The document discusses arithmetic progressions (AP) and geometric progressions (GP). It defines an AP as a sequence where the difference between consecutive terms is constant, and provides the formulas for the nth term, sum of n terms, and examples of finding terms and sums. A GP is defined as a sequence where each term is the previous term multiplied by a common ratio, and the formulas for the nth term and sum of n terms are given, along with examples. Various word problems demonstrate calculating terms and sums for APs and GPs.

1. UNIT-3
ARITHMETIC AND GEOMETRIC
PROGRESSIONS

2. Arithmetic Progression
(AP) is a sequence of numbers in order in
which the difference of any two consecutive
numbers is a constant value.
For example, the series of natural numbers:
1, 2, 3, 4, 5, 6,… is an AP, which has a
common difference between two successive
terms (say 1 and 2) equal to 1 (2 -1).

3. In AP, we will come across three main
terms, which are denoted as:
● Common difference (d)
● nth Term (an)
● Sum of the first n terms (Sn)
Where “d” is a common difference. It
can be positive, negative or zero.

4. FORMULA
• General Form of AP- a, a + d, a + 2d, a + 3d, .
.
• The nth term of AP - tn = a + (n – 1) × d
• Sum of n terms in AP- Sn = n/2[2a +(n − 1)× d]

5. CONTS…
•nth term of A.P.
tn = a + (n – 1) × d
a – first terms
d – common difference (2nd term – 1st term)
•For eg. A.p. 2,4,6,8….
4 -2 =2
6 - 2=2
8 - 6=2

6. Where
a = First term
d = Common difference
n = number of terms
an = nth term

7. Type 1
Example 1: Find the
20th term for the given
AP:3, 5, 7, 9, ……

8. Solution: Given,
3, 5, 7, 9, ……
a = 3, d = 5 – 3 = 2, n = 20
an = a + (n − 1) × d
a20 = 3 + (20 − 1) × 2
a20 = 3 + 38
⇒a20 = 41

9. Example 2: Find the 100th
term for the given AP:5, 10,
15, ……

10. Solution: Given,
a = 5, d = 5 , n = 100
an = a + (n − 1) × d
a100 = 5 + (100 − 1) × 5
⇒a100= 500

11. Example 3:
Find the 17th term of the
Ap with first term five
and common differences
two

12. Solution: Given,
a = 5, d = 2 , n = 17
an = a + (n − 1) × d
a17 = 5 + (17 − 1) × 2
⇒a17 = 37

13. Example 4:
Find the 10th term of the
Ap with first term 8 and
common differences 3

14. Solution: Given,
a = 8, d = 3 , n = 10
an = a + (n − 1) × d
a10 = 8 + (10 − 1) × 3
⇒a 10 = 35

15. Type 2
Example 1:
Find the value of n. If a =
10, d = 5, last term= 95.

16. Solution: Given, a = 10, d = 5, an = 95
From the formula of general term, we have:
an = a + (n − 1) × d
95 = 10 + (n − 1) × 5
(n − 1) × 5 = 95 – 10 = 85
(n − 1) = 85/ 5
(n − 1) = 17
n = 17 + 1
n = 18

17. Example 2:
Find the value of n. If a
= 5, d = 5, last term=
500.

18. Solution: Given, a = 5, d = 5, an = 500
From the formula of general term, we
have:
an = a + (n − 1) × d
500 = 5 + (n − 1) × 5
n = 100

19. Example 3 :
Find the value of n. If a
= 8, d = 3, last term= 62.

20. Solution: Given, a = 8, d = 3, an =
62
From the formula of general term,
we have:
an = a + (n − 1) × d
62 = 8 + (n − 1) × 3
n = 19

21. Example 4 :
Find the value of
n. If a = 8, d = 3,
last term= -19.

22. Solution: Given, a = 8, d = -3, an = -
19
From the formula of general term, we
have:
an = a + (n − 1) × d
-19 = 8 + (n − 1) × -3
n = 10

23. Type 3
Sum of N Terms of AP
Consider an AP consisting “n”
terms.
Example 1 : Let us take the
example of adding natural
numbers up to 15 numbers.

24. AP = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
Given, a = 1, d = 2-1 = 1 and an = 15
Now, by the formula we know;
S = n/2[2a + (n − 1) × d]
= 15/2[2x1+(15-1)x1]
S = 15/2[2+14] = 15/2 [16] = 15 x 8
S = 120
Hence, the sum of the first 15 natural numbers is 120.

25. Example 2 : Find the sum of the first 50
term of the sequence 1,3,5,7,9…...
Given, a = 1, d = 2 and an = 50
Now, by the formula we know;
S = n/2[2a + (n − 1) × d]
= 50/2[2x1+(50-1)x2]
S = 2500

26. Example 3 : Find the sum
of the first 23 term of the
sequence 4,-3,-10,…...

27. Given, a = 4, d = - 7and an = 23
Now, by the formula we know;
S = n/2[2a + (n − 1) × d]
= 23/2[2x4+(23-1)x-7]
S = - 1679

28. Example 4 : Find the sum
of the first 20 term of the
sequence, first term four and
common differences half.

29. Given, a = 4, d = 1/2 and an = 20
Now, by the formula we know;
S = n/2[2a + (n − 1) × d]
= 20/2[2x4+(20-1)x1/2]
S = 175

30. Type 4
Sum of AP when the Last Term is Given
First we need to find out n term
Second as per follow concern formula
Example 1 :1+3.5+6+8+5+....+101

