NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3
Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3: In Exercise 5.3 of Chapter 5, "Arithmetic Progressions," we will look into the application of arithmetic sequences to solve various realworld problems. This exercise builds upon our understanding of the fundamental concepts of arithmetic progressions (APs) by focusing on the practical application of these concepts.
Students will explore how to use the common difference and general terms of an AP to find missing values, solve problems involving sequences of numbers, and understand the relevance of APs in everyday scenarios. Through a series of illustrative problems and exercises, this chapter aims to enhance problemsolving skills and deepen comprehension of arithmetic progressions. Get the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 from the article below.
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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3
Check out the Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 NCERT Solutions below:
1. Find the sum of the following AP’s.
(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 ,1/12,1/10to 11 terms
Answer:
(i) 2, 7, 12… to 10 terms
Here First term = a = 2, Common difference = d = 7 – 2 = 5 and n = 10
Applying formula,to find sum of n terms of AP,
(ii) –37, –33, –29… to 12 terms
Here First term = a = –37, Common difference = d = –33 – (–37) = 4
And n = 12
Applying formula,to find sum of n terms of AP,
(iii) 0.6, 1.7, 2.8… to 100 terms
Here First term = a = 0.6, Common difference = d = 1.7 – 0.6 = 1.1
And n = 100
Applying formula,to find sum of n terms of AP,
(iv) to 11 terms
Here First tern =a = 1/15Common difference =
Applying formula,to find sum of n terms of AP,
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2. Find the sums given below
(i) 7 + 21/2+ 14 + .................. +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Answer:
(i)
Here First term = a = 7, Common difference = d =
And Last term = l = 84
We do not know how many terms are there in the given AP.
So, we need to find n first.
Using formula , to find nth term of arithmetic progression,
[7 + (n − 1) (3.5)] = 84
⇒ 7 + (3.5) n − 3.5 = 84
⇒ 3.5n = 84 + 3.5 – 7
⇒ 3.5n = 80.5
⇒ n = 23
Therefore, there are 23 terms in the given AP.
It means n = 23.
Applying formula, to find sum of n terms of AP,
(ii) 34 + 32 + 30 + … + 10
Here First term = a = 34, Common difference = d = 32 – 34 = –2
And Last term = l = 10
We do not know how many terms are there in the given AP.
So, we need to find n first.
Using formula , to find nth term of arithmetic progression,
[34 + (n − 1) (−2)] = 10
⇒ 34 – 2n + 2 = 10
⇒ −2n = −26⇒ n = 13
Therefore, there are 13 terms in the given AP.
It means n = 13.
Applying formula, to find sum of n terms of AP,
(iii) −5 + (−8) + (−11) + … + (−230)
Here First term = a = –5, Common difference = d = –8 – (–5) = –8 + 5 = –3
And Last term = l = −230
We do not know how many terms are there in the given AP.
So, we need to find n first.
Using formula , to find nth term of arithmetic progression,
[−5 + (n − 1) (−3)] = −230
⇒ −5 − 3n + 3 = −230
⇒ −3n = −228 ⇒ n = 76
Therefore, there are 76 terms in the given AP.
It means n = 76.
Applying formula, to find sum of n terms of AP,
3. 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.
Answer:
(i) Given a = 5, d = 3, an = 50,find n and Sn.
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
⇒ 50 = 5 + (n − 1) (3)
⇒ 50 = 5 + 3n − 3
⇒ 48 = 3n⇒ n = 16
Applying formula,to find sum of n terms of AP,
Therefore, n = 16 and Sn = 440
(ii) Given a = 7, a13 = 35, find d and S13..
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
an = a + (n1)d
a13= 7 + (13 − 1) (d)
⇒ 35 = 7 + 12d
⇒ 28 = 12d⇒ d = 28/12 = 7/3
Applying formula,to find sum of n terms of AP,
Therefore, d = 7/3 and S13 = 273
(iii) Given a12 = 37, d = 3, find a and S12..
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
an = a + (n1)d
a12 = a + (12 − 1) 3
⇒ 37 = a + 33 ⇒ a = 4
Applying formula,to find sum of n terms of AP,
Therefore, a = 4 and S12 = 246
(iv) Given a3 = 15, S10 = 125, find d and a10..
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
an = a + (n1)d
