NCERT exercise solutions are presented here to ease your Class 10 board exam preparation. This page covers all the answers to the questions provided in Exercise 5.2 of the NCERT Class 10 Maths textbook. NCERT Solutions for Class 10 Maths Chapter 5 â€“ Arithmetic Progression Exercise 5.2 will give you a comprehensive methodology to solve nth term questions of arithmetic progression. NCERT solutions will assist you in attaining perfection in all the methodologies provided in the textbook.

The highly experienced faculty at BYJUâ€™S design the answers in such a way that no minute detail is missing from the latest syllabus. NCERT Class 10 Solutions for Maths Chapter 5 Exercise 5.2 can be downloaded in PDF format below.

## Topics Covered in Exercise 5.2

- nth term of arithmetic progression
- Solved examples
- Alternate solutions

### How it is helpful

- Helps you practise different types of questions.
- Helps you to know different methodologies to solve arithmetic progression.
- Boosts your confidence in solving arithmetic progression problems.
- NCERT Class 10 Maths Solution is the best study material to start your exam preparation.

## NCERT solutions for Class 10 Maths chapter 5 Arithmetic Progression Exercise 5.2

### Access other exercise solutions of Class 10 Maths Chapter 5 â€“ Arithmetic Progression

Students can access the exercise-wise NCERT Class 10 Solutions for Maths Chapter 5 below.

Exercise 5.1â€“ 4 questions (1 MCQ and 3 descriptive type questions)

Exercise 5.2- 20 questions (1 fill in the blanks, 2 MCQs, 7 Short answer questions and 10 Long answer questions)

Exercise 5.3â€“ 20 Questions (3 fill in the blanks, 4 daily life examples, and 13 descriptive-type questions)

Exercise 5.4 5 Questions (5 Long answer questions)

### Access Answers of Maths NCERT Class 10 Chapter 5 â€“ Arithmetic Progression Exercise 5.2

**1. Fill in the blanks in the following table, given thatÂ aÂ is the first term,Â dÂ the common difference andÂ a_{n}Â theÂ n^{th}Â term of the A.P.**

**Solutions:**

**(i)Â **Given,

First term, *a*Â = 7

Common difference,Â *d*Â = 3

Number of terms,Â *n*Â = 8,

We have to find the nth term, *a _{n}*Â = ?

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

Putting the values,

=> 7+(8 âˆ’1) 3

=> 7+(7) 3

=> 7+21 = 28

Hence,*Â a _{n}*Â = 28

**(ii)** Given,

First term, *a*Â = -18

Common difference,Â *d*Â = ?

Number of terms,Â *n*Â = 10

Nth term, *a _{n}*Â = 0

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

Putting the values,

0 = âˆ’ 18 +(10âˆ’1)*d*

18 = 9*d*

*d*Â = 18/9 = 2

Hence, common difference,Â *dÂ *= 2

**(iii)** Given,

First term, *a*Â = ?

Common difference,Â *d*Â = -3

Number of terms,Â *n*Â = 18

Nth term, *a _{n}*Â = -5

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

Putting the values,

âˆ’5 =Â *a*+(18âˆ’1) (âˆ’3)

âˆ’5 =Â *a*+(17) (âˆ’3)

âˆ’5 =Â *a*âˆ’51

*a*Â = 51âˆ’5 = 46

Hence,Â *a*Â = 46

**(iv)**Â Given,

First term, *a*Â = -18.9

Common difference,Â *d*Â = 2.5

Number of terms,Â *n*Â = ?

Nth term, *a _{n}*Â = 3.6

As we know, for an A.P.,

*a _{n}*Â =Â

*a*Â +(

*n*Â âˆ’1)

*d*

Putting the values,

3.6 = âˆ’ 18.9+(*n*Â âˆ’1)2.5

3.6 + 18.9 = (*n*âˆ’1)2.5

22.5 = (*n*âˆ’1)2.5

(*n*Â â€“ 1) = 22.5/2.5

*n*Â â€“ 1 = 9

*n*Â = 10

Hence,Â *n*Â = 10

**(v)**Â Given,

First term, *a*Â = 3.5

Common difference,Â *d*Â = 0

Number of terms,Â *n*Â = 105

Nth term, *a _{n}*Â = ?

