NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2
NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2 is important in understanding the idea of Arithmetic Progression (AP) and determining the nth term of an AP. This exercise lays a solid foundation for the study of the properties of APs and for solving questions related to the topic in examinations. It helps students understand both theoretical and practical concepts of the subject. The given exercises are as per the new CBSE syllabus and exam guidelines to have a proper grasp of the subject. Now, let's dive into this necessary topic for Class 10.
1.0Download NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.2
2.0Introduction Nth Term of an Arithmetic Progression (AP)
Before getting an insight into the nth term of an AP, let’s get a quick look at AP Itself. An Arithmetic Progression (AP) is a series of numbers where the common difference between two consecutive terms remains the same. The common difference is denoted by d. The nth term of an AP can be obtained by using a formula which assists students in determining any term from the series without wasting time.
The nth term of an Arithmetic Progression refers to the general term that represents the value of the term at the n-th position in the sequence. In other words, the n-th term is the unknown term of an AP sequence which can be termed as a term at n–th position. For instance, if you are asked to find the 5th term of a sequence, then n will be 5, or if you need to find the 10th term, then n = 10.
3.0Class 10 Maths Chapter 5 Exercise 5.2 Overview: Key Concepts
Formula to Find nth Term of an AP
The nth term of an AP is denoted with the letter an. The general formula to calculate the nth term can be expressed as:
Here,
- an = nth term
- a = first term
- d = common difference
- n = number of terms in an AP sequence.
Note that if there are n terms in a sequence of arithmetic progression, then an represents the last term of the sequence, which is sometimes denoted with “l”.
Common Difference
A common difference is a constant value in an Arithmetic Progression that is added to each term to get the subsequent terms of the sequence. The common difference of an AP defines the sequence of an AP, meaning for a sequence to be classified as an AP, the difference between consecutive terms must remain constant. Mathematically, common difference can denoted with the letter d and has a formula:
. . . .
Here, a1, a2, a3, and a4 … are the terms of the AP.
The nth Term
Exercise 5.2 uses the formula for the nth term of an AP for checking the validity of certain terms in a sequence. Note that the value of n always comes out to be a non-negative integer, meaning a term can neither be negative nor in decimal expansion. If any number gives the value of n which is either negative or in decimal fraction then that number is not a part of the AP.
Real-Life Problem Solving with AP
In the exercise, the nth-term formula was predominantly used for solving real-life problems like predicting coming terms in an AP, calculating the instalment amount for each instalment, the growth pattern of certain areas, etc.
The concepts mentioned earlier show the versatile nature of arithmetic progression and its nth term making it even more important to master this chapter. You can do this by frequent practice with NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2.
4.0NCERT Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.2 : Detailed Solutions
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n, the nth term of the AP.
Solution
(i) a = 7, d = 3, n = 8
a₈ = a + 7d = 7 + 7 × 3 = 28.
Hence, a₈ = 28.
(ii) a = -18, n = 10, aₙ = 0, d = ?
aₙ = a + (n - 1)d
0 = -18 + (10 - 1)d
18 = 9d => d = 18/9 = 2
Hence, d = 2.
(iii) d = -3, n = 18, aₙ = -5
aₙ = a + (n - 1)d
-5 = a + (18 - 1)(-3)
-5 = a + (17)(-3)
-5 = a - 51
a = 51 - 5 = 46
Hence, a = 46.
(iv) a = -18.9, d = 2.5, aₙ = 3.6
=> a + (n - 1)d = 3.6
=> -18.9 + (n - 1) × (2.5) = 3.6
=> (n - 1) × (2.5) = 3.6 + 18.9 = 22.5
=> n - 1 = 22.5 / 2.5 = 225 / 25 = 9
=> n = 10
(v) a = 3.5, d = 0, n = 105
Then a₁₀₅ = a + 104d = 3.5 + 0 = 3.5
2. Choose the correct option in the following and justify:
(i) 30th term of the AP : 10, 7, 4, ... is
(A) 97
(B) 77
(C) -77
(D) -87
(ii) 11th term of the AP : -3, -1/2, 2, ... is
(A) 28
(B) 22
(C) -38
(D) -48 1/2
Solution
(i) a = 10, d = -3
t₃₀ = a + 29d = 10 + 29 × (-3)
= 10 - 87 = -77
Hence, the correct option is (C).
(ii) a = -3, d = 5/2
t₁₁ = a + 10d = -3 + 10 × (5/2) = 22
Hence, the correct option is (B).
