** According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.*

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 are prepared by highly experienced teachers at BYJU’S according to the latest CBSE Syllabus 2023-24 to help the students in attempting challenging questions with utmost confidence. Using the NCERT Solutions provided here, students can learn new methods of solving a particular problem in an expeditious time to improve their performance in the Class 11 Maths exam. These solutions have been designed after undertaking extensive research on each question and their problem-solving methods.

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

### Access exercise-wise NCERT Solutions for Class 11 Maths Chapter 9 from the below links

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 – 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. a _{n} = n (n + 2)Â **

**Solution:**

Given,

n^{th} term of a sequence a_{n} = n (n + 2)**Â **

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

a_{1} = 1(1 + 2) = 3

a_{2} = 2(2 + 2) = 8

a_{3} = 3(3 + 2) = 15

a_{4} = 4(4 + 2) = 24

a_{5} = 5(5 + 2) = 35

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

**2.Â a _{n} = n/n+1**

**Solution:**

Given the n^{th} term, a_{n} = n/n+1

On substitutingÂ *n*Â = 1, 2, 3, 4, 5, we get

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

**3. a_{n}Â = 2^{n}**

**Solution: **

Given the n^{th} term, *a _{n}*Â = 2

^{n}On substituting*Â n*Â = 1, 2, 3, 4, 5, we get

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 32

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

4. Â *a _{n}*Â = (2n – 3)/6

**Solution:**

Given the n^{th} term, *a _{n}*Â = (2n – 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. a _{n} = (-1)^{n-1} 5^{n+1}**

**Solution:**

Given the n^{th} term, a_{n} = (-1)^{n-1} 5^{n+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 *n* = 1, 2, 3, 4, 5, we get the 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 n ^{th} terms are:**

**7. a _{n} = 4n â€“ 3; a_{17}, a_{24}**

**Solution:**

Given,

The* n*^{th}Â term of the sequence is a_{n} = 4n â€“ 3

On substitutingÂ *n*Â = 17, we get

a_{17} = 4(17) â€“ 3 = 68 â€“ 3 = 65

Next, on substitutingÂ *n*Â = 24, we get

a_{24} = 4(24) â€“ 3 = 96 â€“ 3 = 93

**8. a _{n}Â =Â n^{2}/2^{n}Â ;Â a^{7}**

**Solution:**

Given,

The* n*^{th}Â term of the sequence is a_{n} = n^{2}/2^{n}

Now, on substitutingÂ *n*Â = 7, we get

a_{7} = 7^{2}/2^{7} = 49/ 128

**9. a _{n} = (-1)^{n-1} n^{3}; a_{9}**

**Solution: **

Given,

The* n*^{th}Â term of the sequence is a_{n} = (-1)^{n-1} n^{3}

On substitutingÂ *n*Â = 9, we get

a_{9} = (-1)^{9-1} (9)^{3} = 1 x 729 = 729

**10. **

**Solution:**

On substitutingÂ *n*Â = 20, we get

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

**11. a _{1} = 3, a_{n} = 3a_{n-1} + 2 for all n > 1**

**Solution: **

Given, a_{n} = 3a_{n-1} + 2 and a_{1} = 3

Then,

a_{2} = 3a_{1} + 2 = 3(3) + 2 = 11

a_{3} = 3a_{2} + 2 = 3(11) + 2 = 35

a_{4} = 3a_{3} + 2 = 3(35) + 2 = 107

a_{5} = 3a_{4} + 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. a _{1} = -1, a_{n} = a_{n-1}/n, n â‰¥ 2**

**Solution:**

Given,

a_{n} = a_{n-1}/n and a_{1} = -1

Then,

a_{2} = a_{1}/2 = -1/2

a_{3} = a_{2}/3 = -1/6

a_{4} = a_{3}/4 = -1/24

a_{5} = a_{4}/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. a _{1} = a_{2 }= 2, a_{n} = a_{n-1} â€“ 1, n > 2**

**Solution: **

Given,

a_{1} = a_{2}, a_{n} = a_{n-1} â€“ 1

Then,

a_{3} = a_{2} â€“ 1 = 2 â€“ 1 = 1

a_{4} = a_{3} â€“ 1 = 1 â€“ 1 = 0

a_{5} = a_{4} 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 = a _{1} = a_{2} and a_{n} = a_{n – 1 }+ a_{n – 2}, n > 2**

**Find a _{n+1}/a_{n}, for n = 1, 2, 3, 4, 5Â **

**Solution: **

Given,

1 = a_{1} = a_{2 }

a_{n} = a_{n – 1 }+ a_{n – 2}, n > 2

So,

a_{3} = a_{2} + a_{1} = 1 + 1 = 2

a_{4} = a_{3} + a_{2} = 2 + 1 = 3

a_{5} = a_{4} + a_{3} = 3 + 2 = 5

a_{6} = a_{5} + a_{4} = 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, *a*Â = 1

The common difference, *d*Â = 2

Now,

a + (n -1)d = 2001

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

2n â€“ 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_{n} = 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, *a*Â = 105

The common difference, *d*Â = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n â€“ 5 = 995

5n = 995 â€“ 105 + 5 = 895

n = 895/5

n = 179

We know,

S_{n} = 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 the 20 ^{th}Â term is â€“112.**

**Solution: **

Given,

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

Letâ€™s assumeÂ *d* is the common difference of the A.P.

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

Then,

Sum of first five terms = 10 + 10*d*

Sum of next five terms = 10 + 35*d*

From the question, we have

10 + 10d = Â¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a_{20} = a + (20 – 1)d = 2 + (19) (-6) = 2 â€“ 114 = -112

Therefore, the 20^{th}Â 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 *n*Â terms of the given A.P. as â€“25.

We known that,

S_{n} = n/2 [2a + (n-1)d]

whereÂ *n*Â = number of terms,Â *a*Â = first term, andÂ *d*Â = common difference

So here,Â *a*Â = â€“6

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

Thus, we have

**5. In an A.P., ifÂ p^{th}Â term isÂ 1/q andÂ q^{th}Â 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, the sum of a certain number of terms of the A.P. is 116.

The number of terms is n.

So, we have

S_{n} = 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€“ 3n^{2}

3n^{2} â€“ 53n + 232 = 0

3n^{2} â€“ 24n â€“ 29n+ 232 = 0

3n(n – 8) â€“ 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

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

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

a_{8} = 25 + (8 – 1)(-3)

= 25 â€“ 21

= 4

**7. Find the sum toÂ nÂ terms of the A.P., whoseÂ k^{thÂ }term is 5kÂ + 1.**

**Solution:**

Given, theÂ *k*^{th}Â term of the A.P. is 5*k*Â + 1.