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*Â âˆ’1)

*d*

Putting the values,

*a _{n}*Â = 3.5+(105âˆ’1) 0

*a _{n}*Â = 3.5+104Ã—0

*a _{n}*Â = 3.5

Hence,Â *a _{n}*Â = 3.5

**2. Choose the correct choice in the following and justify:
(i) 30 ^{th}Â term of the A.P: 10,7, 4, â€¦, is
(A)Â 97 (B)Â 77 (C)Â âˆ’77 (D) âˆ’87**

(ii) 11^{thÂ }term of the A.P. -3, -1/2, ,2 â€¦. is

(A) 28 (B) 22 (C) â€“ 38 (D)

**Â **

**Solutions:**

(i) Given here,

A.P. = 10, 7, 4, â€¦

Therefore, we can find,

First term,Â *a*Â = 10

Common difference,Â *d*Â =Â *a*_{2}Â âˆ’Â *a*_{1Â }= 7âˆ’10 = âˆ’3

As we know, for an A.P.,

*a _{n}*Â =Â

*a*Â +(

*n*âˆ’1)

*d*

Putting the values;

*a*_{30}Â = 10+(30âˆ’1)(âˆ’3)

*a*_{30}Â = 10+(29)(âˆ’3)

*a*_{30}Â = 10âˆ’87 = âˆ’77

Hence, the correct answer is optionÂ C.

(ii) Given here,

A.P. =Â -3, -1/2, ,2 â€¦

Therefore, we can find,

First termÂ *a*Â = â€“ 3

Common difference,Â *d*Â =Â *a*_{2}Â âˆ’Â *a*_{1}Â = (-1/2) -(-3)

â‡’(-1/2)Â + 3 = 5/2

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

Putting the values;

*a*_{11}Â = -3+(11-1)(5/2)

*a*_{11}Â = -3+(10)(5/2)

*a*_{11}Â = -3+25

*a*_{11}Â = 22

Hence, the answer is option B.

**3. In the following APs find the missing term in the boxes.**

**Solutions:**

(i)Â For the given A.P., 2,2 , 26

The first and third term are;

*a*Â = 2

*a*_{3}Â = 26

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*Â âˆ’1)

*d*

Therefore, putting the values here,

*a*_{3}Â = 2+(3-1)*d*

26 = 2+2*d*

24 = 2*d*

*d*Â = 12

*a*_{2}Â = 2+(2-1)12

= 14

Therefore, 14 is the missing term.

**(ii)Â For the given A.P., , 13, ,3**

*a*_{2}Â = 13 and

*a*_{4}Â = 3

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)Â

*d*

Therefore, putting the values here,

*a*_{2}Â =Â *a*Â +(2-1)*d*

13 =Â *a*+*d*Â â€¦â€¦â€¦â€¦â€¦â€¦.Â **(i)**

*a*_{4}Â =Â *a*+(4-1)*d*

3 =Â *a*+3*d*Â â€¦â€¦â€¦â€¦..Â **(ii)**

On subtracting equationÂ **(i)**Â fromÂ **(ii)**, we get,

â€“ 10 = 2*d*

*d*Â = â€“ 5

From equationÂ **(i)**, putting the value of d,we get

13 =Â *a*+(-5)

*a*Â = 18

*a*_{3}Â = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

Therefore, the missing terms are 18 and 8, respectively.