a3 = a + (3 − 1) (d)
⇒ 15 = a + 2d
⇒ a = 15 − 2d… (1)
Applying formula,to find sum of n terms of AP,
⇒ 125 = 5 (2a + 9d) = 10a + 45d
Putting (1) in the above equation,
125 = 5 [2 (15 − 2d) + 9d] = 5 (30 − 4d + 9d)
⇒ 125 = 150 + 25d
⇒ 125 – 150 = 25d
⇒ −25 = 25d⇒ d = −1
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
an = a + (n1)d
a10= a + (10 − 1) d
Putting value of d and equation (1) in the above equation,
a10= 15 − 2d + 9d = 15 + 7d
= 15 + 7 (−1) = 15 – 7 = 8
Therefore, d = −1 and a10= 8
(v) Given d = 5,S9 = 75, find a and a9..
Applying formula,to find sum of n terms of AP,
⇒ 150 = 18a + 360
⇒ −210 = 18a
⇒ a =
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
a9 = + (9 − 1) (5)
a9 =
Therefore, a = and a9 =
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
Applying formula,to find sum of n terms of AP,
⇒ 2n (n − 5) + 9 (n − 5) = 0
⇒ (n − 5) (2n + 9) = 0
⇒ n = 5,−9/2
We discard negative value of n because here n cannot be in negative or fraction.
The value of n must be a positive integer.
Therefore, n = 5
Using formula an = a + (n1)d ,to find nth term of arithmetic progression,
a5 = 2 + (5 − 1) (8) = 2 + 32 = 34
Therefore, n = 5 and a5 = 34
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
Using formula an = a + (n1)d, to find nth term of arithmetic progression,
62 = 8 + (n − 1) (d) = 8 + nd – d
⇒ 62 = 8 + nd − d
⇒ nd – d = 54
⇒ nd = 54 + d… (1)
Applying formula,to find sum of n terms of AP,
⇒ n = 6
Putting value of n in equation (1),
6d = 54 + d ⇒ d =
Therefore, n = 6 and d =
(viii) Given Given an = 4, d = 2, Sn = − 14, find n and a
Using formula an = a + (n1)d , to find nth term of arithmetic progression,
4 = a + (n − 1) (2) = a + 2n − 2
⇒ 4 = a + 2n – 2
⇒ 6 = a + 2n
⇒ a = 6 − 2n… (1)
Applying formula,to find sum of n terms of AP,
⇒ (n + 2) (n − 7) = 0
⇒ n = −2, 7
Here, we cannot have negative value of n.
Therefore, we discard negative value of n which means n = 7.
Putting value of n in equation (1), we get
a = 6 − 2n = 6 – 2 (7) = 6 – 14 = −8
Therefore, n = 7 and a = −8
(ix)Given a = 3, n = 8, S = 192, find d.
Using formula, an = a + (n1)d to find sum of n terms of AP, we get
192 = [6 + (8 − 1) d] = 4 (6 + 7d)
⇒ 192 = 24 + 28d
⇒ 168 = 28d ⇒ d = 6
(x) Given l = 28, S = 144, and there are total of 9 terms. Find a.
Applying formula, , to find sum of n terms, we get
144 = [a + 28]
⇒ 288 = 9 [a + 28]
⇒ 32 = a + 28⇒ a = 4
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4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Answer:
First term = a = 9, Common difference = d = 17 – 9 = 8, Sn = 636
Applying formula,to find sum of n terms of AP, we get
636 =[18 + (n − 1) (8)]
⇒ 1272 = n (18 + 8n − 8)
⇒
We discard negative value of n here because n cannot be in negative, n can only be a positive integer.
Therefore, n = 12
Therefore, 12 terms of the given sequence make sum equal to 636.
5. 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.
Answer.
First term = a = 5, Last term = l = 45,
Applying formula, to find sum of n terms of AP, we get
⇒ n = 16
Applying formula,to find sum of n terms of AP and putting value of n, we get
We get,
45 = 5 + (16  1)d
45 = 5 + (15)d
45  5 = 15d
6. 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?
Answer:
First term = a = 17, Last term = l = 350 and Common difference = d = 9
Using formula , to find nth term of arithmetic progression, we get
350 = 17 + (n − 1) (9)
⇒ 350 = 17 + 9n − 9
⇒ 342 = 9n ⇒ n = 38
Applying formula,to find sum of n terms of AP and putting value of n, we get
=19 (34 + 333) = 6973
Therefore, there are 38 terms and their sum is equal to 6973.
7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer:
It is given that 22nd term is equal to 149
Using formula an = a + (n1)d, to find nth term of arithmetic progression, we get
an = a + (n1)d
149 = a + (22 − 1) (7)
⇒ 149 = a + 147⇒ a = 2
Applying formula,to find sum of n terms of AP and putting value of a, we get
= 11 (4 + 147)
⇒ 11(151)= 1661
Therefore, sum of first 22 terms of AP is equal to 1661.
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
It is given that second and third term of AP are 14 and 18 respectively.
Using formula an = a + (n1)d, to find nth term of arithmetic progression, we get
14 = a + (2 − 1) d
⇒ 14 = a + d … (1)