3. In the following APs, find the missing terms :
(i) 2, ..., 26
(ii) ...., 13, ...., 3
(iii) 5, ....., ......, 9 1/2
(iv) -4, ..., ..., ..., ..., 6
(v) ..., 38, ..., ..., ..., -22
Solution
(i) a = 2, a + 2d = 26
=> 2 + 2d = 26
=> 2d = 26 - 2 = 24
=> d = 12
Then the missing term t2 = a + d = 2 + 12 = 14
(ii) a + d = 13
a + 3d = 3
Subtracting (1) from (2), we get
(a + 3d) - (a + d) = 3 - 13
=> 2d = -10
=> d = -5
from (1), a - 5 = 13
=> a = 18
Therefore, the first missing term is 18
The next missing term t3 = t2 + d = 13 + (-5) = 8
(iii) a = 5
a4 = 9 1/2 = 19/2 = a + 3d
=> 19/2 = 5 + 3d
=> 19/2 - 5 = 3d
=> 9/2 = 3d
d = 3/2
a2 = a + d = 5 + 3/2 = 13/2
a3 = a + 2d = 5 + 2(3/2) = 8
Therefore, the missing terms are 13/2 and 8 respectively.
(iv) a = -4
a6 = 6
a + 5d = 6
6 = -4 + 5d
10 = 5d
d = 2
a2 = a + d = -4 + 2 = -2
a3 = a + 2d = -4 + 2(2) = 0
a4 = a + 3d = -4 + 3(2) = 2
a5 = a + 4d = -4 + 4(2) = 4
Therefore, the missing terms are -2, 0, 2, and 4 respectively.
(v) a2 = 38
a6 = -22
38 = a + d
-22 = a + 5d
On subtracting equation (1) from (2), we obtain
-22 - 38 = 4d
-60 = 4d
d = -15
a = a2 - d = 38 - (-15) = 53
a3 = a + 2d = 53 + 2(-15) = 23
a4 = a + 3d = 53 + 3(-15) = 8
a5 = a + 4d = 53 + 4(-15) = -7
Therefore, the missing terms are 53, 23, 8, and -7 respectively.
4. Which term of the AP : 3, 8, 13, 18, ... is 78?
Solution
a = 3, d = 5
Let tn = 78
=> a + (n - 1)d = 78
=> 3 + (n - 1) × 5 = 78
=> 5n - 2 = 78
=> 5n = 80
=> n = 16
Hence, t16 = 78
5. Find the number of terms in each of the following AP's:
(i) 7, 13, 19, ..., 205
(ii) 18, 15 1/2, 13, ..., -47
Solution
(i) a = 7, d = 6, tn = 205
=> a + (n - 1)d = 205
=> 7 + (n - 1) × 6 = 205
=> 6n + 1 = 205
=> 6n = 204
=> n = 34
Hence, 34 terms.
(ii) a = 18
d = a2 - a1 = 15 1/2 - 18
d = (31 - 36) / 2 = -5 / 2
Let there are n terms in this AP.
Therefore, an = -47 and we know that
an = a + (n - 1)d
-47 = 18 + (n - 1)(-5 / 2)
-47 = 18 + (n - 1)(-5 / 2)
-65 = (n - 1)(-5 / 2)
(n - 1) = -130 / -5
(n - 1) = 26
n = 27
Therefore, this given AP has 27 terms in it.
6. Check whether -150 is a term of the AP : 11, 8, 5, 2, ...
Solution:
a = 11, d = -3
Let if possible tn = -150
=> a + (n - 1)d = -150
=> 11 + (n - 1) × (-3) = -150
=> 11 - 3n + 3 = -150
=> 14 - 3n = -150
=> 3n = 14 + 150 = 164
=> n = 164 / 3 = 54 2/3
It is not possible because n has to be a natural number.
Hence, -150 cannot be a term of the AP.
7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution
Given that,
a₁₁ = 38
a₁₆ = 73
We know that,
an = a + (n - 1)d
a₁₁ = a + (11 - 1)d
38 = a + 10d ...(i)
Similarly,
a₁₆ = a + (16 - 1)d
73 = a + 15d ...(ii)
On subtracting (i) from (ii), we obtain
35 = 5d, d = 7
From equation (i),
38 = a + 10 × (7)
38 - 70 = a
a = -32
a₃₁ = a + (31 - 1)d
= -32 + 30(7)
= -32 + 210
= 178
Hence, 31st term is 178.
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution
t₃ = 12, t₅₀ (last term) = 106
=> a + 2d = 12 ...(i)
And a + 49d = 106 ...(ii)
Subtracting (i) from (ii), we get
47d = 106 - 12 = 94 => d = 2
From (i), a + 2 × 2 = 12 => a = 8
t₂₉ = a + 28d = 8 + 28 × 2 = 64
9. If the 3rd and 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution
Given that,
a₃ = 4, a₉ = -8
We know that,
an = a + (n - 1)d
a₃ = a + (3 - 1)d
4 = a + 2d ...(i)
a₉ = a + (9 - 1)d
-8 = a + 8d ...(ii)
On subtracting equation (i) from (ii), we obtain
-12 = 6d, d = -2
From equation (i), we obtain
4 = a + 2(-2)
4 = a - 4
a = 8
Let nth term of this AP be zero.
an = a + (n - 1)d
0 = 8 + (n - 1)(-2)
0 = 8 - 2n + 2
2n = 10
n = 5
Hence, 5th term of this AP is 0.
10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution
a₁₇ - a₁₀ = 7
=> (a + 16d) - (a + 9d) = 7
=> 7d = 7
=> d = 1
Therefore, the common difference is 1.
