In this presentation, you will learn sequences and arithmetic progression. How to find the terms, common differences, etc. I have given detailed solutions to each problem.

1. Chapter
Arithmetic Progression
© S S CLASSES Founder: Satish Pandit

2. Let’s discuss
1, 4, 9, 16, 25, 36,…?
2, 4, 6, 8, 10,…..?
1,8, 27, 64,….?
4, 8, 12, 16, 20,….?
Natural numbers
Even numbers
Squares of natural numbers
Cubes of natural numbers
Multiples of 4
1, 2, 3, 4, 5,…..?

3. Sequence
A set of numbers where the numbers are arranged in a definite order, like the
natural numbers, is called a sequence.
• Terms in Sequence
In a sequence, ordered terms are represented as 𝑡1, 𝑡2, 𝑡3 … … 𝑡𝑛
In general sequence is written as { 𝑡𝑛 }.
If the sequence is infinite, for every positive integer , there is a term 𝑡𝑛.
Example; 7, 14, 21, 28, 35….?
Here as 𝑡1 = 7, 𝑡2 = 14, 𝑡3 = 21, 𝑡4 = 28

4. Arithmetic Progression
A sequence in which the difference between any two consecutive terms is
constant then that sequence is known as arithmetic progression.
𝑡1 𝑡2 𝑡3 𝑡4 𝑡5 𝑡6
d d d d d
d = 𝑡𝑛+1 − 𝑡𝑛
In the general,
d = 𝑡2 − 𝑡1 d = 𝑡3 − 𝑡2 d = 𝑡4 − 𝑡3
d= common difference
a= first term

5. Terms in A.P
𝑡2 = 𝑡1 + 𝑑
𝑡3 = 𝑡2 + 𝑑
𝑡4 = 𝑡3 + 𝑑
𝑡𝑛 = 𝑡𝑛−1 + 𝑑
𝑡1 = 𝑎
𝑡2 = 𝑎 + 𝑑
𝑡3 = 𝑎 + 2𝑑
𝑡4 = 𝑎 + 3𝑑
𝑡𝑛 = 𝑎 + (𝑛 − 1)𝑑

6. 1. Which of the following sequences are A.P. ? If they are A.P. find the common
difference?
(1) 2, 4, 6, 8, . . .
Solution : From given sequence ,
𝑡1 = 2 , 𝑡2 = 4 , 𝑡3 = 6 , 𝑡4 = 8
∴ 𝑡2 − 𝑡1 = 4 − 2 = 2
∴ 𝑡3 − 𝑡2 = 6 − 4 = 2
𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐 = 𝒕𝟒 − 𝒕𝟑
Since the difference between each term is common.
Therefore the given sequence is an A.P.
∴ 𝐜𝐨𝐦𝐦𝐨𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐜𝐞 (𝐝) = 𝟐
∴ 𝑡4 − 𝑡3 = 8 − 6 = 2

7. 2 . 𝟐 ,
𝟓
𝟐
, 𝟑 ,
𝟕
𝟐
… … . .
Solution: From given sequence ,
𝑡1 = 2 , 𝑡2 =
5
2
, 𝑡3 = 3 , 𝑡4 =
7
2
∴ 𝑡2 − 𝑡1 =
5
2
− 2 =
1
2
∴ 𝑡3 − 𝑡2 = 3 −
7
2
=
1
2
𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐 = 𝒕𝟒 − 𝒕𝟑
Since the difference between each term
is common. Hence, the given sequence
is an A.P.
Common difference ∴ 𝒅 =
𝟏
𝟐
∴ 𝑡4 − 𝑡3 =
7
2
− 3 =
1
2

8. (3) -10, -6, -2, 2, . . .
Solution : From given sequence ,
𝑡1 = −10 , 𝑡2 = −6 , 𝑡3 = −2 , 𝑡4 = 2
∴ 𝑡2 − 𝑡1 = −6 − −10 = 4
∴ 𝑡3 − 𝑡2 = −2 − (−6) = 4
𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐
Since the difference between each term is common.
Hence, the given sequence is an A.P.
Common difference 𝒅 = 𝟒

9. (4) 0.3, 0.33, .0333, . . .
Solution : From given sequence ,
𝑡1 = 0.3 , 𝑡2 = 0.33 , 𝑡3 = 0.0333
∴ 𝑡2 − 𝑡1 = 0.33 − 0.3 = 0.03
∴ 𝑡3 − 𝑡2 = 0.0333 − 0.33 = −0.2967
𝒕𝟐 − 𝒕𝟏≠ 𝒕𝟑 − 𝒕𝟐
Since the difference between each term is not common.
Hence , the given sequence is not an A.P.

