**NCERT Solutions for Class 10 Maths Exercise 5.2 Chapter 5 Arithmetic Progression**

NCERT Solutions for Class 10 Maths Exercise 5.2 Chapter 5 Arithmetic Progression are presented here by Future Study Point for helping class 10 maths students of CBSE to boost their preparation for the class 10 CBSE Board exam Term 2. NCERT Solutions for Class 10 Maths Exercise 5.2 Chapter 5 Arithmetic Progression are the best study inputs for clearing the concepts on Arithmetic Progression. The questions on Arithmetic Progression are based on our everyday life problems on different types of topics where we need to predict physical quantities. All NCERT Solutions for Class 10 Maths Exercise 5.2 Chapter 5 Arithmetic Progression are created here by an experienced CBSE Maths teacher by a step-by-step method as per the norms of CBSE, therefore class 10 students can easily understand the solutions.

**NCERT Solutions for Class 10 Maths Exercise 5.2 Chapter 5 Arithmetic Progression**

**Q1.Fill in the blanks in the following table, given that ****a**** is the first term, ****d**** the common difference and ****a _{n}**

**the**

**n**

^{th}**term of the A.P.**

a | d | n | a_{n} | |

(i) | 7 | 3 | 8 | …….. |

(ii) | -18 | …….. | 10 | 0 |

(iii) | ……….. | -3 | 18 | -5 |

(iv) | -18.9 | 2.5 | ……… | 3.6 |

(v) | 3.5 | 0 | 105 | ……… |

(i) a =7,d=3,n =8

**n**^{th}** term of an AP is given by**

a_{n}= a +(n -1)d

a_{n}= 7 +(8-1)3

=7 +7 ×3

= 7 +21

=28

(ii) a =-18,d=?,n =10, a_{n}=0

**n**^{th}** term of an AP is given by**

a_{n}= a +(n -1)d

0= -18 +(10-1)d

9d = 18

d = 2

(iii) a =?,d=-3,n =18, a_{n}=-5

**n**^{th}** term of an AP is given by**

a_{n}= a +(n -1)d

-5= a +(18-1)×-3

-5= a +17×-3

a =-5 +51 =46

(iv) a =-18.9,d=2.5,n =?, a_{n}=3.6

**n**^{th}** term of an AP is given by**

a_{n}= a +(n -1)d

3.6= -18.9+(n-1)×2.5

3.6 = -18.9 +2.5n -2.5

3.6 = 2.5n -21.4

2.5n =3.6 +21.4 =25

n = 10

(v) a =3.5,d=0,n =105, a_{n}=?

**n**^{th}** term of an AP is given by**

a_{n}=3.5 +(n -1)×0

a_{n}=3.5

**Q2.Choose the correct choice in the following and justify:**

(i) 30^{th} term of the A.P: 10,7, 4, …, is

(A) 97 (B) 77 (C) -77 (D) -87

**Ans.**(C) -77

First term of the AP ,a =10,Common difference,d=7-10 = =-3,n=30

n^{th} term of AP is given by

a_{n}= a +(n -1)d

30^{th} term of the AP =

a_{30 }=10 +(30-1)×-3

=10 +29×-3

a_{30}=10 -87 = -77

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**(ii) 11 ^{th }term of the A.P. -3, -1/2, ,2 …. is**

**(A) 28 (B) 22 (C) -38 (D)-48&1/2**

Ans. The given AP is -3, -1/2, ,2 ….

a = -3, d = (-1/2 +3) =(-1+6)/2=5/2

n^{th} term of AP is given by

a_{n}= a +(n -1)d

11^{th} term of the AP =

a_{11 }=-3 +(11-1)×5/2

=-3 +10×5/2

a_{11 = }-3 +25 = 22

**Q3.In the following APs find the missing term in the boxes.**

Ans. a =2, a_{3 }= 26,a_{2 }= ?

n^{th} term of AP is given by

a_{n}= a +(n -1)d

a_{3}= a +(3 -1)d

26 =2 +2d

d = 12

**a _{2}= 2 +(2 -1)12= 2 +12 =14**

(ii) a_{2}=13, a_{4}=3

n^{th} term of AP is given by

a_{n}= a +(n -1)d

a_{2}= a +(2 -1)d

a +d = 13……(i)

a_{4}= a +(4 -1)d

a +3d =3…….(ii)

Substracting equation (i) from equation (ii)

