.

They are: 13, 17, â€¦ 97.

As itâ€™s seen that this series forms an A.P. with first term (a) 13 and common difference (d) 4.

LetÂ nÂ be the number of terms of the A.P.

We know that, theÂ nthÂ term of an A.P. is given by,

anÂ =Â aÂ + (nÂ â€“1)Â d

So, 97 = 13 + (nÂ â€“1) (4)

4 (nÂ â€“1) = 84

nÂ â€“ 1 = 21

nÂ = 22

Now, the sum ofÂ nÂ terms of an A.P. is given by,

Therefore, the required sum is 1210.

7. IfÂ fÂ is a function satisfyingÂ f(x + y) = f(x) f(y) for all x, y âˆˆ N such that , find the value ofÂ n.

Solution:

Given that,

fÂ (xÂ +Â y) =Â fÂ (x) Ã—Â fÂ (y) for allÂ x,Â yÂ âˆˆ N â€¦ (1)

fÂ (1) = 3

TakingÂ xÂ =Â yÂ = 1 in (1), we have

fÂ (1 + 1) =Â fÂ (2) =Â fÂ (1)Â fÂ (1) = 3 Ã— 3 = 9

Similarly,

fÂ (1 + 1 + 1) =Â fÂ (3) =Â fÂ (1 + 2) =Â fÂ (1)Â fÂ (2) = 3 Ã— 9 = 27

And, fÂ (4) =Â fÂ (1 + 3) =Â fÂ (1)Â fÂ (3) = 3 Ã— 27 = 81

Thus, fÂ (1),Â fÂ (2),Â fÂ (3), â€¦, that is 3, 9, 27, â€¦, forms a G.P. with the first term and common ratio both equal to 3.

We know that sum of terms in G.P is given by,

And itâ€™s given that,

Hence, the sum of terms of the function is 120.

Therefore, the value ofÂ nÂ is 4.

8. 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.

Solution:

Given that the sum of some terms in a G.P is 315.

Let the number of terms be n.

We know that, sum of terms is

Given that the first termÂ aÂ is 5 and common ratioÂ rÂ is 2.

Hence, the last term of the G.P = 6thÂ term =Â ar6 â€“ 1Â = (5)(2)5Â = (5)(32) = 160

Therefore, the last term of the G.P. is 160.

9. 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.

Solution:

Letâ€™s considerÂ aÂ andÂ rÂ to be the first term and the common ratio of the G.P. respectively.

Given, aÂ = 1

a3Â =Â ar2Â =Â r2

a5Â =Â ar4Â =Â r4

Then, from the question we have

r2Â +Â r4Â = 90

r4Â +Â r2Â â€“ 90 = 0

Therefore, the common ratio of the G.P. is Â±3.

10. 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.

Solution:

Letâ€™s consider the three numbers in G.P. to be asÂ a,Â ar, andÂ ar2.

Then from the question, we have

aÂ +Â ar +Â ar2Â = 56

aÂ (1 +Â r +Â r2) = 56

Â â€¦ (1)

Also, given

aÂ â€“ 1,Â arÂ â€“ 7,Â ar2Â â€“ 21 forms an A.P.

So, (arÂ â€“ 7) â€“ (aÂ â€“ 1) = (ar2Â â€“ 21) â€“ (arÂ â€“ 7)

arÂ â€“Â aÂ â€“ 6 =Â ar2Â â€“Â arÂ â€“ 14

ar2Â â€“ 2arÂ +Â aÂ = 8

ar2Â â€“Â arÂ â€“Â arÂ +Â aÂ = 8

a(r2Â + 1 â€“ 2r) = 8

aÂ (rÂ â€“ 1)2Â = 8 â€¦ (2)

7(r2Â â€“ 2rÂ + 1) = 1 +Â rÂ +Â r2

7r2Â â€“ 14Â rÂ + 7 â€“ 1 â€“Â rÂ â€“Â r2Â = 0

6r2Â â€“ 15rÂ + 6 = 0

6r2Â â€“ 12rÂ â€“ 3rÂ + 6 = 0

6rÂ (rÂ â€“ 2) â€“ 3 (rÂ â€“ 2) = 0

(6rÂ â€“ 3) (rÂ â€“ 2) = 0

r = 2, 1/2

WhenÂ rÂ = 2,Â aÂ = 8

When r = Â½, a = 32

Thus,

WhenÂ rÂ = 2, the three numbers in G.P. are 8, 16, and 32.

