**NCERT Solutions for Class 10 Maths Exercise 5.3 Chapter 5 Arithmetic Progression**

NCERT Solutions for Class 10 Maths Exercise 5.3 of Chapter 5 Arithmetic Progression are introduced here by Future Study Point for helping class 10 maths students of CBSE to help their groundwork preparations for the class 10 CBSE Board exam Term 2. NCERT Solutions for Class 10 Maths Exercise 5.3 Chapter 5 Arithmetic Progression are the best review inputs for clearing the ideas on Arithmetic Progression. The questions on Arithmetic Progression are based on our day-to-day existence issues on various sorts of points where we really want to predict actual amounts of physical quantity which are ordered in a sequence. All NCERT Solutions for Class 10 Maths Exercise 5.3 Chapter 5 Arithmetic Progression are made here by an accomplished CBSE Maths instructor by a bit by bit strategy according to the standards of CBSE, in this manner, class 10 maths students can undoubtedly comprehend the solutions.

**NCERT Solutions for Class 10 Maths Exercise 5.3 Chapter 5 Arithmetic Progression**

** Q1.Find the sum of the following APs.**

**(i) 2, 7, 12 ,…., to 10 terms.****(ii) – 37, -33, -29 ,…, to 12 terms****(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms****(iv) 1/15, 1/12, 1/10, …… , to 11 terms**

Ans.** (i) The given AP is 2, 7, 12 ,…., to 10 terms.**

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where n = 10, first term,a = 2, common difference,d = 7 -2 = 5

S_{10}= 10/2[2×2 + (10 – 1)d]

**S _{10}= 5[4 + 9×5]= 5(4 +45) =5×49 =245**

**(ii) The given AP is** **– 37, -33, -29 ,…, to 12 terms**

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where n = 12, first term,a = -37, common difference,d = -33 +37 = 4

S_{10}= 12/2[2×-37 + (12 – 1)×4]

**S _{10}= 6[-74 + 11×4]= 6(-74 +44) =6×-30 =-180**

**(iii) The given AP is** **0.6, 1.7, 2.8 ,…….., to 100 terms**

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where n = 100, first term,a = 0.6, common difference,d = ** 1.7** -0.6 = 1.1

S_{10}= 100/2[2×0.6 + (100 – 1)×1.1]

**S _{10}= 50[1.2 + 99×1.1]= 50(1.2 +108.9) =50×110.1 =5505**

**(iv) The given AP is 1/15, 1/12, 1/10, …… , to 11 terms**

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where n = 11, first term,a = 1/15, common difference,d =1/12 -1/15= (5-4)/60 =1/60

S_{11}= 11/2[2×1/15 + (11 – 1)×1/60]

**S _{10}=** 11/2[2/15 + 10/60]= 11/2(2/15 +1/6) =11/2(4+5)/30) =(11×9)/60=99/60

**=33/20**

**Hence the sum of 11 terms of the given AP is 33/20**

**Q2.Find the sums given below**:

**(ii) 34 + 32 + 30 + ……….. + 10**

(iii) – 5 + (− 8) + (- 11) + ………… + (- 230)

Ans. The given AP is

Where a = 7,

n^{th} term of a AP is given by

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

7 +(n -1)×7/2 = 84

(n -1)×7/2 = 84 – 7 = 77

n -1 = 77×2/7 =22

n = 23

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

= 23/2[2×7 + (23 – 1)×7/2]

= 23/2[14 + 22×7/2]

= 23/2[14 + 77]

= 23/2 ×91

=2093/2

**(ii)The given AP is 34 + 32 + 30 + ……….. + 10**

n^{th} term of a AP is given by

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

Where a = 34, d = 32 – 34 = -2 and a_{n}=10

34 +(n -1)×-2 = 10

-2n + 2 = 10 – 34 = -24

-2n = -26

n = 13

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

= 13/2[2×34 + (13 – 1)×-2]

= 13/2[68 -12×2]

=13/2[68 -24]

=13/2×44

=13× 22 =286

**Hence sum of the given AP is 286**

(iii) The given AP is – 5 + (− 8) + (- 11) + ………… + (- 230)

n^{th} term of a AP is given by

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

Where a = -5, d = -8 +5= -3 and a_{n}=-230

-230= -5 + (n -1)×-3

(n -1)×-3 = -235+5 =-225

n -1 = -225/-3 = 75

n = 75+1 =76

The sum of n terms of the AP is given by

S_{n}= 76/2[2×-5 + (76 – 1)×-3]

= 76/2[-10 + 75×-3]

= 76/2[-10-225]= 38×-235=-8930

**S _{n}=-8930**

**Q3.In an AP**

(i) Given *a* = 5, *d* = 3, *a _{n}* = 50, find

*n*and

*S*.

