NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3
The NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.3 is a crucial exercise in calculating the sum of the first in terms of an Arithmetic Progression (AP). This NCERT Solutions exercise is useful for learning the concepts related to sequence and series, which help solve complex problems in mathematics and real life as well.
The breakdown of this exercise will guide students in learning these key concepts and applying them to practical problems. The information mentioned here is as per the CBSE syllabus and examination purposes. Let’s dive deeper into this important topic.
1.0Download NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 : Free PDF
2.0Introduction to an Arithmetic Progressions
In mathematics, Arithmetic Progression (AP) refers to a list of numbers such that the difference between any two consecutive terms remains constant. The common difference is the name of this constant, which is represented by d. To understand, for the sequence 5, 8, 11, 14,..., the common difference is 3 since each term is derived from the previous one by adding 3.
Insight Into Sum of First 'n' Terms of an AP
When dealing with Arithmetic Progressions, a useful idea is how to calculate the sum of the first n terms in the series. The total of the first n terms assists in a quick calculation of the overall value of a series of terms without necessarily adding them one by one.
For instance, for real-life applications such as saving cash or interest calculation, we frequently work with sequences where we need to understand the overall sum within a defined period without necessarily adding the terms one by one.
3.0Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.3 Overview: Key Concepts
The formula for the sum of the first n terms of an AP
The sum of n terms of an AP is denoted with the letter “Sn” and can be calculated with the formula given as:
sn=2n[2a+(n−1)d]
Here,
- Sn = Sum of the first n terms
- a = first term of the AP
- d = common difference between consecutive terms of an AP. (d = a2 – a1)
- n = Number of terms
The Relation Between nth Term and Sum of n terms of an AP
The nth term is the value at a given position, and the sum of n terms is the sum of the values up to that term, which represents the total growth of the sequence. According to the exercise, if the last or nth term of an AP is given, then the formula for the sum of the n term of an AP can be simply expressed as:
Sn=2n(a+an)
Remark: From the formula of the sum of n terms, we can conclude that the nth term of an AP is simply the difference between the sum of the first n terms and the sum of the first (n-1) terms. This can be mathematically expressed as:
an=Sn−Sn−1
The sum of Positive Integers
The sum of the first n positive integers is an important formula for deriving the sum of positive integers without the summation of each term one by one. In the exercise, the formula is derived from the formula of the sum of n-terms of a series. The formula can be written as:
Sn=2n(n+1)
Handling series with conditions
Where the sum of terms is given in a particular form (e.g., as a sum of multiples or the sum of odd numbers), you can use the principles of AP to determine the total number of terms or the value for each term.
Real-life uses
Real-life problems, like salary increases, savings trends, and fines, can be solved with the use of arithmetic progressions, for which the formula for the sum is used to find the amount over a specific duration.
By understanding these ideas, students can solve many problems related to the number of terms, the sum of terms, and the identification of specific terms for AP. Practice NCERT Solutions Class 10 Maths - Chapter 5 Exercise 5.3 to grasp these concepts from today, from here.
4.0NCERT Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 : Detailed Solutions
1. 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) a = 2, d = 5
S10 = (10/2)(2a + 9d)
(Since Sn = (n/2)[2a + (n-1)d])
S10 = 5 × (2 × 2 + 9 × 5) = 5 × 49 = 245
(ii) a = -37, d = 4
S12 = (12/2)(2a + 11d)
= 6 × [2(-37) + 11 × 4]
= 6 × (-74 + 44) = -180
(iii) a = 0.6
d = a2 - a1 = 1.7 - 0.6 = 1.1
n = 100
S100 = (100/2)[2(0.6) + (100-1)1.1]
= 50[1.2 + (99) × (1.1)]
= 50(110.1)
= 5505
(iv) a = 1/15
n = 11
d = a2 - a1 = 1/12 - 1/15 = (5-4)/60 = 1/60
We know that,
Sn = (n/2)[2a + (n-1)d]
S11 = (11/2)[2(1/15) + (11-1)(1/60)]
= (11/2)[2/15 + 10/60]
= (11/2)[2/15 + 1/6] = (11/2)[(4+5)/30]
= (11/2)(9/30) = 33/20
2. Find the sums given below:
(i) 7 + 10 1/2 + 14 + ... + 84
(ii) 34 + 32 + 30 + ... + 10
(iii) (-5) + (-8) + (-11) + ... + (-230)
Solution
(i) a = 7, d = 10 1/2 - 7 = 7/2
l = tn = 84 => a + (n - 1)d = 84
=> 7 + (n - 1) × 7/2 = 84
=> (n - 1) × 7/2 = 77
=> n - 1 = 77 × 2/7 = 22
=> n = 23
The sum = (n/2)(a + tn) = (23/2)(7 + 84)
= (23/2) × 91 = 2093/2 = 1046 1/2
(ii) 34 + 32 + 30 + ... + 10
a = 34
d = a2 - a1 = 32 - 34 = -2
l = 10
Let 10 be the nth term of this AP.
