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Class 10 Maths Chapter 5 Exercise 5.2 - Arithmetic Progression NCERT Solutions
Get a clear understanding of Class 10 Maths Chapter 5 with our easy-to-follow NCERT Solutions for Exercise 5.2. This exercise covers Arithmetic Progression and is important for doing well in your exams and competitive entrance tests. The Class 10 Chapter 5 Maths Exercise 5.2 Solutions explain each step in a simple way to help you understand the concepts easily.
These Class 10 Maths Chapter 5 Exercise 5.2 Solutions are created by an experienced CBSE Maths expert, making sure you get accurate and helpful guidance on solving Arithmetic Progression problems.
| a | d | n | an | |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | 28 |
| (ii) | -18 | 2 | 10 | 0 |
| (iii) | 46 | -3 | 18 | -5 |
| (iv) | -18.9 | 2.5 | 10 | 3.6 |
| (v) | 3.5 | 0 | 105 | 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
Solution:
(C) -77
First term of the AP, a = 10, Common difference, d = 7 – 10 = -3, n = 30
The 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
= 10 – 87
= -77
(ii) 4, 10, 16, 22, … is an A.P. If the sum of its first 20 terms is
(A) 2440 (B) 2400 (C) 2480 (D) 2880
Solution:
(D) 2880
First term of the AP, a = 4, Common difference, d = 10 – 4 = 6
Sum of first n terms of AP is given by
Sn = n/2 [2a + (n -1)d]
S20 = 20/2 [2 × 4 + (20 -1) × 6]
S20 = 10 × [8 + 19 × 6]
S20 = 10 × [8 + 114]
S20 = 10 × 122
S20 = 1220
Q3. Find the sum of the first 15 terms of the AP whose nth term is given by an = 3 + 5(n – 1).
Solution:
The nth term of the AP is given by an = a +(n -1)d
Comparing with an = 3 + 5(n – 1), we have
a = 3, d = 5
Sum of first n terms of AP is given by
Sn = n/2 [2a + (n -1)d]
S15 = 15/2 [2 × 3 + (15 -1) × 5]
S15 = 15/2 [6 + 14 × 5]
S15 = 15/2 [6 + 70]
S15 = 15 × 38
S15 = 570
(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.
Solution:
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)
Subtracting 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?
Solution:
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 16th 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
Let there be nth terms in the given AP.
The nth terms of the AP is given by:
an=a+(n−1)dan = a + (n – 1)dan
The given AP is 7, 13, 19, …, 205
Where a=7a = 7, d=13−7=6d = 13 – 7 = 6, and an=205an = 205an
Substitute these values into the formula:
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.
(ii) 18, 15½, 13, …, -47
The given AP is 18, 15½, 13, …, -47
Where a=18a = 18, an=−47an = -47an
Find the common difference dd:
d=15½−18=312−18=−52=−2.5d = 15½ – 18 = 31/2 – 18 = -5/2 = -2.5
The nth term of the AP is given by:
an=a+(n−1)dan = a + (n – 1)dan
Substitute these values into the formula:
-47 = 18 + (n – 1) × (-2.5)
(n – 1) × (-2.5) = -47 – 18 = -65
n – 1 = -65 / -2.5 = 26
n = 26 + 1 = 27
Hence, there are 27 terms in the given AP.
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.
Solution:
It is given that 3rd term of the AP is 4 and 9th 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)
Subtracting 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.
Solution:
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?
Solution:
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?
Solution:
Let two AP’s be 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)
Subtracting equation (ii) from equation (i)
a100 – a’100 = a + 99d – (a’ + 99d) = a – a’
100 = a – a’
a – a’ = 100…..(iii)
Let’s find 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’
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?
Solution:
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 7, we get the remainder 5
Subtracting 5 from 999 (999 – 5 = 994) is the last three-digit number divisible by 7
The AP is formed according to the 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?
Solution:
Dividing 10 by 4, we get remainder 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 remainder is 2
The last number between 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?
Solution:
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 APs are equal.
Q16.Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
nth term of AP is given by
an= a + (n – 1)d
3rd 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 the question
7th term = 5th term + 12
a + 6d = a + 4d + 12
2d = 12
d = 6
Putting the 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.
Solution:
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 the 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.
Solution:
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 the 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 the second condition of the question
a + 6d + a + 9d = 44
2a + 15d = 44
a + 7.5d = 22…….(ii)
Subtract equation (i) from equation (ii)
a + 7.5d – (a + 5d) = 22 – 12
2.5d = 10
d = 4
Putting the value of d in equation (i)
a + 5 × 4 = 12
a + 20 = 12
a = -8
Therefore first three terms of the AP are -8, -4, 0
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?
Solution:
Let the salary of Subba Rao reach Rs 7000 after n years, therefore last term is 7000.
Since it is given that his initial salary was Rs 5000 in 1995, therefore first term of the AP is 5000.
The increment is given as 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
Subba Rao will achieve his salary after 11 years and his income will 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.
Solution:
Let the 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 = 9
n = 9 + 1 = 10
Hence saving of Ramkali reaches Rs 20.75 in 10 weeks.
Understanding Class 10 Maths Chapter 5 Exercise 5.2 is key to getting good at Arithmetic Progression (AP). Our easy-to-follow Class 10 Maths Chapter 5 Exercise 5.2 Solutions will guide you through each problem step-by-step. With these solutions, Arithmetic Progression Exercise 5.2 will seem much easier. Be sure to download the PDF for Class 10 Chapter 5 Maths Exercise 5.2 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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