_{n}(ii) Given

*a*= 7,

*a*

_{13}= 35, find

*d*and

*S*

_{13}.

(iii) Given

*a*

_{12}= 37,

*d*= 3, find

*a*and

*S*

_{12}.

(iv) Given

*a*

_{3}= 15,

*S*

_{10}= 125, find

*d*and

*a*

_{10}.

(v) Given

*d*= 5,

*S*

_{9}= 75, find

*a*and

*a*

_{9}.

(vi) Given

*a*= 2,

*d*= 8,

*S*= 90, find

_{n}*n*and

*a*.

_{n}(vii) Given

*a*= 8,

*a*= 62,

_{n}*S*= 210, find

_{n}*n*and

*d*.

(viii) Given

*a*= 4,

_{n}*d*= 2,

*S*= − 14, find

_{n}*n*and

*a*.

(ix) Given

*a*= 3,

*n*= 8,

*S*= 192, find

*d*.

(x) Given

*l*= 28,

*S*= 144 and there are total 9 terms. Find

*a*.

Ans. (i) We are given *a* = 5, *d* = 3, *a _{n}* = 50

n^{th} term of a AP is given by

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

Where a = 5, d = 3 and a_{n}=50

50 = 5 + (n -1)3

(n -1)3 = 50 – 5 =45

n -1 = 15

n = 15 +1=16

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

= 16/2[2×5 + (16 – 1)3]

= 8(10 +15×3)

=8(10+45) = 8×55 =440

S_{n}= 440

Ans. (ii) We are given *a* = 7, *a*_{13} = 35

n^{th} term of a AP is given by

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

Where a = 7, and a_{13}=35⇒n =13

a_{13}= a + (13 -1)d

35 = 7 + 12d

12d = 35 -7 =28

d = 28/12 =7/3

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

= 13/2[2×7 + (13 – 1)×7/3]

= 13/2(14 +12×7/3)

=13/2(14+28) = 13/2×42 =13×21 =273

S_{n}= 273

**(iii)Given a_{12} = 37, d = 3,**

n^{th} term of a AP is given by

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

Where d = 3 and a_{12}=37⇒n=12

37 = a + (12 -1)3

a+11×3 = 37

a+33= 37

a = 37 -33=4

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

⇒n/2(a +l) where l = a + (n -1)d

=12/2(4 +37)

=6×41 =246

**NCERT Solutions for Class 10 Maths Exercise 5.3 Chapter 5 Arithmetic Progression**

**(iv)Given a_{3} = 15, S_{10} = 125,**

n^{th} term of a AP is given by

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

Where and a_{3}=15⇒n=3

15= a + (3 -1)d

a +2d = 15…..(i)

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where* S*_{10} = 125⇒n =10

S_{10}= 10/2[2a + (10 – 1)d]

125 =5(2a +9d)

2a +9d =25……(ii)

Multiplying equation (i) by 2 and subtracting it from equation (ii)

5d =-5

d = -1

Putting d =-1 in equation (i)

a +2×-1 = 15

a -2 = 15

a = 15 +2 =17

a_{10}= 17 + (10 -1)×-1 =17 +9×-1 =17 -9 =8

**(v)Given d = 5, S_{9} = 75,**

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where* S*_{9} = 75⇒n =9

S_{9}= 9/2[2a + (9 – 1)5]

75 =9/2(2a +40)

2a +40 =150/9 =50/3

2a =50/3 -40 =(50 -120)/3=-70/3

a = -35/3

n^{th} term of a AP is given by

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

a_{9}= -35/3 + (9 -1)5 =-35/3 +40 =(-35 +120)/3 =85/3

**(vi)Given a = 2, d = 8, S_{n} = 90**

n^{th} term of a AP is given by

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

Where a=2, d = 8

a_{n}= 2 + (n -1)8….(ii)

It is given that

S_{n} = 90

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

90=n/2(a +a_{n}) …….(ii) ,where a_{n}= a + (n -1)d

From equation (i)

90=n/2[2 +2 + (n -1)8]

n[2 +2 + (n -1)8] =180

4n +n(n -1)8 =180

8n² -8n +4n =180

8n² – 4n -180 =0

2n² – n – 45 =0

2n² – 10n+9n – 45 =0

2n(n – 5) + 9(n – 5) =0

(n – 5)(2n +9) =0

n =5,n = -9/2(canceling it since -9/2 is not a natural no.)

