Class 10 Maths Chapter 5 Exercise 5.3 - Arithmetic Progression NCERT Solutions
Class 10 Maths Chapter 5 Exercise 5.3 - Arithmetic Progression NCERT Solutions with PDF
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 are provided here by Future Study Point to assist students in their preparations for the CBSE Board exam Term 2. These solutions for Class 10 Maths Exercise 5.3 of Chapter 5 Arithmetic Progression offer the best revision materials, helping students thoroughly understand the concepts of Arithmetic Progression.
The problems in Exercise 5.3 are based on real-life scenarios where students need to predict quantities arranged in a sequence. All solutions are crafted by an experienced CBSE Maths teacher, following a step-by-step method according to CBSE guidelines. This ensures that students can easily grasp the solutions for Class 10 Chapter 5 Maths Exercise 5.3.
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Class 10 Chapter 5 Maths - Arithmetic Progression: Find Links to All Exercises NCERT Solutions
Class 10 Maths Chapter 5 Exercise 5.3 - Arithmetic Progression NCERT Solutions
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
Solution:
(i) The given AP is 2, 7, 12 ,…., to 10 terms.
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
Where n = 10, first term,a = 2, common difference,d = 7 -2 = 5
S10= 10/2[2×2 + (10 – 1)d]
S10= 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
Sn= n/2[2a + (n – 1)d]
Where n = 12, first term,a = -37, common difference,d = -33 +37 = 4
S10= 12/2[2×-37 + (12 – 1)×4]
S10= 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
Sn= n/2[2a + (n – 1)d]
Where n = 100, first term,a = 0.6, common difference,d = 1.7 -0.6 = 1.1
S10= 100/2[2×0.6 + (100 – 1)×1.1]
S10= 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
Sn= 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
S11= 11/2[2×1/15 + (11 – 1)×1/60]
S10= 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)
Solution:
The given AP is
Where a = 7,
nth term of a AP is given by
an= 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
Sn= 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
nth term of a AP is given by
an= a + (n -1)d
Where a = 34, d = 32 – 34 = -2 and an=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
Sn= 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)
nth term of a AP is given by
an= a + (n -1)d
Where a = -5, d = -8 +5= -3 and an=-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
Sn= 76/2[2×-5 + (76 – 1)×-3]
= 76/2[-10 + 75×-3]
= 76/2[-10-225]= 38×-235=-8930
Sn=-8930
Q3.In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find 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.
Solution:
(i) We are given a = 5, d = 3, an = 50
nth term of a AP is given by
an= a + (n -1)d
Where a = 5, d = 3 and an=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
Sn= n/2[2a + (n – 1)d]
= 16/2[2×5 + (16 – 1)3]
= 8(10 +15×3)
=8(10+45) = 8×55 =440
Sn= 440
(ii) We are given a = 7, a13 = 35
nth term of a AP is given by
an= a + (n -1)d
Where a = 7, and a13=35⇒n =13
a13= 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
Sn= 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
Sn= 273
(iii) Given a12 = 37, d = 3,
nth term of a AP is given by
an= a + (n -1)d
Where d = 3 and a12=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
Sn= n/2[2a + (n – 1)d]
⇒n/2(a +l) where l = a + (n -1)d
=12/2(4 +37)
=6×41 =246
(iv) Given a3 = 15, S10 = 125,
nth term of a AP is given by
an= a + (n -1)d
Where and a3=15⇒n=3
15= a + (3 -1)d
a +2d = 15…..(i)
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
Where S10 = 125⇒n =10
S10= 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
a10= 17 + (10 -1)×-1 =17 +9×-1 =17 -9 =8
(v) Given d = 5, S9 = 75,
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
Where S9 = 75⇒n =9
S9= 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
nth term of a AP is given by
an= a + (n -1)d
a9= -35/3 + (9 -1)5 =-35/3 +40 =(-35 +120)/3 =85/3
(vi) Given a = 2, d = 8, Sn = 90
nth term of a AP is given by
an= a + (n -1)d
Where a=2, d = 8
an= 2 + (n -1)8….(ii)
It is given that
Sn = 90
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
90=n/2(a +an) …….(ii) ,where an= 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.)
an= 2 + (n -1)8 =2 +(5 -1)8 =2 +4×8 = 2+32 =34
(vii) Given that a = 8, an = 62, Sn = 210
nth term of a AP is given by
an= a + (n -1)d
Where a=8, an = 62
62 = 8 + (n -1)d
(n -1)d = 54….(i)
The sum of n terms of an AP is given by
Sn= 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 an = 4, d = 2, Sn = − 14
nth term of a AP is given by
an= a + (n -1)d
Where d=2, an = 4
4 = a + (n -1)2
2n+a = 6….(i)
The sum of n terms of an AP is given by
Sn = 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
Sn = 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
Sn= 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?