*k*^{th}Â term =Â *a _{k}*Â =Â

*aÂ*+ (

*k*Â â€“ 1)

*d*

And,

*aÂ *+ (*k*Â â€“ 1)*d*Â = 5*k*Â + 1

*a*Â +Â *kd*Â â€“Â *d*Â = 5*k*Â + 1

On comparing the coefficient ofÂ *k*, we getÂ *d*Â = 5

*aÂ *â€“Â *dÂ *= 1

*a*Â â€“ 5 = 1

â‡’Â *a*Â = 6

**8. If the sum ofÂ nÂ terms of an A.P. is (pnÂ +Â qn^{2}), whereÂ pÂ andÂ qÂ are constants, find the common difference.**

**Solution:**

We know that,

S_{n} = n/2 [2a + (n-1)d]

From the question, we have

On comparing the coefficients ofÂ *n*^{2}Â on both sides, we get

d/2 = q

Hence, *d*Â = 2*q*

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

**9. The sums ofÂ nÂ terms of two arithmetic progressions are in the ratio 5nÂ + 4: 9nÂ + 6. Find the ratio of their 18^{th}Â terms.**

**Solution:**

LetÂ *a*_{1},Â *a*_{2}, andÂ *d*_{1},Â *d*_{2Â }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 the first 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 *a*Â andÂ *d* to be the first term and the common difference of the A.P., respectively.

Then, it is 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 *a*_{1}Â andÂ *d* 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 m^{2}:Â n^{2}. Show that the ratio of the m^{thÂ }and the n^{th}Â term is (2mÂ â€“ 1): (2nÂ â€“ 1).**

**Solution: **

Letâ€™s consider thatÂ *a*Â andÂ *b* are 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 3n^{2} + 5n and itsÂ m^{thÂ }term is 164, find the value ofÂ m.**

**Solution:**

Letâ€™s considerÂ *a*Â andÂ *b* to be the first term and the common difference of the A.P., respectively.

*a _{m}*Â =Â

*a*Â + (

*m*Â â€“ 1)

*d*Â = 164 â€¦ (1)

The sum of the terms is given by,

S_{n} = 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 A_{1}, A_{2}, A_{3}, A_{4}, and A_{5}Â to be five numbers between 8 and 26 such that 8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26 are in an A.P.

Here, we have,

*aÂ *= 8,Â *bÂ *= 26,Â *n*Â = 7

So,

26 = 8 + (7 â€“ 1)Â *d*

6*d*Â = 26 â€“ 8 = 18

*dÂ *= 3

Now,

A_{1}Â =Â *a*Â +Â *d*Â = 8 + 3 = 11

A_{2}Â =Â *a*Â + 2*d*Â = 8 + 2 Ã— 3 = 8 + 6 = 14

A_{3}Â =Â *a*Â + 3*d*Â = 8 + 3 Ã— 3 = 8 + 9 = 17

A_{4}Â =Â *a*Â + 4*dÂ *= 8 + 4 Ã— 3 = 8 + 12 = 20

A_{5}Â =Â *a*Â + 5*d*Â = 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 7^{th}Â and (mÂ â€“ 1)^{th}Â numbers is 5: 9. Find the value ofÂ m.**

**Solution: **

Letâ€™s consider a_{1}, a_{2}, â€¦ a* _{m}*Â beÂ

*m*Â numbers such that 1, a

_{1}, a

_{2}, â€¦ a

*, 31 is an A.P.*

_{m}And here,

*a*Â = 1,Â *b*Â = 31,Â *n*Â =Â *m*Â + 2

So, 31 = 1 + (*m*Â + 2 â€“ 1) (*d*)

30 = (*m*Â + 1)Â *d*

d = 30/ (m + 1) â€¦â€¦. (1)

Now,

a_{1}Â =Â *a*Â +Â *d*

a_{2}Â =Â *a*Â + 2*d*

a_{3}Â =Â *a*Â + 3*d*Â â€¦

Hence, a_{7}Â =Â *a*Â + 7*d*

a_{m}_{â€“1}Â =Â *a*Â + (*m*Â 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 the first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount will he pay in the 30 ^{th}Â 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 then, A.P. is 100, 105, 110, â€¦

Where the first term, *a*Â = 100

Common difference,Â *d*Â = 5

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

A_{30}Â =Â *a*Â + (30 â€“ 1)*d*

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30^{th}Â 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 a common difference *d*Â = 5Â° and first termÂ *a* = 120Â°.

And we know that the sum of all angles of a polygon with *n*Â sides is 180Â° (*n*Â 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 20 ^{th}Â andÂ n^{th }terms 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 *a*Â = First term =Â 5/2

*r*Â = Common ratio =Â (5/4)/(5/2) = Â½

Thus, the 20^{th} term and n^{th} term

**2. Find the 12 ^{th}Â term of a G.P. whose 8^{th} term is 192, and the common ratio is 2.**

**Solution:**

Given,

The common ratio of the G.P.,Â *r*Â = 2

And, letÂ *a*Â be the first term of the G.P.

Now,

*a*_{8}Â =Â *ar*Â ^{8â€“1}Â =Â *ar*^{7}

*ar*^{7}Â = 192

*a*(2)^{7}Â = 192

*a*(2)^{7}Â = (2)^{6}Â (3)

**3. The 5 ^{th}, 8^{th}Â and 11^{th}Â terms of a G.P. areÂ p,Â qÂ andÂ s, respectively. Show thatÂ q^{2}Â =Â ps.**

**Solution:**

Letâ€™s takeÂ *a*Â to be the first term and*Â r*Â to be the common ratio of the G.P.

Then, according to the question, we have

*a*_{5}Â =Â *a*Â *r*^{5â€“1Â }=Â *a*Â r^{4}Â =Â *p*Â â€¦ (i)

*a*_{8Â }=Â *a*Â *r*^{8â€“1Â }=Â *a*Â *r*^{7}Â =Â *q*Â â€¦ (ii)

*a*_{11}Â = aÂ *r*^{11â€“1Â }=Â *a*Â *r*^{10}Â =Â *sÂ *â€¦ (iii)

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

**4. The 4 ^{th} term of a G.P. is the square of its second term, and the first term is â€“3. Determine its 7^{th}Â term.**

**Solution:**

Letâ€™s considerÂ *a*Â to be the first term andÂ *r*Â to be the common ratio of the G.P.