And, 18 = a + (3 − 1) d
⇒ 18 = a + 2d … (2)
These are equations consisting of two variables.
Using equation (1), we get, a = 14 − d
Putting value of a in equation (2), we get
18 = 14 – d + 2d
⇒ d = 4
Therefore, common difference d = 4
Putting value of d in equation (1), we get
18 = a + 2 (4)
⇒ a = 10
Applying formula,to find sum of n terms of AP, we get
Therefore, sum of first 51 terms of an AP is equal to 5610.
9. 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.
Answer:
It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.
Applying formula,to find sum of n terms of AP, we get
⇒ 98 = 7 (2a + 6d)
⇒ 7 = a + 3d ⇒ a = 7 − 3d … (1)
And, s17= 289
⇒ 578 = 17 (2a + 16d)
⇒ 34 = 2a + 16d
⇒ 17 = a + 8d
Putting equation (1) in the above equation, we get
17 = 7 − 3d + 8d
⇒ 10 = 5d ⇒ d = 2
Putting value of d in equation (1), we get
a = 7 − 3d = 7 – 3 (2) = 7 – 6 = 1
Again applying formula,to find sum of n terms of AP, we get
10. 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.
Answer:
(i)
is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Therefore, sum of first 15 terms of AP is equal to 525.
(ii)
is same every time. Therefore, this is an AP with common difference as 5 and first term as 4.
Therefore, sum of first 15 terms of AP is equal to –465.
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11. 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.
Answer:
It is given that the sum of n terms of an AP is equal to
It means
So here, we can find the first term by substituting n = 1 ,
=41=3
Similarly, the sum of first two terms can be given by,
= 8  4
= 4
Now, as we know,
So,
= 4  3
= 1
Now, using the same method we have to find the third, tenth and nth term of A.P.
So, for the third term,
=
= (12  9)  (8  4)
= 3  4
=  1
Also, for the tenth term,
=
= (40  100)  (36  81)
=  60 + 45
= 15
So, for the nth term,
=
=
=
= 5  2n
Therefore,=
12. Find the sum of the first 40 positive integers divisible by 6.
Answer:
The first positive integers that are divisible by 6 are 6, 12, 18, 24 … 40 terms.
Therefore, we want to find sum of 40 terms of sequence of the form:
6, 12, 18, 24 … 40 terms
Here, first term = a = 6 and Common difference = d = 12 – 6 = 6, n = 40
Applying formula, to find sum of n terms of AP, we get
a = 6
d = 6
S40 = ?
=
= 20 [12 + (39) (6 )]
= 20 (12 + 234)
= 20 x 246
= 4920
13. Find the sum of the first 15 multiples of 8.
Answer:
The first 15 multiples of 8 are 8,16, 24, 32 … 15 times
First term = a = 8 and Common difference = d = 16 – 8 = 8, n = 15
Applying formula, to find sum of n terms of AP, we get
I = a + (n1) d
= 8 + (15  1) 8
= 120
Reruired Sum =
=
Hence, the required sum is 960.
14. Find the sum of the odd numbers between 0 and 50.
Answer:
The odd numbers between 0 and 50 are 1, 3, 5, 7 … 49
It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a = 1, Common difference = 3 – 1 = 2, Last term = l = 49
We do not know how many odd numbers are present between 0 and 50.
Therefore, we need to find n first.
Using formula an = a + (n − 1) d, to find nth term of arithmetic progression, we get
49 = 1 + (n − 1) 2
⇒ 49 = 1 + 2n − 2
⇒ 50 = 2n ⇒ n = 25
Applying formula, to find sum of n terms of AP, we get
=
=
= 625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Answer:
Penalty for first day = Rs 200, Penalty for second day = Rs 250
Penalty for third day = Rs 300
It is given that penalty for each succeeding day is Rs 50 more than the preceding day.
It makes it an arithmetic progression because the difference between consecutive terms is constant.
We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
So, we have an AP of the form200, 250, 300, 350 … 30 terms
First term = a = 200, Common difference = d = 50, n = 30
Applying formula, to find sum of n terms of AP, we get
⇒
⇒ 15 (400 + 1450)
= 15 x 1850
= 27750
Therefore, penalty for 30 days is Rs. 27750.
16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If, each prize is Rs 20 less than its preceding term, find the value of each of the prizes.
Answer:
It is given that sum of seven cash prizes is equal to Rs 700.
And, each prize is R.s 20 less than its preceding term.
Let value of first prize = Rs. a
Let value of second prize =Rs (a − 20)
Let value of third prize = Rs (a − 40)