11. Which term of the AP : 3, 15, 27, 39, ... will be 132 more than its 54th term?
Sol.
a = 3, d = 12
Let us suppose tn = t₅₄ + 132
=> a + (n - 1)d = a + 53d + 132
=> (n - 1)d - 53d = 132
=> (n - 1 - 53)d = 132
=> (n - 54) × 12 = 132
=> n - 54 = 11
=> n = 65
Hence, t₆₅ is 132 more than t₅₄.
12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution
Let the two APs with the same common difference d be:
a, a+d, a+2d, ...
b, b+d, b+2d, ... (a > b)
We are given that:
(100th term of the first AP) - (100th term of the second AP) = 100
=> (a + 99d) - (b + 99d) = 100
=> a - b = 100
Now, (1000th term of the first AP) - (1000th term of the second AP) = (a + 999d) - (b + 999d)
= a - b = 100
13. How many three-digit numbers are divisible by 7?
Solution
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, ...
All are three-digit numbers which are divisible by 7 and thus, all these are terms of an AP having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 - 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows:
105, 112, 119, ..., 994
Let 994 be the nth term of this AP.
a = 105
d = 7
an = 994
n = ?
an = a + (n - 1)d
994 = 105 + (n - 1)7
889 = (n - 1)7
(n - 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Solution
The multiples of 4 between 10 and 250 are 12, 16, 20, 24, ..., 248.
Let these numbers be n.
a = 12, d = 4
tn = 248
=> a + (n - 1)d = 248
=> 12 + (n - 1) × 4 = 248
=> 4n + 8 = 248
=> n = 60
Therefore, 60 multiples of 4 lie between 10 and 250.
15. For what value of n, are the nth terms of two APs 63, 65, 67, ... and 3, 10, 17, ... equal?
Solution
Two APs are 63, 65, 67, ... and 3, 10, 17, ...
From (i), First term = 63 and common difference = 2.
Its nth term = 63 + (n - 1) × 2 = 2n + 61.
From (ii), First term = 3 and common difference = 7.
Its nth term = 3 + (n - 1) × 7 = 7n - 4.
Putting 7n - 4 = 2n + 61
=> 7n - 2n = 61 + 4
=> 5n = 65
=> n = 13
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution
a3 = 16
a + (3 - 1)d = 16
a + 2d = 16 ...(i)
a7 - a5 = 12
[a + (7 - 1)d] - [a + (5 - 1)d] = 12
(a + 6d) - (a + 4d) = 12
2d = 12
d = 6
From equation (i), we obtain
a + 2(6) = 16
a + 12 = 16
a = 4
Therefore, the AP will be
4, 10, 16, 22, ...
17. Find the 20th term from the last term of the AP 3, 8, 13, ..., 253.
Solution
The AP is 3, 8, 13, ..., 253.
Its first term = 3 and the common difference = 5.
Now, the AP in the reverse order will have the first term = 253 and the common difference = -5.
The 20th term from the end of the AP = The 20th term of the AP in the reverse order.
= a + 19d
= 253 + 19 × (-5) = 253 - 95 = 158.
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution
t4 + t8 = 24; t6 + t10 = 44
=> (a + 3d) + (a + 7d) = 24; (a + 5d) + (a + 9d) = 44
=> 2a + 10d = 24; 2a + 14d = 44
We have a + 5d = 12 ...(i)
and a + 7d = 22 ...(ii)
Subtracting (i) from (ii), we get
2d = 10 => d = 5
From (i) a + 5 × 5 = 12, a = -13
t1 = -13, t2 = -8, t3 = -3
19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution
It can be observed that the incomes that Subba Rao obtained in various years are in AP as every year, his salary is increased by ₹ 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, ...
Here, a = 5000
d = 200
Let after nth year, his salary be ₹ 7000.
Therefore, an = a + (n - 1)d
7000 = 5000 + (n - 1)200
200(n - 1) = 2000
(n - 1) = 10
n = 11
Therefore, in the 11th year, his salary will be ₹ 7000.
20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution
= ₹ 5 (savings in the 1st week)
= ₹ 5 + ₹ 1.75 = ₹ 6.75 (savings in the 2nd week)
= ₹ 6.75 + ₹ 1.75 = ₹ 8.50 (savings in the 3rd week)
= ₹ 20.75
=> a + (n - 1)d = 20.75
=> 5 + (n - 1) × 1.75 = 20.75
=> (n - 1) × 1.75 = 15.75
=> n - 1 = 15.75 / 1.75 = 1575 / 175 = 9
=> n = 10
Hence, in the 10th week, Ramkali's savings will be ₹ 20.75.
5.0Benefits of Studying Class 10 Maths Chapter 5 Exercise 5.2
- Helps in logical thinking and analytical reasoning.
- Helps in identifying and analyzing numerical patterns.
- Encourages quicker and more accurate mathematical calculations.
- Questions on AP frequently appear in various entrance tests.
- Useful for higher studies in mathematics and science.