Given, *a*Â = â€“3

And we know that,

*a _{n}*Â =Â

*ar*

^{n}^{â€“1}

So, *a*_{4Â }=Â *ar*^{3}Â = (â€“3)Â *r*^{3}

*a*_{2}Â =Â *a r*^{1}Â = (â€“3)Â *r*

Then, from the question, we have

(â€“3)Â *r*^{3}Â = [(â€“3)Â *r*]^{2}

â‡’ â€“3*r*^{3}Â = 9Â *r*^{2}

â‡’Â *r*Â = â€“3

*a*_{7}Â =Â *a*Â *r*Â ^{7â€“1Â }=Â *a*Â *r*^{6}Â = (â€“3) (â€“3)^{6}Â = â€“ (3)^{7}Â = 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 n^{th} term of this sequence as 128, we have

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

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

We have,

a = âˆš3 and r = 3/âˆš3 = âˆš3

Taking the n^{th} term of this sequence to be 729, we have

Therefore, the 12^{th} 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 n^{th} term of this sequence to be 1/19683, we have

Therefore, the 9^{th} 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, for*Â x*Â = 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 *a*Â = 0.15 and r = 0.015/0.15 = 0.1**Â **

**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, a^{2}, -a^{3} â€¦. (if a â‰ -1)**

**Solution:**

The given G.P. isÂ 1, -a, a^{2}, -a^{3} â€¦.

Here, the first term =Â *a*_{1}Â = 1

And the common ratio =Â *r*Â = â€“Â *a*

We know that,

**10. Find the sum toÂ nÂ terms in the geometric progression x^{3}, x^{5}, x^{7}, â€¦ (if x â‰ Â±1 )**

**Solution:**

Given G.P. isÂ x^{3}, x^{5}, x^{7}, â€¦

Here, we haveÂ *a*Â =Â *x*^{3}Â andÂ *r*Â =Â *x*^{5}/x^{3} = x^{2}

**11. Evaluate: **

**Solution: **

**12. The sum of the first three terms of a G.P. is 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

a^{3} = 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 + r^{2})/r = 39/10

10 + 10r + 10r^{2} = 39r

10r^{2} â€“ 29r + 10 = 0

10r^{2} â€“ 25r â€“ 4r + 10 = 0

5r(2r – 5) â€“ 2(2r â€“ 5) = 0

(5r – 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, 3 ^{2}, 3^{3}, â€¦ are needed to give the sum 120?**

**Solution: **

Given G.P. is 3, 3^{2}, 3^{3}, â€¦

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

We know that,

Here,Â *a*Â = 3 andÂ *r*Â = 3

Equating the exponents, we get *n*Â = 4

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

**14. The sum of the first three terms of a G.P. is 16, and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to Â nÂ terms of the G.P.**

**Solution: **

Letâ€™s assume the G.P. to beÂ *a*,Â *ar*,Â *ar*^{2},Â *ar*^{3}, â€¦

Then, according to the question, we have

*aÂ *+Â *ar*Â +Â *ar*^{2}Â = 16 andÂ *ar*^{3Â }+Â *ar*^{4}Â +Â *ar*^{5Â }= 128

*a*Â (1 +Â *r*Â +Â *r*^{2}) = 16 â€¦ (1) and,

*ar*^{3}(1 +Â *r*Â +Â *r*^{2}) = 128 â€¦ (2)

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

r^{3} = 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 7^{th}Â term 64, determine S_{7}.**

**Solution: **

Given,

*a*Â = 729 and *a*_{7}Â = 64

LetÂ *r*Â be the common ratio of the G.P.

Then, we know that, *a _{n}*Â =Â

*a r*

^{n}^{â€“1}

*a*_{7}Â =Â *ar*^{7â€“1}Â = (729)*r*^{6}

â‡’ 64 = 729Â *r*^{6}

*r*^{6} = 64/729

*r*^{6} = (2/3)^{6}

r = 2/3

And we know that

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

**Solution: **

ConsiderÂ *a*Â to be the first term andÂ *r*Â to be the common ratio of the G.P.

Given, S_{2} = -4

Then, from the question, we have

And,

a_{5} = 4 x a_{3}

ar^{4} = 4ar^{2}

r^{2} = 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 4 ^{th}, 10^{th}Â and 16^{th}Â terms of a G.P. areÂ x, yÂ andÂ z, respectively. Prove thatÂ x,Â y, and zÂ are in G.P.**

**Solution:
**

LetÂ *a*Â be the first term andÂ *r*Â be the common ratio of the G.P.

According to the given condition,

*a*_{4}Â =Â *a*Â *r*^{3}Â =Â *x*Â â€¦ (1)

*a*_{10}Â =Â *a*Â *r*^{9}Â =*Â y*Â â€¦ (2)

*a*_{16}^{Â }=*Â a r*^{15Â }=Â *z*Â â€¦ (3)

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

**18. Find the sum to Â 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

*S _{n}*Â = 8 + 88 + 888 + 8888 + â€¦â€¦â€¦â€¦â€¦.. toÂ

*n*Â 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 + 1/2^{2}]

Now, itâ€™s seen that

4, 2, 1,Â Â½, 1/2^{2} is a G.P.

With the first term, *a*Â = 4

Common ratio,Â *r*Â =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, ar ^{2}, â€¦ar^{n-1} and A, AR, AR^{2}, â€¦ AR^{n-1}Â form a G.P, and find the common ratio.**

**Solution:**

To be proved: The sequence,Â *aA*,Â *arAR*,Â *ar*^{2}*AR*^{2}, â€¦*ar ^{n}*

^{â€“1}

*AR*

^{n}^{â€“1}, forms a G.P.

Now, we have

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

**21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4 ^{th}Â by 18.**

**Solution: **

ConsiderÂ *a*Â to be the_{Â }first term andÂ *r*Â to be the common ratio of the G.P.

Then,

*a*_{1}Â =Â *a*,Â *a*_{2}Â =Â *ar*,Â *a*_{3}Â =Â *ar*^{2},Â *a*_{4}Â =Â *ar*^{3}

From the question, we have

*a*_{3}Â =Â *a*_{1}Â + 9

*ar*^{2}Â =Â *a*Â + 9 â€¦ (i)

*a*_{2}Â =Â *a*_{4}Â + 18

*arÂ *=Â *ar*^{3}Â + 18 â€¦ (ii)

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

*a*(*r*^{2}Â â€“ 1) = 9 â€¦ (iii)

*arÂ *(1â€“Â *r*^{2}) = 18 â€¦ (iv)

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

-r = 2

r = -2

On substituting the value ofÂ *r*Â in (i), we get

4*aÂ *=Â *a*Â + 9

3*a*Â = 9

âˆ´Â *a*Â = 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 p ^{th}, q^{th} and r^{th}Â terms of a G.P. areÂ a, bÂ andÂ c, respectively. Prove that a^{q-r }b^{r-p }c^{p-q} = 1**

**Solution: **

Letâ€™s take *A*Â to be the first term andÂ *R*Â to be the common ratio of the G.P.

