MATH 10 PPT LESSON

1. 1. Tinapa (smoked fish) is best paired with Atchara
(pickled papaya). Diana, a Tinapa vendor in Salinas,
Rosario, Cavite, decided to sell atchara at her store.
On the first week, she started to sell 7 atchara bottles
and due to high demand, she decided to add 7 more
bottles on each succeeding weeks. Supposed that the
pattern continues, how may week is needed to sell
105 atchara bottles?

2. Which term of the arithmetic sequence
7, 14, 21, 28, .… is 105?
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
Given:
𝒂𝟏=7
d= 7
n=?
𝒂𝒏 = 𝟏𝟎𝟓
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟏𝟎𝟓 = 𝟕 + 𝒏 − 𝟏 𝟕
𝟏𝟎𝟓 = 𝟕 + 𝟕𝒏 − 𝟕
𝟏𝟎𝟓 = 𝟕𝒏
𝟏𝟎𝟓
𝟕
=
𝟕𝒏
𝟕
𝟏𝟓 = 𝒏

4. A Zumba Program calls for 15
minutes dancing each day for a week.
Each week thereafter, the amount of
time spent dancing increases by 5
minutes per day. In how many weeks
will a person be dancing 200 minutes?

5. Which term of the arithmetic sequence
15, 20, 25, 30, .… is 200?
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
Given:
𝒂𝟏=15
d= 5
n = ?
𝒂𝒏 = 𝟐𝟎𝟎
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟐𝟎𝟎 = 𝟏𝟓 + 𝒏 − 𝟏 𝟓
𝟐𝟎𝟎 = 𝟏𝟓 + 𝟓𝒏 − 𝟓
𝟐𝟎𝟎 = 𝟏𝟎 + 𝟓𝒏
𝟐𝟎𝟎 − 𝟏𝟎 = 𝟓𝒏
𝟏𝟗𝟎
𝟓
=
𝟓𝒏
𝟓
𝟑𝟖 = 𝒏

7. The population of a city was 7million and
after 12years the population is now 40
million.
The population growth can be described as
an arithmetic sequence.
Find the annual growth of the population.
7, __, __, __, __, __, __, __, __, __, __, __, 40

8. What is the common difference of the arithmetic sequence if the
first term is 7, last term is 40, and the number of terms is 12?
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
Given:
𝒂𝟏=7
d= ?
𝒂𝒏 = 𝟒𝟎
𝒏 = 𝟏𝟐
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟒𝟎 = 𝟕 + 𝟏𝟐 − 𝟏 𝒅
𝟒𝟎 = 𝟕 + 𝟏𝟏 𝒅
𝟒𝟎 = 𝟕 + 𝟏𝟏𝒅
𝟒𝟎 − 𝟕 = 𝟏𝟏𝒅
𝟑𝟑 = 𝟏𝟏𝒅
𝟑𝟑
𝟏𝟏
=
𝟏𝟏𝒅
𝟏𝟏
𝟑 = 𝒅

10. What is the common difference of the arithmetic sequence if the
first term is 5, last term is 278, and the number of terms is 40?
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
Given:
𝒂𝟏=5
d= ?
𝒂𝒏 = 𝟐𝟕𝟖
𝒏 = 𝟒𝟎
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟐𝟕𝟖 = 𝟓 + 𝟒𝟎 − 𝟏 𝒅
𝟐𝟕𝟖 = 𝟓 + 𝟑𝟗 𝒅
𝟐𝟕𝟖 = 𝟓 + 𝟑𝟗𝒅
𝟐𝟕𝟖 − 𝟓 = 𝟑𝟗𝒅
𝟐𝟕𝟑 = 𝟑𝟗𝒅
𝟐𝟕𝟑
𝟑𝟗
=
𝟑𝟗𝒅
𝟑𝟗
𝟕 = 𝒅

12. The 10th term of an arithmetic sequence is 52 and the 20th
term is 102. Find the common difference.
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
Given: assume 𝑎10 as 𝑎1 &𝑎20 𝑎𝑠 𝑎11
d= ?
𝒂𝟏 = 𝟓𝟐
𝒂𝟏𝟏 = 𝟏𝟎𝟐
n= 𝟏𝟏
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟏𝟎𝟐 = 𝟓𝟐 + 𝟏𝟏 − 𝟏 𝒅
𝟏𝟎𝟐 = 𝟓𝟐 + 𝟏𝟎 𝒅
𝟏𝟎𝟐 = 𝟓𝟐 + 𝟏𝟎𝒅
𝟏𝟎𝟐 − 𝟓𝟐 = 𝟏𝟎𝒅
𝟓𝟎 = 𝟏𝟎𝒅
𝟓𝟎
𝟏𝟎
=
𝟏𝟎𝒅
𝟏𝟎
𝟓 = 𝒅