**(iii)** For the given A.P.,

*aÂ *= 5 and

*a*_{4}Â = 19/2

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

Therefore, putting the values here,

*a*_{4}Â =Â *a*+(4-1)*d*

19/2 =Â *5*+3d

(19/2) â€“ 5 = 3d

3d =Â 9/2

d = 3/2

*a*_{2}Â =Â *a*+(2-1)*d*

*a*_{2}Â =Â *5*+3/2

*a*_{2}Â =Â 13/2

*a*_{3}Â =Â *a*+(3-1)*d*

*a*_{3}Â =Â *5*+2Ã—3/2

*a*_{3}Â =Â *8*

Therefore, the missing terms are 13/2 and 8, respectively.

(iv) For the given A.P.,

*a*Â = âˆ’4 and

*a*_{6}Â = 6

As we know, for an A.P.,

*a _{n}*Â =Â

*a*Â +(

*n*âˆ’1)Â

*d*

Therefore, putting the values here,

a_{6}Â = a+(6âˆ’1)d

6 = âˆ’ 4+5*d*

10 = 5*d*

*d*Â = 2

*a*_{2}Â =Â *a*+*d*Â = âˆ’ 4+2 = âˆ’2

*a*_{3}Â =Â *a*+2*d*Â = âˆ’ 4+2(2) = 0

*a*_{4}Â =Â *a*+3*d*Â = âˆ’ 4+ 3(2) = 2

*a*_{5}Â =*Â a*+4*d*Â = âˆ’ 4+4(2) = 4

Therefore, the missing terms are âˆ’2, 0, 2, and 4 respectively.

**(v)** For the given A.P.,

*a*_{2}Â = 38

*a*_{6}Â = âˆ’22

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*Â âˆ’1)

*d*

Therefore, putting the values here,

*a*_{2}Â =Â *a*+(2âˆ’1)*d*

38 =Â *a*+*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â **(i)**

*a*_{6}Â =Â *a*+(6âˆ’1)*d*

âˆ’22 =Â *a*+5*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦.**Â (ii)**

On subtracting equationÂ **(i)**Â fromÂ **(ii)**, we get

âˆ’ 22 âˆ’ 38 = 4*d*

âˆ’60 = 4*d*

*d*Â = âˆ’15

*a*Â =Â *a*_{2}Â âˆ’Â *d*Â = 38 âˆ’ (âˆ’15) = 53

*a*_{3}Â =Â *aÂ *+ 2*dÂ *= 53 + 2 (âˆ’15) = 23

*a*_{4}Â =Â *a*Â + 3*d*Â = 53 + 3 (âˆ’15) = 8

*a*_{5}Â =Â *a*Â + 4*d*Â = 53 + 4 (âˆ’15) = âˆ’7

Therefore, the missing terms are 53, 23, 8, and âˆ’7, respectively.

**4. Which term of the A.P. 3, 8, 13, 18, â€¦ is 78?**

**Solutions:**

Given the A.P. series as3, 8, 13, 18, â€¦

First term, aÂ = 3

Common difference, dÂ =Â a_{2}Â âˆ’Â a_{1}Â = 8 âˆ’ 3 = 5

LetÂ the *n*^{th}Â term of given A.P. be 78. Now as we know,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*Therefore,*

78 = 3+(*n*Â âˆ’1)5

75 = (*n*âˆ’1)5

(*n*âˆ’1) = 15

*n*Â = 16

Hence, 16^{th}Â term of this A.P. is 78.

**5. Find the number of terms in each of the following A.P.**

**(i) 7, 13, 19, â€¦, 205**

**Solutions:**

**(i) Given, 7, 13, 19, â€¦, 205 is the A.P**

Therefore

*First term, a*Â = 7

*Common difference, d*Â =Â *a*_{2}Â âˆ’Â *a*_{1}Â = 13 âˆ’ 7 = 6

Let there areÂ *n*Â terms in this A.P.

*a _{n}*Â = 205

As we know, for an A.P.,

*a _{n}*Â =Â

*a*Â + (

*n*Â âˆ’ 1)Â

*d*

Therefore, 205 = 7 + (*nÂ *âˆ’ 1) 6

198 = (*n*Â âˆ’ 1) 6

33 = (*n*Â âˆ’ 1)

*n*Â = 34

Therefore, this given series has 34 terms in it.