Then, according to the question, we have

*AR ^{p}*

^{â€“1Â }=Â

*a*

*AR ^{q}*

^{â€“1Â }=Â

*b*

*AR ^{r}*

^{â€“1Â }=Â

*c*

Then,

*a ^{qâ€“r}*

^{Â }

*b*

^{râ€“p}^{Â }

*c*

^{pâ€“q}=Â *A ^{q}*

^{â€“}

*Ã—Â*

^{rÂ }*R*

^{(}

^{p}^{â€“1) (qâ€“r)}Â Ã— A

^{r}^{â€“}

*Â Ã—Â*

^{p}*R*

^{(}

^{q}^{â€“1) (}

^{r}^{–}

^{p}^{)}Â Ã—Â

*A*

^{p}^{â€“}

*Â Ã—Â*

^{q}*R*

^{(}

^{rÂ }^{â€“1)(}

^{p}^{â€“}

^{q}^{)}

=Â *Aq*^{Â â€“Â }^{r}^{Â +Â }^{r}^{Â â€“Â }^{p}^{Â +Â }^{p}^{Â â€“Â }* ^{q}*Â Ã—Â

*R*Â

^{(}

^{pr}^{Â â€“Â }

^{pr}^{Â â€“Â }

^{q}^{Â +Â }

^{r}^{) + (}

^{rq}^{Â â€“}

^{Â rÂ }^{+Â }

^{p}^{Â â€“Â }

^{pq}^{) + (}

^{pr}^{Â â€“Â }

^{p}^{Â â€“Â }

^{qr}^{Â +Â }

^{q}^{)}

=Â *A*^{0}Â Ã—Â *R*^{0}

= 1

Hence proved.

**23. If the first and theÂ n^{th}Â term of a G.P. areÂ a andÂ b, respectively, and ifÂ PÂ is the product ofÂ nÂ terms, prove thatÂ P^{2}Â = (ab)^{n}.**

**Solution:**

Given the first term of the G.P is *a,*Â and the last term isÂ *b*.

Thus,

The G.P. isÂ *a*,Â *ar*,Â *ar*^{2},Â *ar*^{3}, â€¦Â *ar ^{n}*

^{â€“1}, whereÂ

*r*Â is the common ratio.

Then,

*b*Â =Â *ar ^{n}*

^{â€“1}Â â€¦ (1)

*P*Â = Product ofÂ *n*Â terms

= (*a*) (*ar*) (*ar*^{2}) â€¦ (*ar ^{n}*

^{â€“1})

= (*a*Â Ã—Â *a*Â Ã—â€¦*a*) (*r*Â Ã—Â *r*^{2}Â Ã— â€¦*r ^{n}*

^{â€“1})

=Â *a ^{n}*

^{Â }

*r*Â

^{1 + 2 +â€¦(}

^{n}^{â€“1)}Â â€¦ (2)

Here, 1, 2, â€¦(*n*Â 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 the first nÂ terms of a G.P. to the sum of terms fromÂ .**

**Solution:**

LetÂ *a*Â be the first term andÂ *rÂ *be the common ratio of the G.P.

Since there areÂ *n*Â terms from (*n*Â +1)^{th}Â to (2*n*)^{th}Â term,

Sum of terms from(*n*Â + 1)^{th}Â to (2*n*)^{th}Â term

*a*Â ^{nÂ }^{+1}Â =Â *ar*Â ^{n + 1}Â ^{â€“ 1}Â =Â *ar ^{n}*

Thus, the required ratio =

Thus, the ratio of the sum of the first *n*Â terms of a G.P. to the sum of terms from (*n*Â + 1)^{th}Â to (2*n*)^{thÂ }term is^{Â .}

**25. IfÂ a, b, cÂ andÂ d are in G.P., show thatÂ (a^{2} + b^{2} + c^{2})(b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}.**

**Solution: **

Given *a*,Â *b*,Â *c*,Â *d*Â are in G.P.

So, we have

*bc*Â =Â *ad*Â â€¦ (1)

*b*^{2}Â =Â *acÂ *â€¦ (2)

*c*^{2}Â =Â *bd*Â â€¦ (3)

Taking the R.H.S., we have

R.H.S.

= (*ab*Â +Â *bc*Â +Â *cd*)^{2}

= (*ab*Â +Â *adÂ *+Â *cd*)^{2}Â [Using (1)]

= [*ab*Â +Â *d*Â (*a*Â +Â *c*)]^{2}

=Â *a*^{2}*b*^{2}Â + 2*abd*Â (*a*Â +Â *c*) +Â *d*^{2}Â (*a*Â +Â *c*)^{2}

=Â *a*^{2}*b*^{2}Â +2*a*^{2}*bd*Â + 2*acbd*Â +Â *d*^{2}(*a*^{2}Â + 2*ac*Â +Â *c*^{2})

=Â *a*^{2}*b*^{2}Â + 2*a*^{2}*c*^{2}Â + 2*b*^{2}*c*^{2}Â +Â *d*^{2}*a*^{2}Â + 2*d*^{2}*b*^{2}Â +Â *d*^{2}*c*^{2}Â [Using (1) and (2)]

=Â *a*^{2}*b*^{2}Â +Â *a*^{2}*c*^{2}Â +Â *a*^{2}*c*^{2}Â +Â *b*^{2}*c*^{2Â }+Â *b*^{2}*c*^{2}Â +Â *d*^{2}*a*^{2}Â +Â *d*^{2}*b*^{2}Â +Â *d*^{2}*b*^{2}Â +Â *d*^{2}*c*^{2}

=Â *a*^{2}*b*^{2}Â +Â *a*^{2}*c*^{2}Â +Â *a*^{2}*d*^{2Â }+Â *b*^{2Â }Ã—Â *b*^{2}Â +Â *b*^{2}*c*^{2}Â +Â *b*^{2}*d*^{2}Â +Â *c*^{2}*b*^{2}Â +Â *c*^{2Â }Ã—Â *c*^{2}Â +Â *c*^{2}*d*^{2}

=Â *a*^{2}(*b*^{2}Â +Â *c*^{2}Â +Â *d*^{2}) +Â *b*^{2}Â (*b*^{2}Â +Â *c*^{2}Â +Â *d*^{2}) +Â *c*^{2}Â (*b*^{2}+Â *c*^{2}Â +Â *d*^{2})

= (*a*^{2}Â +Â *b*^{2}Â +Â *c*^{2}) (*b*^{2}Â +Â *c*^{2}Â +Â *d*^{2})

= L.H.S.