l = a + (n - 1)d
10 = 34 + (n - 1)(-2)
-24 = (n - 1)(-2)
12 = n - 1
n = 13
Sn = (n/2)(a + l)
= (13/2)(34 + 10)
= (13 × 44)/2 = 13 × 22 = 286
(iii) (-5) + (-8) + (-11) + ... + (-230)
For this AP,
a = -5
l = -230
d = a2 - a1 = (-8) - (-5)
= -8 + 5 = -3
Let -230 be the nth term of this AP.
l = a + (n - 1)d
-230 = -5 + (n - 1)(-3)
-225 = (n - 1)(-3)
(n - 1) = 75
n = 76
And, Sn = (n/2)(a + l)
= (76/2)[(-5) + (-230)]
= 38(-235) = -8930
3. In an AP:
(i) Given a = 5, d = 3, an = 50, find n and S_n.
(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, S_n = 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, Sn = 192, find d.
(x) Given l = 28, Sn = 144, and there are a total of 9 terms. Find a.
Solution
(i) a = 5, d = 3, an = 50
=> a + (n - 1)d = 50
=> 5 + (n - 1)(3) = 50
=> 5 + 3n - 3 = 50 or 3n = 48 or n = 16
Now, S16 = (16/2)(2a + 15d)
= 8(10 + 15 × 3) = 440
(ii) a = 7, a_13 = 35
∴ a13 = a + (13 - 1)d
=> 35 = 7 + 12d
=> 35 - 7 = 12d
=> 28 = 12d
=> d = 7/3
S13 = (n/2)(a + a13)
= (13/2)(7 + 35)
= (13 × 42)/2 = 13 × 21
= 273
(iii) a12 = 37, d = 3
So, a12 = a + (12 - 1)3
=> 37 = a + 33
a = 4
=> Sn = (n/2)(a + a_n)
=> S12 = (12/2)(4 + 37)
=> S12 = 6(41)
=> S12 = 246
(iv) a3 = 15, S_10 = 125
So, a3 = a + (3 - 1)d
15 = a + 2d ...(i)
S_n = (n/2)[2a + (n - 1)d]
S_10 = (10/2)[2a + (n - 1)d]
=> 125 = 5(2a + 9d)
25 = 2a + 9d ...(ii)
On multiplying equation (i) by 2, we obtain
30 = 2a + 4d ...(iii)
On subtracting equation (iii) from (ii), we obtain
-5 = 5d
=> d = -1
From equation (i),
15 = a + 2(-1)
=> 15 = a - 2
=> a = 17
Now, a10 = a + (10 - 1)d
=> a10 = 17 + (9)(-1)
=> a10 = 17 - 9 = 8
(v) d = 5, S_9 = 75
S9 = (9/2)[2a + (9 - 1)5]
=> 75 = (9/2)(2a + 40)
=> 25 = 3(a + 20)
=> 25 = 3a + 60
=> 3a = 25 - 60
=> a = -35/3
a9 = a + (9 - 1)(5)
= -35/3 + 8(5)
= -35/3 + 40
= (-35 + 120)/3 = 85/3
(vi) a = 2, d = 8, S_n = 90
=> (n/2)[2a + (n - 1)d] = 90
=> (n/2)[4 + (n - 1) × 8] = 90
=> (n/2) × [8n - 4] = 90
=> 4n² - 2n - 90 = 0
=> 2n² - n - 45 = 0
=> 2n² - 10n + 9n - 45 = 0
=> 2n(n - 5) + 9(n - 5) = 0