10. Solution : From given sequence ,
𝑡1 = 0 , 𝑡2 = −4 , 𝑡3 = −8 , 𝑡4 = −12
∴ 𝑡2 − 𝑡1 = −4 − 0 = −4
∴ 𝑡3 − 𝑡2 = −8 − −4 = −4
𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐 = 𝒕𝟒 − 𝒕𝟑
Since the difference between each term is common.
Hence, the given sequence is an A.P.
Common difference (d) = -4
(5) 0, -4, -8, -12, . . .
∴ 𝑡4 − 𝑡3 = −12 − −8 = −4

11. Solution : From given sequence ,
𝑡1 = −
1
5
, 𝑡2 = −
1
5
, 𝑡3 = −
1
5
∴ 𝑡2 − 𝑡1 = −
1
5
− (−
1
5
) = 0
∴ 𝑡3 − 𝑡2 = −
1
5
− (−
1
5
) = 0
𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐
Hence , the given sequence is an A.P.
Now,
Common difference ∴ 𝒅 = 𝟎
(6). −
𝟏
𝟓
, −
𝟏
𝟓
, −
𝟏
𝟓
, … …

12. Solution: From given sequence ,
𝑡1 = 3 , 𝑡2 = 3 + 2 , 𝑡3 = 3 + 2 2, 𝑡4 = 3 + 3 2
∴ 𝑡2 − 𝑡1 = 3 + 2 − 3 = 2
∴ 𝑡3 − 𝑡2 = 3 + 2 2 − (3 + 2 ) = 2 2 − 2 = 2
Since 𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐
Hence , the given sequence is an A.P.
Common difference 𝒅 = 𝟐
(7) 3 , 𝟑 + 𝟐 , 𝟑 + 𝟐 𝟐 , 𝟑 + 𝟑 𝟐 , . . .

13. (8) 127, 132, 137, . . .
Solution: From given sequence ,
𝑡1 = 127 , 𝑡2 = 132 , 𝑡3 = 137
∴ 𝑡2 − 𝑡1 = 132 − 127 = 5
∴ 𝑡3 − 𝑡2 = 137 − 132 = 5
Since 𝒕𝟐 − 𝒕𝟏 = 𝒕𝟑 − 𝒕𝟐
Hence, the given sequence is an A.P.
Common difference 𝒅 = 𝟓

14. 2. Write an A.P. whose first term is a and common difference is d in each of
the Following.
(1) a = 10, d = 5
Solution : 𝑎 = 𝑡1 = 10
∴ 𝑡2 = 𝑡1 + 𝑑 = 10 + 5 = 15
∴ 𝑡3 = 𝑡2 + 𝑑 = 15 + 5 = 20
∴ 𝑡4 = 𝑡3 + 𝑑 = 20 + 5 = 25
∴ 𝑡5 = 𝑡4 + 𝑑 = 25 + 5 = 30
Hence , A.P is 10 , 15 , 20 , 25 , 30 ,……….

15. Solution : 𝑎 = 𝑡1 = −3
∴ 𝑡2 = 𝑡1 + 𝑑 = −3 + 0 = −3
∴ 𝑡3 = 𝑡2 + 𝑑 = −3 + 0 = −3
∴ 𝑡4 = 𝑡3 + 𝑑 = −3 + 0 = −3
∴ 𝑡5 = 𝑡4 + 𝑑 = −3 + 0 = −3
Hence , A.P is -3, -3 , -3 , -3 ,……….
(2) a = -3, d = 0

16. Solution : 𝑎 = 𝑡1 = −7
∴ 𝑡2 = 𝑡1 + 𝑑 = −7 +
1
2
=
−14 + 1
2
=
−13
2
= −6.5
∴ 𝑡3= 𝑡2 + 𝑑 =
−13
2
+
1
2
=
−12
2
= −6
∴ 𝑡4 = 𝑡3 + 𝑑 =
−12
2
+
1
2
=
−11
2
= −5.5
Hence , A.P is -7 , -6.5 , -6 , -5.5 ,……….
(3) a = -7, d =1/2