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

Therefore,

(a^{2} + b^{2} + c^{2})(b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}

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

**Solution:**

Letâ€™s assumeÂ *G*_{1}Â andÂ *G*_{2}Â to be two numbers between 3 and 81 such that the series 3,Â *G*_{1},Â *G*_{2}, 81 forms a G.P.

And letÂ *a*Â be the first term andÂ *r*Â be the common ratio of the G.P.

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

81 = (3)Â *(r*)^{3}

*r*^{3}Â = 27

âˆ´Â *r*Â = 3 (Taking real roots only)

ForÂ *r*Â = 3,

*G*_{1}Â =Â *ar*Â = (3) (3) = 9

*G*_{2}Â =Â *ar*^{2}Â = (3) (3)^{2}Â = 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 of Â nÂ so thatÂ may be the geometric mean betweenÂ aÂ andÂ b.**

**Solution: **

We know that,

The G. M. ofÂ *a*Â andÂ *b*Â 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 to be *a*Â 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 that*Â A*Â andÂ *G*Â are A.M. and G.M. between two positive numbers.

And, let these two positive numbers beÂ *a*Â 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 the 2 ^{nd}Â hour, 4^{th}Â hour andÂ n^{th}Â 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 *a*Â = 30 andÂ *r*Â = 2

So,Â *a*_{3}Â =Â *ar*^{2}Â = (30) (2)^{2}Â = 120

Thus, the number of bacteria at the end of 2^{nd}Â hour will be 120.

And, *a*_{5}Â =Â *ar*^{4}Â = (30) (2)^{4}Â = 480

The number of bacteria at the end of 4^{th}Â hour will be 480.

*a _{n}*

_{Â +1Â }=Â

*ar*Â = (30) 2

^{n}

^{n}Therefore, the number of bacteria at the end ofÂ *n*^{th}Â hour will be 30(2)* ^{n}*.

**31. What will Rs 500 amount to in 10 years after its deposit in a bank which pays an 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 2^{nd}Â year, amount = Rs 500 (1.1) (1.1)

At the end of 3^{rd}Â 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 *a*Â andÂ *b*.

Then, we have

We know that,

A quadratic equation can be formed as,

*x*^{2 }â€“Â *x*Â (Sum of roots) + (Product of roots) = 0

*x*^{2}Â â€“Â *x*Â (*a*Â +Â *b*) + (*ab*) = 0

*x*^{2}Â â€“ 16*x*Â + 25 = 0 [Using (1) and (2)]

Therefore, the required quadratic equation isÂ *x*^{2}Â â€“ 16*x*Â + 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,

*n*^{th}Â term,Â *a _{n}*Â =Â

*n*Â (Â

*n*Â + 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,

*n*^{th}Â term,Â *a _{n}*Â =Â

*n*Â (Â

*n*Â + 1) (Â

*n*Â + 2)

= (*n*^{2}Â +Â *n*) (*n*Â + 2)

=*Â n*^{3Â }+ 3*n*^{2Â }+ 2*n*

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

**3. 3 Ã— 1 ^{2}Â + 5 Ã— 2^{2}Â + 7 Ã— 3^{2}Â + â€¦**

**Solution:**

Given series is 3 Ã—1^{2}Â + 5 Ã— 2^{2}Â + 7 Ã— 3^{2}Â + â€¦

Itâ€™s seen that,

*n*^{th}Â term,Â *a _{n}*Â = ( 2

*n*Â + 1)Â

*n*

^{2}Â = 2

*n*

^{3}Â +Â

*n*

^{2}

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 5^{2}Â + 6^{2}Â + 7^{2}Â + â€¦ + 20^{2}**

**Solution:**

Given series is 5^{2}Â + 6^{2}Â + 7^{2}Â + â€¦ + 20^{2}

Itâ€™s seen that,

*n*^{th}Â term,Â *a _{n}*Â = (Â

*n*Â + 4)

^{2}Â =Â

*n*

^{2}Â + 8

*n*Â + 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,

*a _{nÂ }*= (

*n*

^{th}Â term of 3, 6, 9 â€¦) Ã— (

*n*

^{th}Â term of 8, 11, 14, â€¦)

= (3*n*) (3*n*Â + 5)

= 9*n*^{2}Â + 15*n *

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

**7. Find the sum toÂ nÂ terms of the series 1^{2}Â + (1^{2}Â + 2^{2}) + (1^{2}Â + 2^{2}Â + 3^{2}) + â€¦**

**Solution:**

Given series is 1^{2}Â + (1^{2}Â + 2^{2}) + (1^{2}Â + 2^{2Â }+ 3^{2}Â ) + â€¦

Finding the n^{th} term, we have

*a _{n}*Â = (1

^{2}Â + 2

^{2}Â + 3

^{2}Â +â€¦â€¦.+Â

*n*

^{2})

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

**8. Find the sum toÂ nÂ terms of the series whoseÂ n^{th}Â term is given byÂ nÂ (nÂ + 1) (nÂ + 4).**

**Solution:**

Given,

*a _{n}*Â =Â

*n*Â (

*n*Â + 1) (

*n*Â + 4) =Â

*n*(

*n*

^{2Â }+ 5

*n*Â + 4) =Â

*n*

^{3}Â + 5

*n*

^{2}Â + 4

*n*

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

**9. Find the sum toÂ nÂ terms of the series whoseÂ n^{th} term is given byÂ n^{2}Â + 2^{n}**

**Solution:**

Given,

The n^{th} term of the series as

*a _{n}Â = n*

^{2}Â + 2

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

**10. Find the sum toÂ nÂ terms of the series whoseÂ n^{th} term is given by (2nÂ â€“ 1)^{2}**

**Solution:**

Given,

The n^{th} term of the series as:

*a _{n}*Â =Â (2

*n*Â â€“ 1)

^{2}Â = 4

*n*

^{2}Â â€“ 4

*n*Â + 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)^{th}Â and (mÂ â€“Â n)^{th}Â terms of an A.P. is equal to twice theÂ m^{t}^{h}Â term.**

**Solution:**

Letâ€™s takeÂ *a*Â andÂ *d* to be the first term and the common difference of the A.P., respectively.

