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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
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)
Ans. 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.
Ans. (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
Ans. (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
NCERT Solutions for Class 10 Maths Exercise 5.3 Chapter 5 Arithmetic Progression
(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?
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
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.
Ans.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
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.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.
Ans: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.
Ans.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)d a +2d =18……(ii) 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 =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.
Ans.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
Ans.(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
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