WhenÂ r = 1/2, the three numbers in G.P. are 32, 16, and 8.

Therefore in either case, the required three numbers are 8, 16, and 32.

11. 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.

Solution:

Letâ€™s consider the terms in the G.P.to be T1, T2, T3, T4, â€¦ T2n.

The number of terms = 2n

Then, from the question we have

T1Â + T2Â + T3Â + â€¦+ T2nÂ = 5 [T1Â + T3Â + â€¦ +T2nâ€“1]

T1Â + T2Â + T3Â + â€¦ + T2nÂ â€“ 5 [T1Â + T3Â + â€¦ + T2nâ€“1] = 0

T2Â + T4Â + â€¦ + T2nÂ = 4 [T1Â + T3Â + â€¦ + T2nâ€“1] â€¦â€¦ (1)

Now, let the terms in G.P. beÂ a,Â ar,Â ar2,Â ar3, â€¦

Then (1) becomes,

[Using sum of terms in G.P.]

ar = 4a

r = 4

Thus, the common ratio of the G.P. is 4.

12. 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.

Solution:

Letâ€™s consider the terms in A.P. to beÂ a,Â aÂ +Â d,Â aÂ + 2d,Â aÂ + 3d, â€¦Â aÂ + (nÂ â€“ 2)Â d,Â aÂ + (nÂ â€“ 1)d.

From the question, we have

Sum of first four terms =Â aÂ + (aÂ +Â d) + (aÂ + 2d) + (aÂ + 3d) = 4aÂ + 6d

Sum of last four terms = [aÂ + (nÂ â€“ 4)Â d] + [aÂ + (nÂ â€“ 3)Â d] + [aÂ + (nÂ â€“ 2)Â d] + [aÂ +Â nÂ â€“ 1)Â d]

= 4aÂ + (4nÂ â€“ 10)Â d

Then according to the given condition,

4aÂ + 6dÂ = 56

4(11) + 6dÂ = 56 [SinceÂ aÂ = 11 (given)]

6dÂ = 12

dÂ = 2

Hence, 4aÂ + (4nÂ â€“10)Â dÂ = 112

4(11) + (4nÂ â€“ 10)2 = 112

(4nÂ â€“ 10)2 = 68

4nÂ â€“ 10 = 34

4nÂ = 44

nÂ = 11

Therefore, the number of terms of the A.P. is 11.

13. If, then show thatÂ a,Â b,Â cÂ andÂ dÂ are in G.P.

Solution:

Given,

On cross multiplying, we have

Also, given

On cross multiplying, we have

From (1) and (2), we get

b/a = c/b = d/c

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

14. Let S be the sum, P the product and R the sum of reciprocals ofÂ nÂ terms in a G.P. Prove that P2RnÂ = Sn

Solution:

Let the terms in G.P. beÂ a,Â ar,Â ar2,Â ar3, â€¦Â arnÂ â€“ 1â€¦

Form the question, we have

Hence, P2Â RnÂ = Sn

15. TheÂ pth,Â qthÂ andÂ rthÂ terms of an A.P. areÂ a, b, cÂ respectively.

Show that (q â€“ r) a + (r â€“ p) b + (p -q) c = 0

Solution:

Letâ€™s assumeÂ tÂ andÂ dÂ to be the first term and the common difference of the A.P. respectively.