We know that the *k*^{th}Â term of an A. P. is given by

*a _{k}*Â =Â

*a*Â + (

*k*Â â€“1)Â

*d*

So, *a _{m}*

_{Â +Â }

*Â =Â*

_{n}*a*Â + (

*m*Â +Â

*n*Â â€“1)Â

*d*

And, *a _{m}*

_{Â â€“Â }

*Â =Â*

_{n}*a*Â + (

*m*Â â€“Â

*n*Â â€“1)Â

*d*

*a _{m}*

_{Â }=Â

*a*Â + (

*m*Â â€“1)Â

*d*

Thus,

*a _{m}*

_{Â +Â }

*Â +Â*

_{n}*a*

_{m}_{Â â€“Â }

*Â =Â*

_{n}*a*Â + (

*m*Â +Â

*n*Â â€“1)Â

*d*Â +Â

*a*Â + (

*m*Â â€“Â

*n*Â â€“1)Â

*d*

= 2*a*Â + (*m*Â +Â *n*Â â€“1 +Â *m*Â â€“Â *n*Â â€“1)Â *d*

= 2*a*Â + (2*m*Â â€“ 2)Â *d*

= 2*a*Â + 2 (*m*Â â€“ 1)Â *d*

*=*2 [*a*Â + (*m*Â â€“ 1)Â *d*]

= 2*a _{m}*

Therefore, the sum of (*m*Â +Â *n*)^{th}Â and (*m*Â â€“Â *n*)^{th}Â terms of an A.P. is equal to twice theÂ *m ^{t}*

^{h}Â 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 *a*Â â€“Â *d*,Â *a*, andÂ *a*Â +Â *d*.

Then, from the question we have

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

3*a*Â = 24

âˆ´Â *a*Â = 8

And,

(*a*Â â€“Â *d*)Â *a*Â (*a*Â +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 *d*^{2}Â = 55

*d*^{2}Â = 64 â€“ 55 = 9

âˆ´ *dÂ *= Â± 3

Thus,

WhenÂ *d*Â = 3, the numbers are 5, 8, and 11 and

WhenÂ *d*Â = 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 S_{1}, S_{2}Â and S_{3},_{Â }respectively, show that S_{3}Â = 3 (S_{2}â€“ S_{1}).**

**Solution:**

Letâ€™s takeÂ *a*Â andÂ *d* 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 *a*Â = 203

Last term,Â *l*Â = 399 and

Common difference,Â *d*Â = 7

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

Hence, *a _{n}*Â = 399 =Â

*a*Â + (

*n*Â â€“1)Â

*d*

399 = 203 + (*n*Â â€“1) 7

7 (*n*Â â€“1) = 196

*n*Â â€“1 = 28

*n*Â = 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 + (*n*Â â€“1) 2

*n*Â = 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 + (*n*Â â€“1) 5

5*n*Â = 100

*n*Â = 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 + (*n*Â â€“1) (10)

100 = 10*n*

*n*Â = 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 the remainder.**

**Solution:**

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

They are 13, 17, â€¦ 97.

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

LetÂ *n*Â be the number of terms of the A.P.

We know that the *n*^{th} term of an A.P. is given by

*a _{n}*Â =Â

*a*Â + (

*n*Â â€“1)Â

*d*

So, 97 = 13 + (*n*Â â€“1) (4)

4 (*n*Â â€“1) = 84

*n*Â â€“ 1 = 21

*n*Â = 22

Now, the sum ofÂ *n*Â 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 *(*x*Â +Â *y*) =Â *fÂ *(*x*) Ã—Â *fÂ *(*y*) for allÂ *x*,Â *y*Â âˆˆ N â€¦ (1)

*fÂ *(1) = 3

TakingÂ *x*Â =Â *y*Â = 1 in (1), we have

*f*Â (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 *f*Â (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 the terms of the function is 120.

Therefore, the value ofÂ *n*Â 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 the sum of terms is

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

Hence, the last term of the G.P = 6^{th}Â term =Â *ar*^{6 â€“ 1}Â = (5)(2)^{5}Â = (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 *a*Â and*Â r* to be the first term and the common ratio of the G.P., respectively.

Given, *a*Â = 1

*a*_{3}Â =Â *ar*^{2}Â =Â *r*^{2}

*a*_{5}Â =Â *ar*^{4}Â =Â *r*^{4}

Then, from the question we have

*r*^{2}Â +Â *r*^{4}Â = 90

*r*^{4}Â +Â *r*^{2}Â 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, and 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 *a*,Â *ar*, andÂ *ar*^{2}.

Then from the question, we have

*a*Â +Â *ar +*Â *ar*^{2}Â = 56

*a*Â (1 +Â *r +*Â *r*^{2}) = 56

Â â€¦ (1)

Also, given

*a*Â â€“ 1,Â *ar*Â â€“ 7,Â *ar*^{2} â€“ 21 form an A.P.

So, (*arÂ *â€“ 7) â€“ (*a*Â â€“ 1) = (*ar*^{2}Â â€“ 21) â€“ (*ar*Â â€“ 7)

*arÂ *â€“Â *a*Â â€“ 6 =Â *ar*^{2Â }â€“Â *ar*Â â€“ 14

*ar*^{2Â }â€“ 2*ar*Â +Â *a*Â = 8

*ar*^{2Â }â€“Â *ar*Â â€“Â *ar*Â +Â *a*Â = 8

*a*(*r*^{2Â }+ 1 â€“ 2*r*) = 8

*aÂ *(*r*Â â€“ 1)^{2}Â = 8 â€¦ (2)

7(*r*^{2}Â â€“ 2*r*Â + 1) = 1 +Â *r*Â +Â *r*^{2}

7*r*^{2}Â â€“ 14Â *r*Â + 7 â€“ 1 â€“Â *r*Â â€“Â *r*^{2}Â = 0

6*r*^{2}Â â€“ 15*r*Â + 6 = 0

6*r*^{2}Â â€“ 12*r*Â â€“ 3*r*Â + 6 = 0

6*r*Â (*r*Â â€“ 2) â€“ 3 (*r*Â â€“ 2) = 0

(6*r*Â â€“ 3) (*r*Â â€“ 2) = 0

r = 2, 1/2

WhenÂ *r*Â = 2,Â *a*Â = 8

When r = Â½, a = 32

Thus,

WhenÂ *r*Â = 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 T_{1}, T_{2}, T_{3}, T_{4}, â€¦ T_{2}* _{n}*.