First term, aÂ = 18

Common difference, d = a_{2}-a_{1 }=

d = (31-36)/2 = -5/2

Let there are n terms in this A.P.

a_{n}Â = -47

As we know, for an A.P.,

a_{n}Â =Â a+(nâˆ’1)d

-47 = 18+(n-1)(-5/2)

-47-18 = (n-1)(-5/2)

-65 = (n-1)(-5/2)

(n-1) = -130/-5

(n-1) =Â 26

nÂ = 27

Therefore, this given A.P. has 27 terms in it.

**6.Â Check whether -150 is a term of the A.P. 11, 8, 5, 2, â€¦**

**Solution:**

For the given series, A.P. 11, 8, 5, 2..

*First term, a*Â = 11

*Common difference, d*Â =Â *a*_{2}âˆ’*a*_{1}Â = 8âˆ’11 = âˆ’3

Let âˆ’150 be theÂ *n*^{th}Â term of this A.P.

As we know, for an A.P.,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

-150 = 11+(*n*Â -1)(-3)

-150 = 11-3*n*Â +3

-164 = -3*n*

*n*Â = 164/3

Clearly,Â *n*Â is not an integer but a fraction.

Therefore, â€“ 150 is not a term of this A.P.

**7. Find the 31 ^{st}Â term of an A.P. whose 11^{th}Â term is 38 and the 16^{th}Â term is 73.**

**Solution:**

Given that,

11^{th} term, a_{11}Â = 38

and 16^{th} term, a_{16}Â = 73

We know that,

a_{n}Â =Â a+(nâˆ’1)d

a_{11}Â =Â a+(11âˆ’1)d

38 =Â a+10dÂ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â **(i)**

In the same way,

*a*_{16}Â =Â *a*Â +(16âˆ’1)*d*

73 =Â *a*+15*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â **(ii)**

On subtracting equation**Â (i)**Â from**Â (ii)**, we get

35 = 5*d*

*d*Â = 7

From equationÂ **(i)**, we can write,

38 =Â *a*+10Ã—(7)

38 âˆ’ 70 =Â a

aÂ = âˆ’32

a_{31}Â =Â aÂ +(31âˆ’1)Â d

= âˆ’ 32 + 30 (7)

= âˆ’ 32 + 210

= 178

Hence, 31^{st}Â term is 178.

**8. An A.P. consists of 50 terms of which 3 ^{rd}Â term is 12 and the last term is 106. Find the 29^{th}Â term.**

**Solution: **Given that,

3^{rd} term, a_{3}Â = 12

50^{th} term, a_{50}Â = 106

We know that,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*a*_{3}Â =Â *a*+(3âˆ’1)*d*

12 =Â *a*+2*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â **(i)**

In the same way,

*a*_{50Â }=Â *a*+(50âˆ’1)*d*

106 =Â *a*+49*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â **(ii)**

On subtracting equationÂ **(i)**Â fromÂ **(ii)**, we get

94 = 47*d*

*d*Â = 2 = common difference

From equationÂ **(i)**, we can write now,

12 =Â *a*+2(2)

*a*Â = 12âˆ’4 = 8

*a*_{29}Â =Â *a*+(29âˆ’1)Â *d*

*a*_{29}Â = 8+(28)2

*a*_{29}Â = 8+56 = 64

Therefore, 29^{th}Â term is 64.

**9. If the 3 ^{rd}Â and the 9^{th} terms of an A.P. are 4 and âˆ’ 8, respectively. Which term of this A.P. is zero?**

Solution:

Given that,

*3 ^{rd} term, a*

_{3}Â = 4

*and 9 ^{th} term, a*

_{9}Â = âˆ’8

We know that,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*Therefore,*

*a*_{3}Â =Â *a*+(3âˆ’1)*d*

4 =Â *a*+2*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â **(i)**

*a*_{9}Â =Â *a*+(9âˆ’1)*d*

âˆ’8 =Â *a*+8*d*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â **(ii)**

On subtracting equationÂ **(i)**Â fromÂ **(ii)**, we will get here,

âˆ’12 = 6*d*

*d*Â = âˆ’2

From equationÂ **(i)**, we can write,

4 =Â *a*+2(âˆ’2)

4 =Â *a*âˆ’4

*a*Â = 8

LetÂ *n*^{th}Â term of this A.P. be zero.