Then theÂ nthÂ term of the A.P. is given by,Â anÂ =Â tÂ + (nÂ â€“Â 1)Â d

Thus,

apÂ =Â tÂ + (pÂ â€“Â 1)Â dÂ =Â aÂ  â€¦ (1)

aqÂ =Â tÂ + (qÂ â€“Â 1) dÂ =Â bÂ  â€¦ (2)

arÂ =Â tÂ + (rÂ â€“Â 1)Â dÂ =Â cÂ  â€¦ (3)

On subtracting equation (2) from (1), we get

(pÂ â€“ 1 â€“Â qÂ + 1)Â dÂ =Â aÂ â€“Â b

(pÂ â€“Â q)Â dÂ =Â aÂ â€“Â b

On subtracting equation (3) from (2), we get

(qÂ â€“ 1 â€“Â rÂ + 1)Â dÂ =Â bÂ â€“Â c

(qÂ â€“Â r)Â dÂ =Â bÂ â€“Â c

Equating both the values ofÂ dÂ obtained in (4) and (5), we get

Therefore, the given result is proved.

16. IfÂ aare in A.P., prove thatÂ a, b, cÂ are in A.P.

Solution:

17. IfÂ a,Â b,Â c,Â dÂ are in G.P, prove thatÂ (anÂ +Â bn), (bnÂ +Â cn), (cnÂ +Â dn)Â are in G.P.

Solution:

Given,Â a,Â b,Â c,andÂ dÂ are in G.P.

So, we have

âˆ´b2Â =Â acÂ â€¦ (i)

c2Â =Â bdÂ â€¦ (ii)

Required to prove (anÂ +Â bn), (bnÂ +Â cn), (cnÂ +Â dn) are in G.P. i.e.,

(bnÂ +Â cn)2Â = (anÂ +Â bn) (cnÂ +Â dn)

Taking L.H.S.

(bnÂ +Â cn)2Â =Â b2nÂ + 2bncnÂ +Â c2n

= (b2)n+ 2bncnÂ + (c2)Â n

= (ac)nÂ + 2bncnÂ + (bd)nÂ  [Using (i) and (ii)]

=Â anÂ cnÂ +Â bncn+Â bnÂ cnÂ +Â bnÂ dn

=Â anÂ cnÂ +Â bncn+Â anÂ dnÂ +Â bnÂ dnÂ  [Using (iii)]

=Â cnÂ (anÂ +Â bn) +Â dnÂ (anÂ +Â bn)

= (anÂ +Â bn) (cnÂ +Â dn)

= R.H.S.

Therefore, (anÂ +Â bn), (bnÂ +Â cn), and (cnÂ +Â dn) are in G.P

â€“ Hence proved.

18. IfÂ aÂ andÂ bÂ are the roots ofÂ x2 â€“ 3x + p = 0 and c, dare roots ofÂ x2 â€“ 12x + q = 0, whereÂ a,Â b,Â c,Â d, form a G.P. Prove that (qÂ +Â p): (qÂ â€“Â p) = 17:15.

Solution:

Given,Â aÂ andÂ bÂ are the roots ofÂ x2Â â€“ 3xÂ +Â pÂ = 0

So, we have aÂ +Â bÂ = 3 andÂ abÂ =Â pÂ  â€¦ (i)

Also,Â cÂ andÂ dÂ are the roots ofÂ x2 â€“ 12x + q = 0

So, cÂ +Â dÂ = 12 andÂ cdÂ =Â qÂ  â€¦ (ii)

And given a,Â b,Â c,Â dÂ are in G.P.

Letâ€™s takeÂ aÂ =Â x,Â bÂ =Â xr,Â cÂ =Â xr2,Â dÂ =Â xr3

From (i) and (ii), we get

xÂ +Â xrÂ = 3

xÂ (1 +Â r) = 3

And,

xr2Â +Â xr3Â =12

xr2Â (1 +Â r) = 12

On dividing, we get

When r = 2, x = 3/(1 + 2) = 3/3 = 1

When r = -2, x = 3/(1 â€“ 2) = 3/-1 = -3

Case I:

WhenÂ rÂ = 2 andÂ xÂ =1,

abÂ =Â x2rÂ = 2

cdÂ =Â x2r5Â = 32

Case II:

WhenÂ rÂ = â€“2,Â xÂ = â€“3,

abÂ =Â x2r = â€“18

cdÂ =Â x2r5Â = â€“ 288

Therefore, in both the cases, we get (qÂ +Â p): (qÂ â€“Â p) = 17:15\

19. The ratio of the A.M and G.M. of two positive numbersÂ aÂ andÂ b, isÂ m:Â n. Show thatÂ .

Solution:

Let the two numbers beÂ aÂ andÂ b.