The number of terms = 2*n*

Then, from the question we have

T_{1}Â + T_{2}Â + T_{3}Â + â€¦+ T_{2}* _{n}*Â = 5 [T

_{1}Â + T

_{3}Â + â€¦ +T

_{2}

_{n}_{â€“1}]

T_{1}Â + T_{2}Â + T_{3}Â + â€¦ + T_{2}* _{n}*Â â€“ 5 [T

_{1}Â + T

_{3}Â + â€¦ + T

_{2}

_{n}_{â€“1}] = 0

T_{2}Â + T_{4}Â + â€¦ + T_{2}* _{n}*Â = 4 [T

_{1}Â + T

_{3}Â + â€¦ + T

_{2}

_{n}_{â€“1}] â€¦â€¦ (1)

Now, let the terms in G.P. beÂ *a*,Â *ar*,Â *ar*^{2},Â *ar*^{3}, 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 *a*Â +Â *d*,Â *a*Â + 2*d*,Â *a*Â + 3*d*, …Â *a*Â + (*n*Â â€“ 2)Â *d*,Â *a*Â + (*n*Â â€“ 1)*d*.

From the question, we have

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

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

= 4*a*Â + (4*n*Â â€“ 10)Â *d*

Then, according to the given condition,

4*a*Â + 6*d*Â = 56

4(11) + 6*d*Â = 56 [SinceÂ *a*Â = 11 (given)]

6*d*Â = 12

*d*Â = 2

Hence, 4*a*Â + (4*n*Â â€“10)Â *d*Â = 112

4(11) + (4*n*Â â€“ 10)2 = 112

(4*n*Â â€“ 10)2 = 68

4*n*Â â€“ 10 = 34

4*n*Â = 44

*n*Â = 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 P^{2}R^{n}^{Â }= S^{n}**

**Solution:**

Let the terms in G.P. beÂ *a*,Â *ar*,Â *ar*^{2},Â *ar*^{3}, â€¦Â *ar ^{n}*

^{Â â€“ 1}â€¦

From the question, we have

Hence, P^{2}Â R* ^{n}*Â = S

^{n}**15. TheÂ p^{th},Â q^{th}Â andÂ r^{th}Â terms of an A.P. areÂ a, b, c,Â respectively. **

**Show that (q – r) a + (r – p) b + (p -q) c = 0**

**Solution:**

Letâ€™s assumeÂ *t*Â andÂ *d* to be the first term and the common difference of the A.P., respectively.

Then theÂ *n*^{th} term of the A.P. is given byÂ *a _{n}*

_{Â }=Â

*tÂ*+ (

*n*Â â€“Â

*1*)Â

*d*

Thus,

*a _{p}*Â =

*Â tÂ*+ (

*p*Â â€“Â

*1*)Â

*d*Â =Â

*a*Â â€¦ (1)

*a _{q}*Â =Â

*t*Â + (

*q*Â â€“Â

*1*)

*d*Â =Â

*b*Â â€¦ (2)

*a _{r}*Â =Â

*t*Â + (

*r*Â â€“Â

*1*)Â

*d*Â =Â

*c*Â â€¦ (3)

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

(*p*Â â€“ 1 â€“Â *q*Â + 1)Â *d*Â =Â *a*Â â€“Â *b*

(*p*Â â€“Â *q*)Â *d*Â =Â *a*Â â€“*Â b*

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

(*q*Â â€“ 1 â€“Â *r*Â + 1)Â *d*Â =Â *b*Â â€“Â *c*

(*q*Â â€“*Â r*)Â *d*Â =*Â b*Â â€“*Â c*

Equating both the values ofÂ *d*Â 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 (a^{n}Â +Â b^{n}), (b^{n}Â +Â c^{n}), (c^{n}Â +Â d^{n})Â are in G.P.**

**Solution:**

Given *a*,*Â b*,*Â c*, and *d*Â are in G.P.

So, we have

âˆ´*b*^{2}Â =Â *ac*Â â€¦ (i)

*c*^{2}Â =Â *bd*Â â€¦ (ii)

*ad*Â =Â *bc*Â â€¦ (iii)

Required to prove (*a ^{n}*Â +Â

*b*), (

^{n}*b*Â +Â

^{n}*c*), (

^{n}*c*Â +Â

^{n}*d*) are in G.P. i.e.,

^{n}(*b ^{n}*Â +Â

*c*)

^{n}^{2}Â = (

*a*Â +Â

^{n}*b*) (

^{n}*c*Â +Â

^{n}*d*)

^{n}Taking L.H.S.

(*b ^{n}*Â +Â

*c*)

^{n}^{2}Â =Â

*b*

^{2}

*+ 2*

^{nÂ }*b*Â +Â

^{n}c^{n}*c*

^{2}

^{n}= (*b*^{2})* ^{n}*+ 2

*b*Â + (

^{n}c^{n}*c*

^{2})

^{Â n}= (*ac*)* ^{n}*Â + 2

*b*Â + (

^{n}c^{n}*bd*)

*Â [Using (i) and (ii)]*

^{n}=Â *a ^{n}*Â

*c*Â +Â

^{n}*b*+Â

^{n}c^{n}*b*Â

^{n}*c*Â +Â

^{n}*b*Â

^{n}*d*

^{n}=Â *a ^{n}*Â

*c*Â +Â

^{n}*b*+Â

^{n}c^{n}*a*Â

^{n}*d*Â +Â

^{n}*b*Â

^{n}*d*Â [Using (iii)]

^{n}=Â *c ^{n}*Â (

*a*Â +Â

^{n}*b*) +Â

^{n}*d*Â (

^{n}*a*Â +Â

^{n}*b*)

^{n}= (*a ^{n}*Â +Â

*b*) (

^{n}*c*Â +Â

^{n}*d*)

^{n}= R.H.S.

Therefore, (*a ^{n}*Â +Â

*b*), (

^{n}*b*Â +Â

^{n}*c*), and (

^{n}*c*Â +Â

^{n}*d*) are in G.P

^{n}– Hence proved.