a_{n}= 2 + (n -1)8 =2 +(5 -1)8 =2 +4×8 = 2+32 =34

(vii) Given that *a* = 8, a_{n} = 62, *S _{n}* = 210

n^{th} term of a AP is given by

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

Where a=8, a_{n} = 62

62 = 8 + (n -1)d

(n -1)d = 54….(i)

The sum of n terms of an AP is given by

S_{n}= n/2[2a + (n – 1)d]

From equation (i)

210=n/2(2×8 +54)=n/2(16 +54)=n/2(70) =35n

35n = 210

**n =210/35 =6**

Putting the value n =6 in the equation (i)

(6 -1) d =54

5d =54

**d = 54/5 =10.8**

(viii) Given that *a _{n}* = 4,

*d*= 2,

*S*= − 14

_{n}n^{th} term of a AP is given by

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

Where d=2, a_{n} = 4

4 = a + (n -1)2

2n+a = 6….(i)

The sum of n terms of an AP is given by

*S _{n}* = n/2[2a + (n – 1)2]

Putting the value a =6-2n from the equation (i)

− 14= n/2[2(6 -2n) + (n – 1)2]

-14 = n/2[12 -4n+ 2n – 2]

**-14 = n/2(10 – 2n)**

-28 = n(10 -2n)

10n – 2n² +28 = 0

2n² – 10n -28 =0

n² – 5n – 14 =0

n² – 7n +2n- 14 =0

n(n -7) + 2(n – 7) =0

(n -7)(n +2) =0

**n =7,** -2 (neglecting it since -2 is not a natural number)

Putting the value n =7 in the equation (i)

2×7+a = 6

14 +a =6

**a = 6 -14 = -8**

**(ix) Given that a = 3, n = 8, S = 192,**

The sum of n terms of an AP is given by

*S _{n}* = n/2[2a + (n – 1)d]

192 = 8/2[2×3 +(8 -1)d]

4(6 +7d) = 192

6 +7d = 192/4 =48

7d = 48 – 6 =42

**d = 42/7 =6**

**(x) Given that l = 28, S = 144 and there are total 9 terms**

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

= n/2(a +l), where l = a + (n -1)d

144=9/2(a +28)

a +28 = (144 ×2)/9 =16×2 = 32

**a = 32 – 28 = 4**

**Q4.How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?**

Ans.Let there are n terms of the AP. 9, 17, 25 …to give sum of 636

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

Where a =9, d = 17 – 9 =8 and S_{n}= 636

S_{n}= n/2[2×9 +(n -1)8]

636= n/2[18 +(n -1)8]

n/2[18 +(n -1)8] =636

9n + 4n² -4n -636 = 0

4n² + 5n – 636 = 0

4n² + 53n -48n- 636 = 0

n(4n +53) – 12(4n + 53) =0

(4n +53)(n -12) =0

n = -53/4, n = 12

Neglecting n = -53/4, since it is not a natural number

**Hence there are the first 12 terms in the AP to give the sum of 636**

**Q5.The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.**

Ans.The sum of n terms of the AP is given by

S= n/2[2a + (n – 1)d]

400=n/2(a +a_{n}) ,where a_{n}= a + (n -1)d

Where a =5, a_{n}= 45 and S =400

400=n/2(5+45) =n/2(50) =25n

**n = 400/25 =16 **

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

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

45 = 5 + (16 -1)d

(16 -1)d = 40

15d =40

**d = 40/15 =8/3**

**Q6.The first and the last term of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?**

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

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

Where a_{n}= 350, a = 17,d =9

350= 17 + (n -1)9 = 17 +9n – 9

9n + 8 = 350

9n = 350 – 8 = 342

**n = 342/9 = 38**

The sum of n terms of the AP is given by

S= n/2[2a + (n – 1)d]

S=38/2(a +a_{n}) ,where a_{n}= a + (n -1)d

Where a =17, a_{n}= 350

**S=19(17 +350) =19×367 =6973**

**Q7. Find the sum of first 22 terms of an AP in which ****d**** = 7 and 22 ^{nd} term is 149.**

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

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

Where a_{22}= 149,d =7,n =22

149= a + (22 -1)7

149= a + 21×7 = a +147

**a = 149 – 147 = 2**

The sum of n terms of the AP is given by

S= n/2[2a + (n – 1)d]