A.MÂ = (a + b)/ 2Â and G.M. =Â âˆšab

From the question, we have

20. IfÂ a, b, cÂ are in A.P,;Â b, c, dÂ are in G.P andÂ 1/c, 1/d, 1/e are in A.P. prove thatÂ a,Â c,Â eÂ are in G.P.

Solution:

Given a,Â b,Â cÂ are in A.P.

Hence, bÂ â€“Â aÂ =Â cÂ â€“Â bÂ  â€¦ (1)

And, given thatÂ b,Â c,Â d are in G.P.

So, c2Â =Â bdÂ  â€¦ (2)

Also,Â 1/c, 1/d, 1/e are in A.P.

So,

Now, required to prove thatÂ a,Â c,Â eÂ are in G.P. i.e.,Â c2Â =Â ae

From (1), we have

2b = a + c

b = (a + c)/ 2

And from (2), we have

d = c2/ b

On substituting these values in (3), we get

Therefore,Â a,Â c, andÂ eÂ are in G.P.

21. Find the sum of the following series up toÂ nÂ terms:

(i) 5 + 55 + 555 + â€¦ (ii) .6 + .66 + . 666 + â€¦

Solution:

(i) Given, 5 + 55 + 555 + â€¦

Let SnÂ = 5 + 55 + 555 + â€¦..Â up to nÂ terms

(ii) Given, .6 + .66 + . 666 + â€¦

Let SnÂ = 06. + 0.66 + 0.666 + â€¦ up toÂ nÂ terms

22. Find the 20thÂ term of the series 2 Ã— 4 + 4 Ã— 6 + 6 Ã— 8 + â€¦ +Â nÂ terms.

Solution:

Given series is 2 Ã— 4 + 4 Ã— 6 + 6 Ã— 8 + â€¦Â nÂ terms

âˆ´Â nthÂ term =Â anÂ = 2nÂ Ã— (2nÂ + 2) = 4n2Â + 4n

The 20th term,

a20Â = 4 (20)2Â + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Therefore, the 20thÂ term of the series is 1680.

23. Find the sum of the firstÂ nÂ terms of the series: 3 + 7 + 13 + 21 + 31 + â€¦

Solution:

The given series is 3 + 7 + 13 + 21 + 31 + â€¦

S = 3 + 7 + 13 + 21 + 31 + â€¦+Â anâ€“1Â +Â an

S = 3 + 7 + 13 + 21 + â€¦. +Â anÂ â€“ 2Â +Â anÂ â€“ 1Â +Â an

On subtracting both the equations, we get

S â€“ S = [3 + (7 + 13 + 21 + 31 + â€¦+Â anâ€“1Â +Â an)] â€“ [(3 + 7 + 13 + 21 + 31 + â€¦+Â anâ€“1)Â +Â an]

S â€“ S = 3 + [(7 â€“ 3) + (13 â€“ 7) + (21 â€“ 13) + â€¦ + (anÂ â€“Â anâ€“1)]Â â€“Â an

0 = 3 + [4 + 6 + 8 + â€¦ (nÂ â€“1) terms] â€“Â an

anÂ = 3 + [4 + 6 + 8 + â€¦ (nÂ â€“1) terms]

24. If S1, S2, S3Â are the sum of firstÂ nÂ natural numbers, their squares and their cubes, respectively, show thatÂ 9S22 = S3 (1 + 8S1).

Solution:

From the question, we have

Therefore, from (1) and (2), we have 9S22 = S3 (1 + 8S1).