2d =-10

d = -5

Putting d=-5 in equation (ii)

a +3×-5 = 3

a – 15 =3

a = 18

**a _{3}= a +(2-1)d = 18 -5 =13**

(iii) a =5,

a +3d = 19/2

5 +3d = 19/2

3d = 19/2 -5 =(19 – 10)/2 =9/2

d = 9/6 = 3/2

a_{2}= a +d = 5 +3/2 =(10 +3)/2= 13/2

a_{3}=a +2d = 5 +2×3/2 = 5 + 3 = 8

(iv) a = -4, a_{6}= 6

a_{6}= a + 5d = -4 + 5d

-4 + 5d = 6

5d = 6 +4 =10

d = 2

a_{2}= a + d = -4 + 2 = -2

a_{3}= a + 2d = -4 + 2×2 = -4 +4 =0

a_{4}= a + 3d = -4 + 3×3 = -4 +9 = 5

a_{5}= a + 4d = -4 + 4×2 = -4 + 8 = 4

(v) a_{2}=38,a_{6}= -22

a + d = 38 ….(i) and a + 5d = -22……(ii)

Solving both equations

-4d = 60

d = -15

a-15= 38

a = 38 + 15 =53

a_{3}= a + 2d = 53 + 2 ×-15 = 53 -30 = 23

a_{4}= a + 3d = 53 + 3×-15 = 53 -45 = 8

a_{5}= a + 4d = 53 + 4 ×-15 = 53 – 60 = -7

**Q4. Which term of the A.P. 3, 8, 13, 18, … is 78?**

Ans. Let the n^{th} term of the given AP is 78

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a =3, d = 8 -3 = 5

78 = 3 + (n – 1)5

(n – 1)5 = 78 -3 =75

n -1 = 75/5 = 15

n = 15 +1 = 16

**Hence 16 ^{th} term of the given AP is 78**

**Q5.Find the number of terms in each of the following A.P.**

**(i) 7, 13, 19, …, 205**

Ans. Let there are n terms in the given AP

n^{th} term of AP is given by

a_{n}= a +(n -1)d

The given AP is 7, 13, 19, …, 205

Where a = 7, d = 13 – 7 = 6 and a_{n}=205

205 = 7 + (n – 1)6

(n – 1)6 = 205 – 7 = 198

n – 1 = 198/6 = 33

n = 33 +1 =34

**Hence there are 34 terms in the given AP**

Ans. The given AP is

Where

a = 18,a_{n}= -47

n^{th} term of AP is given by

a_{n}= a +(n -1)d

-47 = 18 + (n – 1)×-2.5

(n – 1)×-2.5 = 18 +47 =65

n – 1 = 65/2.5 =26

n = 26 +1 =27

**Hence there are 27 terms in the given AP**

**Q6.Check whether -150 is a term of the A.P. 11, 8, 5, 2, …**

Ans. The given AP is 11, 8, 5, 2, …

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a =11, d = 8 – 11 = -3

Let -150 is n^{th} term of the given AP

-150 = 11 + (n – 1)×-3

(n – 1)×-3 = -11 -150 = -161

n – 1 = 161/3

n = 161/3 +1 =(161 +3)/3 =164/3

**Since n =164/3 is not a natural number therefore -150 can’t be the term of the given AP**

**Q7. Find the 31 ^{st} term of an A.P. whose 11^{th} term is 38 and the 16^{th} term is 73**

Ans.It is given that 11^{th} term of the AP is 38 and 16^{th} term is 73

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a is the first term and d is the common difference

a_{11}= a +(11 -1)d

a +10d = 38…..(i)

a_{16}= a +(16 -1)d

a +15d = 73…..(ii)

Substracting equation (i) from equation (ii)

5d = 35

d = 7

Putting the value of d in equation (i)

a +10×7 = 38

a +70 = 38

a = 38 – 70 =-32

a_{31}= a +(31 -1)d= a +30d

Putting the value of a and d

a_{31}= -32+30×7 = -32 +210 =178

**Hence there are 31 ^{th} term in the given AP is 178**

** Q8.An A.P. consists of 50 terms of which 3 ^{rd} term is 12 and the last term is 106. Find the 29^{th} term.**

Ans. Total terms of the AP are 50 in which

3^{rd} term is 12

a_{3}= a + (3-1)d

a +2d = 12 …….(i)

50^{th} term is 106

a + 49d = 106….(ii)

Substracting equation (i) from the equation (ii)

47d =94

d = 2

Putting the value of d in equation (i)

a +2×2 = 12

a +4 = 12

a = 12 -4 = 8

29^{th} term is a_{29}= ?