17. Solution : 𝑎 = 𝑡1 = −1.25
∴ 𝑡2 = 𝑡1 + 𝑑 = −1.25 + 3 = 1.75
∴ 𝑡3 = 𝑡2 + 𝑑 = 1.75 + 3 = 4.75
∴ 𝑡4 = 𝑡3 + 𝑑 = 4.75 + 3 = 7.75
Hence , A.P is -1.25 , 1.75 , 4.75 , 7.75 ,……….
(4) a = -1.25, d = 3

18. Solution : 𝑎 = 𝑡1 = 6
∴ 𝑡2 = 𝑡1 + 𝑑 = 6 + (−3) = 3
∴ 𝑡3 = 𝑡2 + 𝑑 = 3 + (−3) = 0
∴ 𝑡4 = 𝑡3 + 𝑑 = 0 + −3 = −3
Hence, A.P is 6 , 3 , 0 , -3 ,……….
(5) a = 6, d = -3

19. Solution: 𝑎 = 𝑡1 = −19
∴ 𝑡2 = 𝑡1 + 𝑑 = −19 + −4 = −19 − 4 = −23
∴ 𝑡3 = 𝑡2 + 𝑑 = −23 + −4 = −23 − 4 = −27
∴ 𝑡4 = 𝑡3 + 𝑑 = −27 + −4 = −27 − 4 = −31
Hence , A.P is -19, -23 , -27, -31,……….
(6) a = -19, d = -4

20. 3. Find the first term and common difference for each of the A.P.
(1) 5, 1, -3, -7, . . .
Solution: From given sequence ,
𝑡1 = 5 , 𝑡2 = 1 , 𝑡3 = −3 , 𝑡4 = −7
Now, common difference (d) : 𝑡2 − 𝑡1 = 1 − 5 = −4
𝑡3 − 𝑡2 = −3 − 1 = −4
𝑡1 = 𝑎 = 5
Answer: First term (a) = 5 & common difference d = -4
∴ 𝒂 = 𝟓
∴ 𝒅 = −𝟒

21. Solution: From given sequence ,
𝑡1 = 0.6 , 𝑡2 = 0.9 , 𝑡3 = 1.2 , 𝑡4 = 1.5
Common difference (d) : 𝑡2 − 𝑡1 = 0. 9 − 0.6 = 0.3
𝑡3 − 𝑡2 = 1.2 − 0.9 = 0.3
𝑡1 = 𝑎 = 0.6
Answer: First term (a) = 0.6 & d = 0.3
∴ 𝒂 = 𝟎. 𝟔
∴ 𝒅 = 𝟎. 𝟑
(2) 0.6, 0.9, 1.2, 1.5, . . .

22. (3) 127, 135, 143, 151, . . .
Solution: From given sequence ,
𝑡1 = 127 , 𝑡2 = 135 , 𝑡3 = 143 , 𝑡4 = 151
Now, common difference (d) : 𝑡2 − 𝑡1 = 135 − 127 = 8
𝑡3 − 𝑡2 = 143 − 135 = 8
𝑡1 = 𝑎 = 127
Answer: First term (a) = 127 & d = 8
∴ 𝒂 = 𝟏𝟐𝟕
∴ 𝒅 = 𝟖

23. 𝟒 .
𝟏
𝟒
,
𝟑
𝟒
,
𝟓
𝟒
,
𝟕
𝟒
, … . .
Solution: From given sequence ,
𝑡1 =
1
4
, 𝑡2 =
3
4
, 𝑡3 =
5
4
, 𝑡4 =
7
4
Common difference (d) : 𝑡2 − 𝑡1 =
3
4
−
1
4
=
2
4
=
1
2
𝑡3 − 𝑡2 =
5
4
−
3
4
=
2
4
=
1
2
𝑡1 = 𝑎 =
1
4
Answer: First term (a) =
1
4
& d =
1
2
∴ 𝒂 =
𝟏
𝟒
∴ 𝒅 =
𝟏
𝟐

24. Thank You…