**18. IfÂ aÂ andÂ bÂ are the roots ofÂ x^{2} â€“ 3x + p = 0 and c, dare roots ofÂ x^{2} 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 *a*Â andÂ *b*Â are the roots ofÂ *x*^{2Â }â€“ 3*xÂ *+Â *p*Â = 0

So, we have *a*Â +Â *b*Â = 3 andÂ *ab*Â =Â *p*Â â€¦ (i)

Also,Â *c*Â andÂ *d*Â are the roots ofÂ *x ^{2} â€“ 12x + q = 0*

So, *cÂ *+Â *d*Â = 12 andÂ *cd*Â =Â *q*Â â€¦ (ii)

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

Letâ€™s takeÂ *aÂ *=Â *x*,Â *b*Â =Â *xr*,Â *c*Â =*Â xr ^{2}*,Â

*d*Â =Â

*xr*

^{3}

From (i) and (ii), we get

*x*Â +Â *xr*Â = 3

*x*Â (1 +Â *r*) = 3

And,

*xr*^{2}Â +Â *xr*^{3}Â =12

*xr*^{2Â }(1 +Â *r*) = 12

On dividing, we get

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

When r = -2, x = 3/(1 – 2) = 3/-1 = -3

Case I:

WhenÂ *r*Â = 2 andÂ *x*Â =1,

*ab*Â =Â *x*^{2}*r*Â = 2

*cd*Â =Â *x*^{2}*r*^{5}Â = 32

Case II:

WhenÂ *rÂ *= â€“2,Â *xÂ *= â€“3,

*ab*Â =Â *x*^{2}r = â€“18

*cd*Â =Â *x*^{2}*r*^{5}Â = â€“ 288

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

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

**Solution: **

Let the two numbers beÂ *a*Â 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 *c*Â are in A.P.

Hence, *b*Â â€“Â *a*Â =Â *c*Â â€“Â *b*Â â€¦ (1)

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

So, *c*^{2}Â =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 *e*Â are in G.P. i.e.,Â *c*^{2}Â =Â *ae*

From (1), we have

2b = a + c

b = (a + c)/ 2

And from (2), we have

d = c^{2}/ b

On substituting these values in (3), we get

Therefore,Â *a*,Â *c*, andÂ *e*Â 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 S* _{n}*Â = 5 + 55 + 555 + â€¦..Â up to

*n*Â terms

(ii) Given, .6 + .66 + . 666 + â€¦

Let S* _{n}*Â = 06. + 0.66 + 0.666 + â€¦ up toÂ

*n*Â terms

**22. Find the 20 ^{th}Â 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 *n*Â terms

âˆ´Â *n*^{th}Â term =Â *a _{n}*

_{Â }= 2

*n*Â Ã— (2

*n*Â + 2) = 4

*n*

^{2}Â + 4

*n*

The 20^{th} term,

*a*_{20}Â = 4 (20)^{2}Â + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Therefore, the 20^{th}Â 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 *a _{n}*

_{â€“1Â }+Â

*a*

_{n}S = 3 + 7 + 13 + 21 + â€¦. +Â *a _{n}*

_{Â â€“ 2Â }+Â

*a*

_{nÂ }_{â€“ 1Â }+

*Â a*

_{n}On subtracting both equations, we get

S â€“ S = [3 + (7 + 13 + 21 + 31 + â€¦+Â *a _{n}*

_{â€“1Â }+Â

*a*)] â€“ [(3 + 7 + 13 + 21 + 31 + â€¦+Â

_{n}*a*

_{n}_{â€“1})

_{Â }+Â

*a*]

_{n}S â€“ S = 3 + [(7 â€“ 3) + (13 â€“ 7) + (21 â€“ 13) + â€¦ + (*a _{n}*Â â€“Â

*a*

_{n}_{â€“1})]

_{Â }â€“Â

*a*

_{n}0 = 3 + [4 + 6 + 8 + â€¦ (*n*Â â€“1) terms] â€“Â *a _{n}*

*a _{n}*Â = 3 + [4 + 6 + 8 + â€¦ (

*n*Â â€“1) terms]

**24. If S _{1}, S_{2}, S_{3} are the sum of the firstÂ nÂ natural numbers, their squares and their cubes, respectively, show thatÂ 9S_{2}^{2} = S_{3} (1 + 8S_{1}). **

**Solution: **

From the question, we have

Therefore, from (1) and (2), we have 9S_{2}^{2} = S_{3} (1 + 8S_{1}).

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

**Solution: **

**26. Show thatÂ **

**Solution:**

*n*^{th}Â term of the numerator =Â *n*(*n*Â + 1)^{2}Â =Â *n*^{3}Â + 2*n*^{2}Â +Â *n*

*n*^{th}Â term of the denominator =Â *n*^{2}(*n*Â + 1) =Â *n*^{3}Â +Â *n*^{2}

**27. A farmer buys a used tractor for Rs 12,000. He pays Rs 6,000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will 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 + (*n*Â â€“ 1) 500

1 + (*n*Â â€“ 1) = 12

*n*Â = 12

Now,

The sum of the A.P = 12/2 [2(500) + (12 – 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 22,000. He pays Rs 4,000 cash and agrees to pay the balance in annual instalments of Rs 1,000 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

From 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 form an A.P. with the first term and common difference both equal to 1000.

Let the number of terms be *n*.

So, 18000 = 1000 + (*n*Â â€“ 1) (1000)

*n*Â = 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 it to four different persons with the instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when the 8 ^{th} set of the letter is mailed.**

**Solution:**

Itâ€™s seen that,

The numbers of letters mailed forms a G.P.: 4, 4^{2}, â€¦ 4^{8}

Here, first term = 4 and common ratio = 4

And the number of terms = 8

The sum ofÂ *n* 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 43690 = Rs 43690

Therefore, the amount spent when the 8^{th} set of the letter is mailed will be Rs 43,690.

**30. A man deposited Rs 10,000 in a bank at the rate of 5% simple interest annually. Find the amount in the 15 ^{th}Â 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, the amount in the 15^{th}Â 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 15^{th} year is Rs 17,000, and the total amount after 20 years will be Rs 20,000.

**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 the 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., 4/5^{th }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 5,120.

**32. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day, 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.**

**Solution:**

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

Then, from the question, we have

150*x*Â = 150 + 146 + 142 + â€¦. (*x*Â + 8) terms

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

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

Now, finding the sum of terms

As *x* cannot be negative [Number of days is always a positive quantity]

*x*Â = 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 – Sequences and Series

Listed below 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 2023-24, 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:

- By a sequence, we mean an arrangement of numbers in a 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, ….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.
- Let a
_{1}, a_{2}, a_{3}, … be the sequence, and then the sum expressed as a_{1}+ a_{2}+ a_{3}+ … is called a series. A series is called finite series if it has a finite number of terms. - An arithmetic progression (A.P.) is a sequence in which terms regularly increase or decrease 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 n
^{th}term of the A.P. is given by a_{n}= a + (n â€“ 1) d. - A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is the 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 n
^{th}term of G.P. is given by a_{n}= ar^{n 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., the 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 **

**Dropped Topics –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?

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

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

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