=> (n - 5)(2n + 9) = 0
=> n - 5 = 0 (∵ 2n + 9 ≠ 0)
=> n = 5
an = a5 = a + 4d = 2 + 4 × 8 = 34
=> an = 34
(vii) a = 8, a_n = 62, S_n = 210
So, 210 = (n/2)(8 + 62)
=> 210 = (n/2)(70)
=> n = 6
Now, an = a + (n - 1)d
=> 62 = 8 + (6 - 1)d
=> 62 - 8 = 5d
=> 54 = 5d
d = 54/5
(viii) a_n = 4, d = 2, S_n = -14
Now, a_n = 4 => a + (n - 1)d = 4
=> a + (n - 1)(2) = 4
=> a = 6 - 2n
Given that, S_n = -14
=> (n/2){2a + (n - 1)d} = -14
=> (n/2)[2(6 - 2n) + (n - 1)(2)] = -14 [By (i)]
=> (n/2)(12 - 4n + 2n - 2) = -14
=> (n/2)(10 - 2n) = -14
=> n(n - 5) = 14
=> n^2 - 5n - 14 = 0
=> n^2 - 7n + 2n - 14 = 0
=> n(n - 7) + 2(n - 7) = 0
=> (n - 7)(n + 2) = 0
=> n = 7 (n ≠ -2)
From (i), a = 6 - 2 × 7 = -8
a = -8
(ix) a = 3, n = 8, S = 192
192 = (8/2)[2 × 3 + (8 - 1)d]
192 = 4(6 + 7d)
48 = 6 + 7d
42 = 7d
d = 6
(x) l = 28, i.e., t_n = 28
=> t9 = 28 => a + 8d = 28
Sn = 144, i.e., S9 = 144
Now, S_n = (n/2)(a + l)
=> (9/2)(t1 + t9) = 144 => (9/2)(a + 28) = 144
=> a + 28 = 32 => a = 4
4. 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
a = 5, last term tn = 45 and Sn = 400
Sn = 400 => (n/2)(t1 + tn) = 400
=> (n/2)(5 + 45) = 400
=> (n/2) × 50 = 400
=> n = 16
Now, tn = 45 => t16 = 45
=> a + 15d = 45 => 5 + 15d = 45
=> 15d = 40 => d = 8/3
5. The first and the last terms 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
Given that,
a = 17
l = 350
d = 9
Let there be n terms in the AP.
l = a + (n - 1)d
=> 350 = 17 + (n - 1)9
=> 333 = (n - 1)9
=> (n - 1) = 37
=> n = 38
Now, Sn = (n/2)(a + l)
=> Sn = (38/2)(17 + 350)
= 19(367) = 6973
Thus, this AP contains 38 terms and the sum of the terms of this AP is 6973.
6. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution
d = 7
a22 = 149
S22 = ?
a22 = a + (22 - 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Sn = (n/2)(a + an)
= (22/2)(2 + 149) = 11(151) = 1661
7. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution
t2 = 14, t3 = 18
d = t3 - t2 = 18 - 14 = 4, i.e., d = 4
Now t2 = 14
=> a + d = 14
=> a + 4 = 14
=> a = 10
S51 = (51/2)(2a + 50d)
= (51/2)(2 × 10 + 50 × 4)