*a _{nÂ }*=Â

*a*+(

*n*âˆ’1)

*d*

0 = 8+(*n*âˆ’1)(âˆ’2)

0 = 8âˆ’2*n*+2

2*nÂ *= 10

*n*Â = 5

Hence, 5^{th}Â term of this A.P. is 0.

**10. If 17 ^{th}Â term of an A.P. exceeds its 10^{th}Â term by 7. Find the common difference.**

**Solution: **

We know that, for an A.P series;

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*a*_{17}Â =Â *a*+(17âˆ’1)*d*

*a*_{17}Â =Â *a*Â +16*d*

In the same way,

*a*_{10}Â =Â *a*+9*d*

As it is given in the question,

*a*_{17}Â âˆ’Â *a*_{10}Â = 7

Therefore,

(*a*Â +16*d*)âˆ’(*a*+9*d*) = 7

7*d*Â = 7

*d*Â = 1

Therefore, the common difference is 1.

**11.Â Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54 ^{th}Â term?**

**Solution:**

Given A.P. is 3, 15, 27, 39, â€¦

*first term, aÂ *= 3

*common difference, d*Â =Â *a*_{2}Â âˆ’Â *a*_{1}Â = 15 âˆ’ 3 = 12

We know that,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*Therefore,*

a_{54}Â =Â *a*+(54âˆ’1)*d*

â‡’3+(53)(12)

â‡’3+636 = 639

*a*_{54 }= 639+132=771

We have to find the term of this A.P. which is 132 more than a_{54, }i.e.771.

LetÂ *n*^{th}Â term be 771.

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

771 = 3+(*n*Â âˆ’1)12

768 = (*n*âˆ’1)12

(*n*Â âˆ’1) = 64

*n*Â = 65

Therefore, 65^{th}Â term was 132 more than 54^{th}Â term.

**Or another method is;**

LetÂ *n*^{th}Â term be 132 more than 54^{th}Â term.

*n*Â = 54Â + 132/2

= 54Â + 11 =Â 65^{th}Â term

**12. Two APs have the same common difference. The difference between their 100 ^{th}Â term is 100, what is the difference between their 1000^{th}Â terms?**

**Solution:**

Let, the first term of two APs beÂ *a*_{1}Â andÂ *a*_{2}Â respectively

And the common difference of these APs beÂ *d*.

For the first A.P.,we know,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*Therefore,*

*a*_{100}Â =Â *a*_{1}+(100âˆ’1)*d*

=Â *a*_{1}Â + 99d

*a*_{1000}Â =Â *a*_{1}+(1000âˆ’1)*d*

*a*_{1000}Â =Â *a*_{1}+999*d*

For second A.P., we know,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*Therefore,*

*a*_{100}Â =Â *a*_{2}+(100âˆ’1)*d*

=Â *a*_{2}+99*d*

*a*_{1000}Â =Â *a*_{2}+(1000âˆ’1)*d*

=Â *a*_{2}+999*d*

Given that, difference between 100^{th}Â term of the two APs = 100

Therefore, (*a*_{1}+99*d*) âˆ’ (*a*_{2}+99*d*) = 100

*a*_{1}âˆ’*a*_{2}Â = 100â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â **(i)**

Difference between 1000^{th}Â terms of the two APs

(*a*_{1}+999*d*) âˆ’ (*a*_{2}+999*d*) =Â *a*_{1}âˆ’*a*_{2}

From equationÂ **(i)**,

This difference,Â *a*_{1}âˆ’*a*_{2Â }= 100

Hence, the difference between 1000^{th}Â terms of the two A.P. will be 100.