25. Find the sum of the following series up toÂ nÂ terms:

Solution:

26. Show thatÂ

Solution:

nthÂ term of the numerator =Â n(nÂ + 1)2Â =Â n3Â + 2n2Â +Â n

nthÂ term of the denominator =Â n2(nÂ + 1) =Â n3Â +Â n2

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

Solution:

Given, the farmer pays Rs 6000 in cash.

So, the unpaid amount = Rs 12000 â€“ Rs 6000 = Rs 6000

From the question, the interest paid annually will be

12% of 6000, 12% of 5500, 12% of 5000, â€¦, 12% of 500

Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + â€¦ + 12% of 500

= 12% of (6000 + 5500 + 5000 + â€¦ + 500)

= 12% of (500 + 1000 + 1500 + â€¦ + 6000)

Itâ€™s seen that, the series 500, 1000, 1500 â€¦ 6000 is an A.P. with the first term and common difference both equal to 500.

Letâ€™s take the number of terms of the A.P. to beÂ n.

So, 6000 = 500 + (nÂ â€“ 1) 500

1 + (nÂ â€“ 1) = 12

nÂ = 12

Now,

The sum of the A.P = 12/2 [2(500) + (12 â€“ 1)(500)] = 6 [1000 + 5500] = 6(6500) = 39000

Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 + â€¦ + 6000)

= 12% of 39000 = Rs 4680

Therefore, the tractor will cost the farmer = (Rs 12000 + Rs 4680) = Rs 16680

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

Solution:

Given, Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

So, the unpaid amount = Rs 22000 â€“ Rs 4000 = Rs 18000

Form the question, itâ€™s understood that the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 â€¦ 10% of 1000

Hence, the total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + â€¦ + 10% of 1000

= 10% of (18000 + 17000 + 16000 + â€¦ + 1000)

= 10% of (1000 + 2000 + 3000 + â€¦ + 18000)

Itâ€™s seen that, 1000, 2000, 3000 â€¦ 18000 forms an A.P. with first term and common difference both equal to 1000.

Letâ€™s take the number of terms to beÂ n.

So, 18000 = 1000 + (nÂ â€“ 1) (1000)

nÂ = 18

Now, the sum of the A.P is given by:

Thus,

Total interest paid = 10% of (18000 + 17000 + 16000 + â€¦ + 1000)

= 10% of Rs 171000 = Rs 17100

Therefore, the cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

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

Solution:

Itâ€™s seen that,

The numbers of letters mailed forms a G.P.: 4, 42, â€¦ 48

Here, first term = 4 and common ratio = 4

And the number of terms = 8

The sum ofÂ nÂ terms of a G.P. is given by:

Also, given that the cost to mail one letter is 50 paisa.

Hence, Cost of mailing 87380 letters = Rs 87380 x (50/100) = Rs 43690Â = Rs 43690

Therefore, the amount spent when 8thÂ set of letter is mailed will be Rs 43690.

30. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15thÂ year since he deposited the amount and also calculate the total amount after 20 years.

Solution:

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

Hence, the interest in first year = (5/100) x Rs 10000 = Rs 500

So, Amount in 15thÂ year = Rs

= Rs 10000 + 14 Ã— Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

And, the amount after 20 years =

= Rs 10000 + 20 Ã— Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Therefore, the amount in the 15th year is Rs 17000 and the total amount after 20 years will be Rs 20000.

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

Solution:

Given, the cost of machine = Rs 15625

Also, given that the machine depreciates by 20% every year.

Hence, its value after every year is 80% of the original cost i.e.,Â 4/5th of the original cost.

Therefore, the value at the end of 5 years =

= 5 Ã— 1024 = 5120

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

32. 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.

Solution:

Letâ€™s assumeÂ xÂ to be the number of days in which 150 workers finish the work.

Then from the question, we have

150xÂ = 150 + 146 + 142 + â€¦. (xÂ + 8) terms

The series 150 + 146 + 142 + â€¦. (xÂ + 8) terms is an A.P.