S=n/2(a +a_{n}) ,where a_{n}= a + (n -1)d

Putting the value n =22,a =2 and a_{n}=22

Sum of 22 terms is

**S=22/2(2+149)=11×151 = 1661**

**Q8.Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18, respectively.**

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

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

Where a_{2}= 14, a_{3}= 18 =7

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

*Substracting equation (i) from the equation (ii)*d = 4 Putting the valu3 d =4 in the equation (i) a +4 =14 a = 10 The sum of n terms of the AP is given by S= n/2[2a + (n – 1)d] The sum of 51 terms of the APS = 51/2[2×10 +(51 -1)4]

**S= 51/2[20 +50×4]= 51/2[20 +200]= 51/2×220 =51×110 =5610**

** Q9.If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first ****n**** terms.**

Ans.The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

n = 7, S_{7}= 49, , S_{17}= 289

S_{7}= 7/2[2a + (7- 1)d]

7/2[2a + 6d] = 49

a +3d = 7……(i)

S_{17}= 17/2[2a + (17- 1)d]

17/2[2a +16d] = 289

a +8d = 17……..(ii)

Substracting equation (i) from equation (ii)

5d =10

d =10/5 =2

Putting the value of d in the equation (i)

a +3×2 =7

a = 1

The sum of n terms of the AP is given by

S_{n}= n/2[2×1 + (n – 1)2]

=n/2[2×1 + 2n – 2]

=n/2[2 -2 +2n]

=n/2[2n]

=n²

Q10.Show that a_{1}, a_{2}, ……. a_{n},…… form an AP where a_{n} is defined as below:

(i) a_{n} = 3 + 4n

(ii) a_{n} = 9 – 5n

Also find the sum of the first 15 terms in each case

Ans.(i) n^{th} term of the given AP ,a_{n} is defined as

a_{n} = 3 + 4n

a_{1} = 3 + 4×1 =3 +4 =7

a_{2} = 3 + 4×2 = 3 + 8 =11

a_{3} = 3 + 4×3 = 3 +12 =15

…………. and so on

d = a_{2}-a_{1} =11 -7 =4

a = a_{1} =7

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

The sum of the first 15 terms

S_{15}=15/2[2×7 + (15-1)4]

=15/2[14 +14×4]

=15/2[14 +56]

**S _{15}=15/2(70) =15×35 =525**

(ii) n^{th} term of the given AP ,a_{n} is defined as

a_{n} = 9 – 5n

a_{1} = 9 – 5×1 =9 -5 =4

a_{2} = 9 – 5×2 = 9 – 10 =-1

a_{3} = 9 – 5×3 = 9 -15 =-6

…………. and so on

d = a_{2}-a_{1} =-1 -4 =-5

a = a_{1} =4

The sum of n terms of the AP is given by

S_{n}= n/2[2a + (n – 1)d]

The sum of the first 15 terms

S_{15}=15/2[2×4 + (15-1)×-5]

=15/2[8 +14×-5]

=15/2[8 -70]

**S _{15}=15/2(-62) =15×-31 =-465**

** Q11.If the sum of the first ****n**** terms of an AP is 4****n**** − ****n**^{2}**, what is the first term (that is ****S**_{1}**)? What is the sum of first two terms? What is the second term? Similarly find the 3 ^{rd}, the10^{th} and the **

**n**

^{th}**terms.**

Ans. The sum of the first *n* terms of an AP is 4n − n^{2}

S_{n}=4n − n^{2}

S_{1}=4×1 − 1^{2}= 4 – 1=3, first term,a =3

S_{2}=4×2 − 2^{2}= 8- 4=4,secod term = S_{2}-S_{1}=4 -3 =1

**S _{3}**=4×3 − 3

^{2}= 12- 9=3,

**third term**= S

_{3}-S

_{2}=3 -4 =-1

S_{9}=4×9 − 9^{2}= 36- 81=-45

**S _{10}**=4×10 − 10

^{2}= 40- 100=-60,

**10**

^{th}**term**= S

_{10}-S

_{9}=-60+45 =-15

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

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

Where first term,a =3,d =1-3 =-2

**a _{n}= 3 + (n -1)×-2 =3 -2n +2 =5 -2n**

The NCERT questions of the whole of the chapter 5 Arithmetic Progression are based on the two formulas

The n^{th} term of the AP a ,a +d,a +2d ,……….a_{n }is given as

**a _{n}= a + (n -1)d**

Sum of n terms of the AP a ,a +d,a +2d ,……….a_{n }is given as

**S _{n}=n/2 [2a + (n -1)d]**

Where a =first term of the AP, d = common difference, a_{n}=n^{th} term of the AP, and S_{n} is the sum of n terms of an AP

The sum of n terms of an AP can also be rewritten as

**S _{n}=n/2 [a +l]**,where l = a

_{n},the last term of the AP

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