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# 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 an the nth term of the A.P.

 a d n an (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

nth term of an AP is given by

an= a +(n -1)d

an= 7 +(8-1)3

=7 +7 ×3

= 7 +21

=28

(ii) a =-18,d=?,n =10, an=0

nth term of an AP is given by

an= a +(n -1)d

0= -18 +(10-1)d

9d = 18

d = 2

(iii) a =?,d=-3,n =18, an=-5

nth term of an AP is given by

an= 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 =?, an=3.6

nth term of an AP is given by

an= 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, an=?

nth term of an AP is given by

an=3.5 +(n -1)×0

an=3.5

Q2.Choose the correct choice in the following and justify:
(i) 30th 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

nth term of AP is given by

an= a +(n -1)d

30th term of the AP =

a30 =10 +(30-1)×-3

=10 +29×-3

a30=10 -87 = -77

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(ii) 11th 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

nth term of AP is given by

an= a +(n -1)d

11th term of the AP =

a11 =-3 +(11-1)×5/2

=-3 +10×5/2

a11 = -3 +25 = 22

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

Ans. a =2, a3 = 26,a2 = ?

nth term of AP is given by

an= a +(n -1)d

a3= a +(3 -1)d

26 =2 +2d

d = 12

a2= 2 +(2 -1)12= 2 +12 =14

(ii) a2=13, a4=3

nth term of AP is given by

an= a +(n -1)d

a2= a +(2 -1)d

a  +d = 13……(i)

a4= 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

a3= 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

a2= a +d = 5 +3/2 =(10 +3)/2= 13/2

a3=a +2d = 5 +2×3/2 = 5 + 3 = 8

(iv) a = -4, a6= 6

a6= a + 5d = -4 + 5d

-4 + 5d = 6

5d = 6 +4 =10

d = 2

a2= a + d = -4 + 2 = -2

a3= a + 2d = -4 + 2×2 = -4 +4 =0

a4= a + 3d = -4 + 3×3 = -4 +9 = 5

a5= a + 4d = -4 + 4×2 = -4 + 8 = 4

(v) a2=38,a6= -22

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

Solving both equations

-4d = 60

d = -15

a-15= 38

a = 38 + 15 =53

a3= a + 2d = 53 + 2 ×-15 = 53 -30 = 23

a4= a + 3d = 53 + 3×-15 = 53 -45 = 8

a5= a + 4d = 53 + 4 ×-15 = 53 – 60 = -7

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

Ans. Let the nth term of the given AP is 78

nth term of AP is given by

an= 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

nth term of AP is given by

an= a +(n -1)d

The given AP is 7, 13, 19, …, 205

Where a = 7, d = 13 – 7 = 6 and an=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,an= -47

nth term of AP is given by

an= 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, …

nth term of AP is given by

an= a +(n -1)d

Where a =11, d = 8 – 11 = -3

Let -150 is nth 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 31st term of an A.P. whose 11th term is 38 and the 16th term is 73

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

nth term of AP is given by

an= a +(n -1)d

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

a11= a +(11 -1)d

a +10d = 38…..(i)

a16= 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

a31= a +(31 -1)d= a +30d

Putting the value of a and d

a31= -32+30×7 = -32 +210 =178

Hence there are 31th term in the given AP is 178

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

Ans. Total terms of the AP are 50 in which

3rd term is 12

a3= a + (3-1)d

a +2d = 12 …….(i)

50th 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

29th term is a29= ?

a29= a + (29 -1)d = a + 28d

a29= 8 +28×2

a29= 8 +56 = 64

Hence 29th term is 64

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

Ans.It is given that 9th term of the AP is 4 and9th term is -8

nth term of AP is given by

an= a +(n -1)d

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

a3= a +(3-1)d

a +2d = 4…..(i)

a9= 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 5th term of the AP is 0

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

Ans.nth term of AP is given by

an= a +(n -1)d

17th term of the A.P is

a17= a +(17 -1)d = a + 16d

10th term of the A.P is

a10= a +(10 -1)d = a + 9d

According to the question

a17= a10+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 54th term?

Ans. Let nth term of the given A.P is 132 more than its 54th term

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

nth term of AP is given by

an= a +(n -1)d

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

54th term of the given is

a54= 3 +(54 -1)12

a54= 3 +53×12 = 3 + 636= 639

an= a54+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 100th term is 100, what is the difference between their 1000th terms?

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

nth term of AP is given by

an= a +(n -1)d

100th term of AP’s are

a100= a +(100 -1)d = a + 99d

a100= a+ 99d…..(i)

a’100= a’ +(100 -1)d = a’ + 99d

a’100= a’+ 99d…..(ii)

Substracting equation (ii) from equation (i)

a100– a’100= a +99d -(a’+ 99d) =a – a’

100 =a – a’

a – a’ = 100…..(iii)

Let’s found the difference between their 1000th terms,that is a1000– a’1000= ?

a1000= a+ 999d…..(iv)

a’1000= a’+ 999d…..(v)

a1000– a’1000= a +999d -(a’+ 999d) =a – a’

a1000– a’1000= a – a’

From equation(iii)

a1000– a’1000= 100

The difference between the 1000th 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

nth term of AP is given by

an= a +(n -1)d

Where a =105, d = 112 -105 =7,an= 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

nth term of AP is given by

an= a +(n -1)d

an=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 nth terms of two APs 63, 65, 67, ….and 3, 10, 17, … equal?

Ans. It is given that the nth terms of two APs 63, 65, 67, ….and 3, 10, 17, … equal

nth term of AP is given by

an= a +(n -1)d

The nth term of first series,in which a =63,d =65-63 =2

an= 63 +(n -1)2

The nth term of second series,in which a =3,d =10-3 =7

a’n= 3 +(n -1)7

an= 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 13th term of both AP’s are equal

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

Ans.nth term of AP is given by

an= a +(n -1)d

3 rd term of AP= 16

a3= a +(3 -1)d = a +2d

a +2d = 16….(i)

7th term is

a7= a +(7-1)d= a + 6d

5th term is

a5= a +(5-1)d= a + 4d

According to question

7th term = 5th 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 20th term from the last term of the A.P. 3, 8, 13, …, 253.

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

nth term of AP is given by

an= a +(n -1)d

Since we have to find  20th 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, an= 3

a20= 253 +(20 -1)×-5 =253 +19×-5 =253 – 95 =158

Hence 20th term from the last term of the given A.P is 158

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

Ans.nth term of AP is given by

an= a +(n -1)d

4th term of the AP is

a4= a +(4 -1)d = a +3d

8th term of the AP is

a8= 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)

6th term of the AP is

a6= a +(6 -1)d = a +5d

10th term of the AP is

a10= 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

nth term of AP is given by

an= a +(n -1)d

Where an=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 nth 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

nth term of AP is given by

an= a +(n -1)d

Where an=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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