**13. How many three digit numbers are divisible by 7?**

**Solution:**

First three-digit number that is divisible by 7 are;

First number = 105

Second number = 105+7 = 112

Third number = 112+7 =119

Therefore, 105, 112, 119, â€¦

All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

As we know, the largest possible three-digit number is 999.

When we divide 999 by 7, the remainder will be 5.

Therefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.

Now the series is as follows.

105, 112, 119, â€¦, 994

Let 994 be theÂ nth term of this A.P.

first term, aÂ = 105

common difference, dÂ = 7

a_{n}Â = 994

nÂ = ?

As we know,

a_{n}Â =Â a+(nâˆ’1)d

994 = 105+(nâˆ’1)7

889 = (nâˆ’1)7

(nâˆ’1) = 127

nÂ = 128

Therefore, 128 three-digit numbers are divisible by 7.

**14. How many multiples of 4 lie between 10 and 250?**

**Solution: **

The first multiple of 4 that is greater than 10 is 12.

Next multiple will be 16.

Therefore, the series formed as;

12, 16, 20, 24, â€¦

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 âˆ’ 2 = 248 is divisible by 4.

The series is as follows, now;

12, 16, 20, 24, â€¦, 248

Let 248 be theÂ *n*^{th}Â term of this A.P.

*first term, a*Â = 12

*common difference, d*Â = 4

*a*_{n}Â =Â 248

As we know,

a_{n}Â =Â a+(nâˆ’1)d

248 = 12+(*n*-1)Ã—4

236/4 =Â n-1

59 Â =Â n-1

nÂ = 60

Therefore, there are 60 multiples of 4 between 10 and 250.

**15. For what value ofÂ n, are theÂ n^{th}Â terms of two APs 63, 65, 67, and 3, 10, 17, â€¦ equal?**

**Solution: **

Given two APs as; 63, 65, 67,â€¦ and 3, 10, 17,â€¦.

**Taking first AP,**

63, 65, 67, â€¦

First term, aÂ = 63

Common difference, dÂ =Â a_{2}âˆ’a_{1}Â = 65âˆ’63 = 2

We know, n^{th}Â term of this A.P. =Â a_{n}Â =Â a+(nâˆ’1)d

*a _{n}*= 63+(

*n*âˆ’1)2 = 63+2

*n*âˆ’2

*a _{n}*Â = 61+2

*n*Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â

**(i)**

**Taking second AP,**

3, 10, 17, â€¦

First term, aÂ = 3

Common difference, dÂ =Â a_{2}Â âˆ’Â a_{1}Â = 10 âˆ’ 3 = 7

We know that,

*n*^{th}Â term of this A.P. = 3+(*n*âˆ’1)7

*a _{n}*Â = 3+7

*n*âˆ’7

*a _{n}*Â = 7

*n*âˆ’4 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â

**(ii)**

Given,Â *n*^{th}Â term of these A.P.s are equal to each other.

Equating both these equations, we get,

61+2*n*Â = 7*n*âˆ’4

61+4 = 5*n*

5*n*Â = 65

*n*Â = 13

Therefore, 13^{th}Â terms of both these A.P.s are equal to each other.

Â

**16. Determine the A.P. whose third term is 16 and the 7 ^{th}Â term exceeds the 5^{th}Â term by 12.**

**Solutions: **

Given,

*Third term, a*_{3}Â = 16

As we know,

*a*Â +(3âˆ’1)*d*Â = 16

*a*+2*d*Â = 16 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.Â (i)

It is given that, 7^{th}Â term exceeds the 5^{th}Â term by 12.