= (51/2) × 220 = 51 × 110 = 5610
8. If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of n terms.
Solution
S7 = 49
=> (7/2)(2a + 6d) = 49
=> a + 3d = 7 ...(i)
S17 = 289
=> (17/2)(2a + 16d) = 289
=> a + 8d = 17 ...(ii)
Subtracting (i) from (ii), we get
5d = 17 - 7 = 10
=> d = 2
From (i),
a + 3 × 2 = 7
=> a = 1
So, Sn = (n/2)[2a + (n - 1)d]
= (n/2)[2 × 1 + (n - 1) × 2]
= (n/2)(2n) = n²
Hence, Sn = n²
9. 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) an = 3 + 4n
Putting n = 1, 2, 3, 4, ... in (i), we get
a1 = 3 + 4 = 7, a2 = 3 + 8 = 11,
a3 = 3 + 12 = 15, a4 = 3 + 16 = 19, ...
Thus, the sequence (list of numbers) 7, 11, 15, 19, ...
Here, a2 - a1 = 11 - 7 = 4
a3 - a2 = 15 - 11 = 4,
a4 - a3 = 19 - 15 = 4
Therefore, the sequence forms an AP in which a = 7 and d = 4.
S15 = (15/2)(2a + 14d)
= (15/2)(2 × 7 + 14 × 4)
= (15/2) × 70 = 15 × 35 = 525
(ii) an = 9 - 5n
So, a1 = 9 - 5 × 1 = 9 - 5 = 4
a2 = 9 - 5 × 2 = 9 - 10 = -1
a3 = 9 - 5 × 3 = 9 - 15 = -6
and a4 = 9 - 5 × 4 = 9 - 20 = -11
It can be observed that
a2 - a1 = -1 - 4 = -5
a3 - a2 = -6 - (-1) = -5
a4 - a3 = -11 - (-6) = -5
Therefore, this is an AP with a common difference as -5 and first term as 4.
S15 = (15/2)[2a + (n - 1)d]
= (15/2)[8 + 14(-5)]
= (15/2)(8 - 70)
= (15/2)(-62) = 15(-31) = -465
10. 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 the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution
Sn = 4n - n²
Putting n = 1, we get S₁ = 4 - 1 = 3
i.e., t₁ = 3
S₂ = 4(2) - (2)² = 8 - 4 = 4, i.e., S₂ = 4
=> t₁ + t₂ = 4 => 3 + t₂ = 4 => t₂ = 1
t₂ - t₁ = 1 - 3 = -2 => d = -2
Then t₃ = t₂ + d = 1 - 2 = -1, i.e., t₃ = -1
t₁₀ = t₁ + 9d = 3 + 9(-2) (∵ t₁ = a)
=> t₁₀ = -15
tn = t₁ + (n - 1)d = 3 + (n - 1) × (-2)
i.e., tn = 5 - 2n
11. Find the sum of the first 40 positive integers divisible by 6.
Solution
The positive integers that are divisible by 6 are 6, 12, 18, 24...
It can be observed that these are making an AP whose first term is 6 and common difference is 6.
So, a = 6
And d = 6
S₄₀ = ?
Sn = (n/2)[2a + (n - 1)d]
S₄₀ = (40/2)[2(6) + (40 - 1)6]
= 20[12 + (39)(6)]
= 20(12 + 234)
= 20 × 246
= 4920
12. Find the sum of the first 15 multiples of 8.
Solution
The multiples of 8 are 8, 16, 24, 32...
These are in an AP, having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S₁₅ = ?
Sn = (n/2)[2a + (n - 1)d]
= (15/2)[2(8) + (15 - 1)8]
= (15/2)[16 + 14(8)]
= (15/2)(16 + 112)
= (15(128))/2 = 15 × 64 = 960
13. Find the sum of the odd numbers between 0 and 50.
Solution
1, 3, 5, 7 ..., 49
a = 1, d = 2
l = tn = 49
=> a + (n - 1)d = 49
=> 1 + (n - 1)(2) = 49
=> 1 + 2n - 2 = 49
=> 2n = 50 or n = 25
The sum = (25/2)(a + l) = (25/2)(1 + 49) = (25/2) × 50 = 625
14. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days?
Solution
It can be observed that these penalties are in an AP having first term as 200 and common difference as 50.