a_{29}= a + (29 -1)d = a + 28d

a_{29}= 8 +28×2

a_{29}= 8 +56 = 64

Hence 29^{th} term is 64

**Q9.If the 3 ^{rd} and the 9^{th} terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.**

Ans.It is given that 9^{th} term of the AP is 4 and9^{th} term is -8

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a is the first term and d is the common difference

a_{3}= a +(3-1)d

a +2d = 4…..(i)

a_{9}= a +(9 -1)d

a +8d = -8…..(ii)

Substracting equation (i) from equation (ii)

6d = -12

d = -2

Putting the value of d in equation (i)

a +2×-2 = 4

a -4 = 4

a = 4 +4 =8

0= 8 +(n -1)×-2

Putting the value of a and d

(n -1)×-2 = -8

n -1 = 4

n = 5

**Hence 5 ^{th} term of the AP is 0**

**Q10.If 17 ^{th} term of an A.P. exceeds its 10^{th} term by 7. Find the common difference.**

Ans.n^{th} term of AP is given by

a_{n}= a +(n -1)d

17^{th} term of the A.P is

a_{17}= a +(17 -1)d = a + 16d

10^{th} term of the A.P is

a_{10}= a +(10 -1)d = a + 9d

According to the question

a_{17}= a_{10}+7

a + 16d = a + 9d +7

16d -9d = 7

7d = 7

d = 1

**Hence common difference of the AP is 1**

**Q11.Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54 ^{th} term?**

Ans. Let n^{th} term of the given A.P is 132 more than its 54^{th} term

The given A.P is 3, 15, 27, 39,..

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a =3, d = 15 -3 =12

**54 ^{th} term of the given is**

a_{54}= 3 +(54 -1)12

a_{54}= 3 +53×12 = 3 + 636= 639

a_{n}= a_{54}+132 =639 +132 =771

a +(n -1)d =771

3 + (n -1)12 = 771

(n -1)12 = 771 -3 = 768

n -1 = 64

**n = 64 +1 = 65**

**Q12.Two APs have the same common difference. The difference between their 100 ^{th} term is 100, what is the difference between their 1000^{th} terms?**

Ans. Let two AP’s are a , a +d, a +2d…….. and a’, a’ +d, a’ +2d…….

n^{th} term of AP is given by

a_{n}= a +(n -1)d

**100 ^{th} term of AP’s are**

a_{100}= a +(100 -1)d = a + 99d

a_{100}= a+ 99d…..(i)

a’_{100}= a’ +(100 -1)d = a’ + 99d

a’_{100}= a’+ 99d…..(ii)

Substracting equation (ii) from equation (i)

a_{100}– a’_{100}= a +99d -(a’+ 99d) =a – a’

100 =a – a’

a – a’ = 100…..(iii)

Let’s found the difference between their 1000^{th} terms,that is a_{1000}– a’_{1000}= ?

a_{1000}= a+ 999d…..(iv)

a’_{1000}= a’+ 999d…..(v)

a_{1000}– a’_{1000}= a +999d -(a’+ 999d) =a – a’

a_{1000}– a’_{1000}= a – a’

From equation(iii)

a_{1000}– a’_{1000}= 100

**The difference between the 1000 ^{th} terms of both AP’s is 100**

**Q13.How many three-digit numbers are divisible by 7?**

Ans. Let there are n three-digit numbers divisible by 7

To find the first three-digit number divisible by 7, let’s divide three smallest 3 digit numbers (i.e 100) by 7,we get remainder 2 then we have to add(7-2=5) to dividend(i.e 100) that is 105

The largest three-digit number is 999

The last three-digit number divisible number is computed as

Dividing 999 by, we get the remainder 5

Substracting 5 from 999(999 -5=994) is the last three digit number divisible by 7

The AP is formed according to question is

105,112,119……994

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a =105, d = 112 -105 =7,a_{n}= 994

994 = 105 + (n-1)7

(n-1)7 = 994 -105 = 889

n-1 = 889/7 =127

n = 127 +1 =128

**Hence there are 128 three-digit numbers that are divisible by 7?**

**Q14.How many multiples of 4 lie between 10 and 250?**

Ans. Dividing 10 by 4,we get remiender 2

Therefore first multiple of 4 is = 10 +(4 -2) =10+2 =12

To get the last number between 10 and 250,divisible by 4,dividing 250 by 4,the remiender is 2