*a*_{7}Â âˆ’Â *a*_{5}Â = 12

*a*+(7âˆ’1)

*d*]âˆ’[

*aÂ*+(5âˆ’1)

*d*]= 12

(*a*+6*d*)âˆ’(*a*+4*d*) = 12

2*d*Â = 12

*d*Â = 6

From equationÂ (i), we get,

*a*+2(6) = 16

*a*+12 = 16

*a*Â = 4

Therefore, A.P. will be4, 10, 16, 22, â€¦

**17. Find the 20 ^{th}Â term from the last term of the A.P. 3, 8, 13, â€¦, 253.**

**Solution: **

Given A.P. is3, 8, 13, â€¦, 253

Common difference, d= 5.

Therefore, we can write the given AP in reverse order as;

253, 248, 243, â€¦, 13, 8, 5

Now for the new AP,

first term, aÂ = 253

and common difference, dÂ = 248 âˆ’ 253 = âˆ’5

nÂ = 20

Therefore, using nth term formula, we get,

*a*_{20}Â =Â *a*+(20âˆ’1)*d*

*a*_{20}Â = 253+(19)(âˆ’5)

*a*_{20}Â = 253âˆ’95

*a*Â = 158

Therefore, 20^{th}Â term from the last term of the AP 3, 8, 13, â€¦, 253**.**is 158.

**18. The sum of 4 ^{th}Â and 8^{th}Â terms of an A.P. is 24 and the sum of the 6^{th}Â and 10^{th}Â terms is 44. Find the first three terms of the A.P**.

**Solution: **

We know that, the nth term of the AP is;

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*a*_{4}Â =Â *a*+(4âˆ’1)*d*

*a*_{4}Â =Â *a*+3*d*

In the same way, we can write,

*a*_{8}Â =Â *a*+7*d*

*a*_{6}Â =Â *a*+5*d*

*a*_{10}Â =Â *a*+9*d*

Given that,

a_{4}+a_{8}Â = 24

a+3d+a+7dÂ = 24

2a+10dÂ = 24

a+5dÂ = 12 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦Â **(i)**

a_{6}+a_{10}Â = 44

aÂ +5d+a+9dÂ = 44

2a+14dÂ = 44

a+7dÂ = 22 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â **(ii)**

On subtracting equationÂ **(i)**Â fromÂ **(ii)**, we get,

2dÂ = 22 âˆ’ 12

2dÂ = 10

*d*Â = 5

From equationÂ **(i)**, we get,

*a*+5*d*Â = 12

*a*+5(5) = 12

*a*+25 = 12

*a*Â = âˆ’13

*a*_{2}Â =Â *a*+*d*Â = âˆ’ 13+5 = âˆ’8

*a*_{3}Â =Â *a*_{2}+*d*Â = âˆ’ 8+5 = âˆ’3

Therefore, the first three terms of this A.P. are âˆ’13, âˆ’8, and âˆ’3.

**19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?**

**Solution: **

It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.

Therefore, after 1995, the salaries of each year are;

5000, 5200, 5400, â€¦

Here,Â first term, *a*Â = 5000

*and common difference, d*Â = 200

Let afterÂ *n*^{th}Â year, his salary be Rs 7000.

Therefore, by the n^{th} term formula of AP,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)Â

*d*

7000 = 5000+(*n*âˆ’1)200

200(*n*âˆ’1)= 2000

(*n*âˆ’1) = 10

*n*Â = 11

Therefore, in 11th year, his salary will be Rs 7000.

**20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in theÂ n^{th}Â week, her weekly savings become Rs 20.75, findÂ n.**

**Solution: **

Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.

Hence,

First term, aÂ = 5

and common difference, dÂ = 1.75

Also given,

*a _{nÂ }*= 20.75

Find, nÂ = ?

As we know, by the n^{th} term formula,

*a _{n}*Â =Â

*a*+(

*n*âˆ’1)

*d*

*Therefore,*

20.75 = 5+(*n*Â -1)Ã—1.75

15.75 = (*n*Â -1)Ã—1.75

(*n*Â -1) = 15.75/1.75 = 1575/175

= 63/7 = 9

*n*Â -1 = 9

*n*Â = 10

Hence,Â *n*Â is 10.

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