So, a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S₃₀
S₃₀ = (30/2)[2(200) + (30 - 1)50]
= 15[400 + 1450]
= 15(1850)
= 27750
So, the contractor has to pay Rs. 27750.
15. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution
Let the 1st prize be of ₹ a.
Then, the 2nd prize will be of ₹ (a-20).
Then the 3rd prize will be of ₹ [(a-20)-20], i.e., ₹ (a-40).
Thus, the seven prizes are of ₹ a, ₹ (a-20), ₹ (a-40), ... (This forms an AP)
Then a + (a-20) + (a-40) + ... to 7 terms = 700
=> (7/2)[2a + 6 × (-20)] = 700 (∵ d = -20)
=> (7/2) × (2a - 120) = 700
=> a - 60 = 100
=> a = 160
Thus, the 7 prizes are of ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40.
16. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
Solution
Here, trees planted by the students form an AP.
1, 2, 3, 4, 5, ..., 12
First term, a = 1
And Common difference, d = 2 - 1 = 1
Sn = (n/2)[2a + (n-1)d]
S12 = (12/2)[2(1) + (12-1)(1)]
= 6(2 + 11) = 6(13) = 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.
17. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,... as shown in fig. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Solution
From the figure,
l1 = π × 0.5, l2 = π × 1, l3 = π × 1.5, l4 = π × 2, and so on.
i.e., l1 = (1/2)π, l2 = π, l3 = (3/2)π, l4 = 2π, ...
Thus, l1, l2, l3, l4, ... form an AP.
∵ l2 - l1 = l3 - l2 = l4 - l3 = ... = (1/2)π
Thus, a = π/2, d = π/2
Length of the spiral = l1 + l2 + ... + l13
= (13/2)(2a + 12d) = (13/2)(2 × π/2 + 12 × π/2)
= (91π/2) cm = (91/2) × (22/7) cm = 143 cm.
18. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?
Solution
It can be observed that the numbers of logs in rows are in an AP: 20, 19, 18...
For this AP, a = 20, d = a2 - a1 = 19 - 20 = -1
Let a total of 200 logs be placed in n rows.
Sn = 200
Sn = n/2 [2a + (n - 1)d]
=> 200 = n/2 [2(20) + (n - 1)(-1)]
=> 400 = n(40 - n + 1)
=> 400 = n(41 - n)
=> 400 = 41n - n^2
=> n^2 - 41n + 400 = 0
=> n^2 - 16n - 25n + 400 = 0
=> n(n - 16) - 25(n - 16) = 0
=> (n - 16)(n - 25) = 0
Either (n - 16) = 0 or n - 25 = 0
=> n = 16 or n = 25
=> an = a + (n - 1)d
=> a16 = 20 + (16 - 1)(-1)
=> a16 = 20 - 15
=> a16 = 5
Similarly,
a25 = 20 + (25 - 1)(-1)
a25 = 20 - 24 = -4
Clearly, the number of logs in the 16th row is 5.
However, the number of logs in the 25th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.
19. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig.). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Solution
Distance run to pick up the 1st potato = 2 x 5 = 10 m
Distance run to pick up the 2nd potato = 2 x (5 + 3) m = 16 m
Distance run to pick up the 3rd potato = 2 x (5 + 3 + 3) m = 22 m
Thus, the sequence becomes 10, 16, 22, ... to 10 terms. It forms an AP.
Here, a = 10, d = 6 and n = 10
Sum = S10 = 10/2 (2a + 9d)
= 5 x (2 x 10 + 9 x 6)
= (5 x 74) m = 370 m
Hence, the total distance run by a competitor = 370 m.
5.0Benefits of Studying NCERT Solutions For Class 10 Chapter 5 Exercise 5.3
- Helps in calculating the sum of n terms in an arithmetic progression.
- Improves logical thinking by solving sum-related problems step by step.
- Useful for advanced topics in algebra, sequences, and series.
- Helps in real life scenarios like saving money, loan payments, and business calculations.
- Frequently appears in exams like NTSE, Olympiads, and entrance tests.