So, we have sequence of the form:
a, a − 20, a − 40, a – 60 …
It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a, Common difference = d = (a – 20) – a = –20
n = 7 (Because there are total of seven prizes)
Applying formula, to find sum of n terms of AP, we get
=
⇒
=
= 700 = 7 (a  60)
On further specification, we get
⇒ 700/7 = a  60
⇒ 100 + 60 = a
⇒ a = 160
Therefore, value of first prize = Rs 160
Value of second prize = 160 – 20 = Rs 140
Value of third prize = 140 – 20 = Rs 120
Value of fourth prize = 120 – 20 = Rs 100
Value of fifth prize = 100 – 20 = Rs 80
Value of sixth prize = 80 – 20 = Rs 60
Value of seventh prize = 60 – 20 = Rs 40
17. 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 two trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Answer:
There are three sections of each class and it is given that the number of trees planted by any class is equal to class number.
The number of trees planted by class I = number of sections
The number of trees planted by class II = number of sections
The number of trees planted by class III = number of sections
Therefore, we have sequence of the form 3, 6, 9 … 12 terms
To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.
First term = a = 3, Common difference = d= 6 – 3 = 3 and n = 12
Applying formula, to find sum of n terms of AP , we get
=
= 6 (2+11)
= 6 x 13
= 78
Therefore, number of trees planted by 1 section of classes = 78
number of trees planted by 3 section of classes = 3 x 78 = 234
Therefore, 234 trees will be planted by the students.
18. A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … What is the total length of such a spiral made up of thirteen consecutive semicircles.
Answer:
Length of Semiperimeter of circle = πr
Length of semicircle of radii 0.5 cm = π(0.5) cm = π/2
Length of semicircle of radii 1.0 cm = π(1.0) cm
Length of semicircle of radii 1.5 cm = π(1.5) cm = 3π/2
Therefore, we have sequence of the form:
π(0.5), π(1.0), π(1.5) … 13 terms {There are total of thirteen semi–circles}.
To find total length of the spiral, we need to find sum of the sequence π(0.5), π(1.0), π(1.5) … 13 terms
First term = a = 0.5, Common difference = 1.0 – 0.5 = 0.5 and n = 13
Applying formula, to find sum of n terms of AP, we get
=
=
= (13/2) (7π)
= (91π / 2)
= ((91 x 22) / 2) x (7)
= (13 x 11)
= 143
Therefore, the lenght of such spiral of 13 consecutive semi  circles will be 143 cm.
19. 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. In how many rows are the 200 logs placed and how many logs are in the top row?
Answer:
The number of logs in the bottom row = 20
The number of logs in the next row = 19
The number of logs in the next to next row = 18
Therefore, we have sequence of the form 20, 19, 18 …
First term = a = 20, Common difference = d = 19 – 20 =  1
We need to find that how many rows make total of 200 logs.
Applying formula, to find sum of n terms of AP, we get
=
⇒ 400 = n (40 – n + 1)
⇒ 400 = n (41  n)
⇒ 400 = 41n  n²
⇒ n²  41n + 400 = 0
⇒ n²  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 16th row is 5. However, the number of logs in 25th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in 16th row is 5.
20. In a potato race, a bucket is placed at the starting point, which is 5 meters from the first potato, and the other potatoes are placed 3 meters apart in a straight line. There are ten potatoes in the line. 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?
Answer:
The distance of potatoes are as follows :
5, 8, 11, 14, .....
It can be observed that these distances are in A.P.
a = 5
d = 8  5 = 3
From the formula, we get
=
= 5 [10 + 9 x 3]
= 5 (10 + 27)
= 5 x 37
= 185
As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.
Therefore, the total distance that the competitor will be run = 2 x 185 = 370m.
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Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 Summary
Exercise 5.3 of Chapter 5, "Arithmetic Progressions," focuses on the application of arithmetic progressions to solve a variety of problems. This exercise helps students solidify their understanding of the key concepts related to APs and apply them to practical scenarios. The exercise covers several important areas:

Finding Missing Terms:

Students are given some terms of an arithmetic progression and are required to find other missing terms. To solve these problems, students use the formula for the nth term of an AP: an=a+(n−1)d, where aaa is the first term, d is the common difference, and n is the term number.

Example problems may involve finding the 10th term of an AP if the first term and common difference are known, or finding any specific term when some terms in the sequence are missing.

Determining the Common Difference:

In problems where some terms of the AP are provided, students need to determine the common difference d. This involves using the relationship between consecutive terms: d=an+1−anâ€‹.

Examples include finding the common difference when given two nonconsecutive terms of the AP.

Sum of Terms:

The exercise involves calculating the sum of the first n terms of an arithmetic progression using the formula, where l is the last term of the sequence.

Problems may involve finding the total sum when the number of terms, the first term, and the common difference are known or when the first term and the last term are given.

Application in RealLife Scenarios:

The exercise includes practical problems where arithmetic progressions are applied to reallife contexts, such as calculating the total distance covered in a series of equal steps, financial calculations involving regular deposits or withdrawals, or other scenarios where quantities change linearly.

Examples might involve determining the amount of money saved over time with a regular monthly deposit, or the distance traveled each day if the distance increases by a fixed amount each day.

Verifying Solutions:

Some problems may require students to verify whether a given sequence is an arithmetic progression by checking if the common difference between consecutive terms remains constant.

Complex Problems:

The exercise also presents more complex problems that combine multiple concepts of APs. For example, students may need to solve problems involving a series of APs or find the relationship between different sequences.
Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3 FAQs
Q1. What is an arithmetic progression (AP)?
Ans. An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This difference is called the common difference (d). For example, in sequences 2, 5, 8, and 11, the common difference is 3.
Q2. How do I find the common difference in an AP?
Ans. To find the common difference, subtract the first term from the second term or any term from the next term. For example, if the terms are 7 and 11, the common difference is 11−7=4.
Q3. What is the formula for the nth term of an AP?
Ans. The formula for the nth term of an AP is an=a+(n−1)d, where aaa is the first term, d is the common difference, and n is the term number.
Q4. What should I do if I’m given a few terms and need to find the missing terms in an AP?
Ans. Use the formula for the nth term an=a+(n−1)d and the common difference d to find missing terms. Plug in the known values and solve for the unknowns.
Q5. Can the common difference be negative in an AP?
Ans. Yes, the common difference can be negative. This means the sequence will decrease as you move from one term to the next. For example, in the sequence 10, 7, 4, 1, the common difference is 3.