Solution:
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
Sn= n/2[2a + (n – 1)d]
Where a =9, d = 17 – 9 =8 and Sn= 636
Sn= 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.
Solution:
The sum of n terms of the AP is given by
S= n/2[2a + (n – 1)d]
400=n/2(a +an) ,where an= a + (n -1)d
Where a =5, an= 45 and S =400
400=n/2(5+45) =n/2(50) =25n
n = 400/25 =16
nth term of an AP is given by
an= a + (n -1)d
45 = 5 + (16 -1)d
(16 -1)d = 40
15d =40
d = 40/15 =8/3
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Class 10 Maths Chapter 5 Exercise 5.3 - Arithmetic Progression NCERT Solutions
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?
Solution:
The nth term of an AP is given by
an= a + (n -1)d
Where an= 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 +an) ,where an= a + (n -1)d
Where a =17, an= 350
S=19(17 +350) =19×367 =6973
Q7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
nth term of an AP is given by
an= a + (n -1)d
Where a22= 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 +an) ,where an= a + (n -1)d
Putting the value n =22,a =2 and an=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.
Solution:
nth term of an AP is given by
an= a + (n -1)d
Where a2= 14, a3= 18 =7
a2= a + (2 -1)d = a +d
a + d=14 …..(i)18 =a + (3-1)da +2d =18……(ii)Substracting equation (i) from the equation (ii)d = 4Putting the valu3 d =4 in the equation (i)a +4 =14a = 10The sum of n terms of the AP is given byS= 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 =5610Q9.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.
Solution:
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
n = 7, S7= 49, , S17= 289
S7= 7/2[2a + (7- 1)d]
7/2[2a + 6d] = 49
a +3d = 7……(i)
S17= 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
Sn= 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 a1, a2, ……. an,…… form an AP where an is defined as below:
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also, find the sum of the first 15 terms in each case.
Solution:
(i) nth term of the given AP ,an is defined as
an = 3 + 4n
a1 = 3 + 4×1 =3 +4 =7
a2 = 3 + 4×2 = 3 + 8 =11
a3 = 3 + 4×3 = 3 +12 =15
…………. and so on
d = a2-a1 =11 -7 =4
a = a1 =7
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
The sum of the first 15 terms
S15=15/2[2×7 + (15-1)4]
=15/2[14 +14×4]
=15/2[14 +56]
S15=15/2(70) =15×35 =525
(ii) nth term of the given AP ,an is defined as
an = 9 – 5n
a1 = 9 – 5×1 =9 -5 =4
a2 = 9 – 5×2 = 9 – 10 =-1
a3 = 9 – 5×3 = 9 -15 =-6
…………. and so on
d = a2-a1 =-1 -4 =-5
a = a1 =4
The sum of n terms of the AP is given by
Sn= n/2[2a + (n – 1)d]
The sum of the first 15 terms
S15=15/2[2×4 + (15-1)×-5]
=15/2[8 +14×-5]
=15/2[8 -70]
S15=15/2(-62) =15×-31 =-465
Q11.If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.
Ans. The sum of the first n terms of an AP is 4n − n2
Sn=4n − n2
S1=4×1 − 12= 4 – 1=3, first term,a =3
S2=4×2 − 22= 8- 4=4,secod term = S2-S1=4 -3 =1
S3=4×3 − 32= 12- 9=3,third term = S3-S2=3 -4 =-1
S9=4×9 − 92= 36- 81=-45
S10=4×10 − 102= 40- 100=-60,10th term = S10-S9=-60+45 =-15
nth term of an AP is given by
an= a + (n -1)d
Where first term,a =3,d =1-3 =-2
an= 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 nth term of the AP a ,a +d,a +2d ,……….an is given as
an= a + (n -1)d
Sum of n terms of the AP a ,a +d,a +2d ,……….an is given as
Sn=n/2 [2a + (n -1)d]
Where a =first term of the AP, d = common difference, an=nth term of the AP, and Sn is the sum of n terms of an AP
The sum of n terms of an AP can also be rewritten as
Sn=n/2 [a +l],where l = an,the last term of the AP
Conclusion - Class 10 Maths Chapter 5 Exercise 5.3 - Arithmetic Progression
Understanding Class 10 Maths Chapter 5 Exercise 5.3 is key to getting good at Arithmetic Progression (AP). Our easy-to-follow Class 10 Maths Chapter 5 Exercise 5.3 Solutions will guide you through each problem step-by-step. With these solutions, Arithmetic Progression Exercise 5.3 will seem much easier. Be sure to download the PDF for Class 10 Chapter 5 Maths Exercise 5.3 so you can practice anytime and keep these important concepts handy. Keep practicing, and you’ll do great in your exams!
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