With first term (a) = 150, common difference (d) = â€“4 and number of terms (n) = (xÂ + 8)

Now, finding the sum of terms:

As xÂ cannot be negative. [Number of days is always a positive quantity]

xÂ = 17

Hence, the number of days in which the work should have been completed is 17.

But, due to the dropping out of workers the number of days in which the work is completed

= (17 + 8) = 25

 Also Access NCERT Exemplar for Class 11 Maths Chapter 9 CBSE Notes for Class 11 Maths Chapter 9

### NCERT Solutions for Class 11 Maths Chapter 9 â€“ Sequences and Series

Enlisted beneath are the important concepts of Maths included in the Chapter 9 Sequences and Series:

9.1 Introduction

9.2 Sequences

9.3 Series

9.4 Arithmetic Progression (A.P.)

9.4.1 Arithmetic mean

9.5 Geometric Progression (G. P.)

9.5.1 General term of a G.P

9.5.2. Sum to n terms of a G.P

9.5.3 Geometric Mean (G.M.)

9.6 Relationship Between A.M. and G.M.

9.7 Sum to n Terms of Special Series

## NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series

The chapter Sequences and Series belongs to the unit Algebra under the first term Class 11 Maths CBSE Syllabus 2022-23, which adds up to 30 marks of the total 80 marks. There are 4 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts related to Sequences and Series clearly. Some of the topics discussed in Chapter 9 of NCERT Solutions for Class 11 Maths are as follows:

1. By a sequence, we mean an arrangement of numbers in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3, â€¦.k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence.
2. Let a1, a2, a3, â€¦ be the sequence, then the sum expressed as a1 + a2 + a3 + â€¦ is called series. A series is called finite series if it has a finite number of terms.
3. An arithmetic progression (A.P.) is a sequence in which terms increase or decrease regularly by the same constant. This constant is called the common difference of the A.P. Usually, we denote the first term of A.P. by a, the common difference by d and the last term by l. The general term or the nth term of the A.P. is given by an = a + (n â€“ 1) d.
4. A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nth term of G.P. is given by an = arn â€“ 1

A student who has mastered the chapter Sequences and Series of Class 11 would also have a strong hold on the concepts related to the chapter, namely Sequence and Series, Arithmetic Progression (A. P.), Arithmetic Mean (A.M.), Geometric Progression (G.P.), general term of a G.P., sum of n terms of a G.P., infinite G.P. and its sum, geometric mean (G.M.) and the relation between A.M. and G.M.

Disclaimer â€“Â

Dropped Topics â€“Â

9.4 Arithmetic Progression (A.P.) (up to Exercise 9.2)
9.7 Sum to n terms of Special Series
Examples 21, 22 and 24
Ques. 1â€“6, 12, 15, 16, 20, 23â€“26 (Miscellaneous Exercise)
Point 3 and 4 in the Summary

## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 9

### What is the concept of Arithmetic Progression discussed in Chapter 9 of NCERT Solutions for Class 11 Maths?

Progression is a sequence in which the terms maintain a specific pattern. In Arithmetic Progression, there are two consecutive terms and their difference is constant. To understand these concepts in a better way, students should refer to the NCERT Solutions designed by the expert faculty at BYJUâ€™S. Even though the textbook has enough details, opting for the best study material will help students in understanding their applications.

### How many sections are present in Chapter 9 of NCERT Solutions for Class 11 Maths?

The sections present in Chapter 9 of NCERT Solutions for Class 11 Maths are â€“
1. Introduction
2. Sequences
3. Series
4. Arithmetic Progression
5. Geometric Progression
6. Relationship between AM and GM
7. Sum to n terms of Special Series

### How to secure full marks in Chapter 9 of the NCERT Solutions for Class 11 Maths?

The Chapter 9 Sequence and Series requires a lot of practice to reduce the conceptual errors as it contains lots of difficult topics. The fundamental concepts might be a little tricky in the starting but with the right guidance securing a good score is not that difficult. If your aim is to score good marks in this chapter, you should solve different types of tricky problems as well. The NCERT Solutions has fast solving tips that keep students aware of the type of problems that would appear in the exams.