31. Question 1: Find 10th term in the series
1,3,5,7…
A. 16 B.12 C.19 D.20

32. EXPLANATION
a=1 d=2 t=10
tn = a + (n – 1) d
t10 = 1+(10 – 1) 2
= 1+ (9) 2
= 1 +18
t10 ⇒ 19.
1,3,5,7
3-1 =2
5-3=2
7-5= 2 so, d = 2

33. 2. Find the value of n.
If a = 10, d = 5, an = 95
A. 18 B.16 C.11 D.14

34. EXPLANATION
Given, a = 10, d = 5, an = 95
tn = a + (n − 1) × d
95 = 10 + (n − 1) × 5
(n − 1) × 5 = 95 – 10 = 85
(n − 1) = 85/ 5
(n − 1) = 17
n = 17 + 1
⇒ n = 18

35. 3. Find the 20th term for the given AP:3, 5, 7, 9, ……
A. 11 B.41 C.21 D.26

36. EXPLANATION
a = 3, d = 5 – 3 = 2, n = 20
tn = a + (n − 1) × d
t20 = 3 + (20 − 1) × 2
t20 = 3 + 38
⇒t20 = 41

37. 4. Find 16th term in the series 7,13,19,25…
A. 56 B.76 C.97 D.86

38. EXPLANATION
a=7 d=6 t=16
tn = a + (n – 1) d
t16 = 7+(16 – 1) 6
= 7+ (15) 6
= 7 +90
t16 = 97.

39. 5. Find the no. of terms in the series 8,12,16….72
A. 14 B.17 C.27 D.36

40. EXPLANATION
a=8 d=4l=72 i.e.72-8= 64
tn = a + (n – 1) d
72= 8+(n-1) 4
64=(n-1) 4
(n-1) = 64/4 =16
n-1 =16
n= 16+1= 17
n=17.

41. 6. Find 4 + 7 +10 + 13 + 16 +…up to 20 terms
A. 650 B.630 C.324 D.506

42. EXPLANATION
n = 20 a= 4 d=3
Sn = n/2[2a + (n − 1) × d]
= 20/2 [2(4) + (20-1) 3]
= 10 [8+(19)3]
= 650.

43. 7.Find 6 + 9 +12 + …. up to 30 terms
A. 216 B.130 C.162 D.300

44. EXPLANATION
n= 30 a=6 d=3
tn = a + (n – 1) d
30=6+(n-1)3
24/3 = (n-1)
n = 9.
Sn = n/2(a + l)
S9 = n/2(a + l)
= 9/2(6+ 30)
= 9/2 (36)
S9 = 162.

45. 8. Find the sum of first 30 multiples of 4.
A. 1216 B. 1860 C.1162 D.1300

46. EXPLANATION
a = 4, n = 30, d = 4
Sn = n/2 [2a + (n − 1) × d]
= 30/2[2 (4) + (30 − 1) × 4]
= 15[8 + 116]
= 1860

47. Example 1
:1+3.5+6+8.5+....+101

48. Find out the sum of
this sequence…..
10,15,20,25,30,…………..,1
00

49. n=19
Sn=1045

50. GEOMETRIC PROGRESSION
What is a Geometric Progression?
Geometric Progression (GP) is a type of sequence
where each succeeding term is produced by multiplying
each preceding term by a fixed number, which is called
a common ratio.
Example
The example of GP is: 3, 6, 12, 24, 48, 96,…

51. Find the term
tn = arn-1
Formula for sum of first n terms in an
GP
if r >1
Sn = a (rn -1)/ r-1,
if r < 1
Sn = a (1 - rn )/ 1-r,

52. CONTS…
a = First term
r = common ratio
tn = nth term

53. Question 1: Find the 10th term in the series
2,4,8,16…
A. 1024 B.1030 C.1602 D.1300

54. EXPLANATION
a= 2 r= 4/2= 2 n= 10
tn = arn-1
t10 = 2 x 210-1
= 1024.

55. 2. If 2,4,8,…., is the GP, then find its 10th term.
A. 4 B.3 C.6 D.5

56. EXPLANATION
a = 2 r = 4/2 = 2 n=10
tn = arn-1
t10 = 2 x 22-1
= 2 x 2
= 4

57. Find the sum of the first 5
terms of the following
series. Given that the
series is finite: 3, 6, 12, …

58. if r >1
Sn = a (rn -1)/ r-1,
S5 = 3(25 – 1)/(5 -1) = 93/4

59. 3. Find 4 + 12 + 36 +… up to 6 terms
A. 1216 B. 1860 C.1162 D. 1456

60. EXPLANATION
a= 4 r= 3 n=6
Sn = a[(rn-1)/(r-1)] r > 1
= 4(36-1)/ (3-1)
= 2(36-1)
= 1456.

61. 4. Find 1 + 1/2 + 1/4 + ….up to 5 terms

62. EXPLANATION
a = 1 r = ½ n= 5
Sn = a[(1- rn)/(1- r)] r < 1
S5 = a[(1- rn)/(1- r)]
= 1[1- (1/2)5 ]/ 1-1/2
= (1- 1/32)/ 1/2
= 31/32 / ½
= 31/16
= 1 15/16.

63. 5. Find 1 + 1/2 + 1/4 + 1/8 +….∞
A. 8 B.2 C.3 D.5

64. EXPLANATION
a =1 r= ½
S∞=a/1-r (if -1<r<1)
S∞= 1/(1-1/2) = 2.

65. THANK YOU…