The last number beteen 10 and 250 divisible by 4 is =250 -2 =248

Therefore multiples of 4 lie between 10 and 250 are

12,16,20……..248

n^{th} term of AP is given by

a_{n}= a +(n -1)d

a_{n}=48, a = 12,d = 16-12 =4

248= 12 +(n -1)×4

(n -1)×4 = 248 – 12 =236

n -1 = 236/4=59

n = 60

**Hence there are 60 multiples of 4 lie between 10 and 250.**

**Q15.For what value of ****n****, are the ****n**^{th}** terms of two APs 63, 65, 67, ….and 3, 10, 17, … equal?**

Ans. It is given that the *n*^{th} terms of two APs 63, 65, 67, ….and 3, 10, 17, … equal

n^{th} term of AP is given by

a_{n}= a +(n -1)d

The n^{th} term of first series,in which a =63,d =65-63 =2

a_{n}= 63 +(n -1)2

The n^{th} term of second series,in which a =3,d =10-3 =7

a’_{n}= 3 +(n -1)7

a_{n}= a’_{n}

63 +(n -1)2 = 3 +(n -1)7

(n -1)2 – (n -1)7 = 3 – 63 = -60

2n – 2 -7n + 7 = -60

-5n +5 = -60

-5n = -65

n = 13

**Therefore 13 ^{th} term of both AP’s are equal**

**Q16.Determine the A.P. whose third term is 16 and the 7 ^{th} term exceeds the 5^{th} term by 12.**

Ans.n^{th} term of AP is given by

a_{n}= a +(n -1)d

3 rd term of AP= 16

a_{3}= a +(3 -1)d = a +2d

a +2d = 16….(i)

**7 ^{th} term is**

a_{7}= a +(7-1)d= a + 6d

**5 ^{th} term is**

a_{5}= a +(5-1)d= a + 4d

According to question

**7 ^{th} term = 5^{th} term +12**

a + 6d = a + 4d +12

2d = 12

d =6

Putting value of d in equation (i)

a +2×6 = 16

a +12= 16

a = 4

Therefore AP is

4,10,16…….

**Q17.Find the 20 ^{th} term from the last term of the A.P. 3, 8, 13, …, 253.**

Ans. The given AP is 3, 8, 13, …, 253.

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Since we have to find 20^{th} term from the last term of the A.P,therefore reversing the AP

253,248,243……..13,8,3

Where a = 253, d = 3 -8=-5, a_{n}= 3

a_{20}= 253 +(20 -1)×-5 =253 +19×-5 =253 – 95 =158

**Hence 20 ^{th} term from the last term of the given A.P is 158**

**Q18.The sum of 4 ^{th} and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 44. Find the first three terms of the A.P**.

Ans.n^{th} term of AP is given by

a_{n}= a +(n -1)d

**4 ^{th} term of the AP is**

a_{4}= a +(4 -1)d = a +3d

**8 ^{th} term of the AP is**

a_{8}= a +(8 -1)d = a +7d

According to first condition of the question

a +3d + a +7d =24

2a + 10d = 24

a + 5d = 12…….(i)

**6 ^{th} term of the AP is**

a_{6}= a +(6 -1)d = a +5d

**10 ^{th} term of the AP is**

a_{10}= a +(10 -1)d = a +9d

According to second condition of the question

a +5d + a +9d=44

2a + 14d = 44

a + 7d = 22…….(ii)

Substracting equation (i) from equation (ii)

2d =10

d = 5

Putting the value of d in equation (i)

a +5×5 =12

a +25 =12

a = 12 – 25 = -13

**Hence first three terms of the AP are -13,-8,-3**

**Q19.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?**

Ans.Let the salary of Subha Rao reach Rs 7000 after n years,therefore last term is 7000

Since it is given that his salary initial salary was Rs 5000 in 1995,therefore first term of the AP is 5000

The increment is given Rs 200, therefore common difference is 200

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a_{n}=7000, a =5000, d =200

7000= 5000 +(n -1)200

(n -1)200 = 7000 – 5000 = 2000

n -1 = 2000/200 = 10

n = 10 +1 = 11

**Subha Rao will achieve his salary after 11 years and his income reach Rs 7000 in the year (1995 +11=2006)**

**Q20.Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the ****n**^{th}** week, her weekly savings become Rs 20.75, find ****n.**

Ans.Let saving of Ramkali become Rs 20.75 in n weeks

Ramkali saved Rs 5 in the first week of a year, therefore first term of AP is 5

Her saving increased weekly by Rs 1.75, therefore common difference of AP is 1.75

n^{th} term of AP is given by

a_{n}= a +(n -1)d

Where a_{n}=20.75, a =5, d =1.75

20.75= 5 +(n -1)1.75

(n -1)1.75 = 20.75-5=15.75

n -1 = 15.75/1.75 = 1575/175=9

n = 9 +1 = 10

**Hence saving of Ramkali reaches to Rs 20.75 in 10 weeks**

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