13. The 10th term of an arithmetic
sequence is 52 and the 20th term is
102. Find the first term.

14. The 10th term of an arithmetic sequence is 52 and the
20th term is 102. Find the first term.
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= 5
𝒂𝟏 =?
𝒂𝟏𝟎 = 𝟓𝟐
n= 𝟏𝟎
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟓𝟐 = 𝒂𝟏 + 𝟏𝟎 − 𝟏 𝟓
𝟓𝟐 = 𝒂𝟏 + 𝟗 𝟓
𝟓𝟐 = 𝒂𝟏 + 𝟒𝟓
𝟓𝟐 − 𝟒𝟓 = 𝒂𝟏
𝟕 = 𝒂𝟏

15. 79 is the 13th term of
___, ___, 9, 16, 23, …
Find the first term.

16. 79 is the 13th term of ___, ___, 9, 16, 23, …
Find the first term.
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= 7
𝒂𝟏 =?
𝒂𝟏𝟑 = 𝟕𝟗
n= 𝟏𝟑
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟕𝟗 = 𝒂𝟏 + 𝟏𝟑 − 𝟏 𝟕
𝟕𝟗 = 𝒂𝟏 + 𝟏𝟐 𝟕
𝟕𝟗 = 𝒂𝟏 + 𝟖𝟒
𝟕𝟗 − 𝟖𝟒 = 𝒂𝟏
−𝟓 = 𝒂𝟏

17. Find the first term of
__,__,__,__, 16, 19, 22, 25, 28, 31

18. Find the first term of
__,__,__,__, 16, 19, 22, 25, 28, 31
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= 3
𝒂𝟏 =?
𝒂𝟏𝟎 = 𝟑𝟏
n= 𝟏𝟎
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟑𝟏 = 𝒂𝟏 + 𝟏𝟎 − 𝟏 3
𝟑𝟏 = 𝒂𝟏 + 𝟗 𝟑
𝟑𝟏 = 𝒂𝟏 + 𝟐𝟕
𝟑𝟏 − 𝟐𝟕 = 𝒂𝟏
𝟒 = 𝒂𝟏

19. The second term of an arithmetic
sequence is 24 and the fifth term is
3. Find the common difference and
the first term.

20. The second term of an arithmetic sequence is 24 and the fifth
term is 3. Find the common difference and the first term.
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
Given: assume 𝑎2 as 𝑎1 &𝑎5 𝑎𝑠 𝑎4
d= ?
𝒂𝟏 = 𝟐𝟒
𝒂𝟒 = 𝟑
n= 𝟒
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟑 = 𝟐𝟒 + 𝟒 − 𝟏 𝒅
𝟑 = 𝟐𝟒 + 𝟑 𝒅
𝟑 = 𝟐𝟒 + 𝟑𝒅
𝟑 − 𝟐𝟒 = 𝟑𝒅
−𝟐𝟏 = 𝟑𝒅
−𝟐𝟏
𝟑
=
𝟑𝒅
𝟑
−𝟕 = 𝒅

21. The second term of an arithmetic sequence
is 24 and the fifth term is 3. Find the
common difference and the first term.
d=-7
___, 24, ___, ____, _3_
31 17 10

22. Find 𝒂𝟏 if
𝒂𝟖 =58 and 𝒂𝟗 =60.

23. Find 𝒂𝟏 if
𝒂𝟖 =58 and 𝒂𝟗 =60.
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= 2
𝒂𝟏 =?
𝒂𝟗 = 𝟔𝟎
n= 𝟗
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟔𝟎 = 𝒂𝟏 + 𝟗 − 𝟏 𝟐
𝟔𝟎 = 𝒂𝟏 + 𝟖 𝟐
𝟔𝟎 = 𝒂𝟏 + 𝟏𝟔
𝟔𝟎 − 𝟏𝟔 = 𝒂𝟏
𝟒𝟒 = 𝒂𝟏

24. Give the arithmetic sequence
of 5 terms if the first term is
8 and the last term is 100.

25. 𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= ?
𝒂𝟏 = 𝟖
𝒂𝟓 = 𝟏𝟎𝟎
n= 𝟓
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟏𝟎𝟎 = 𝟖 + 𝟓 − 𝟏 𝒅
1𝟎𝟎 = 𝟖 + 𝟒 𝒅
𝟏𝟎𝟎 = 𝟖 + 𝟒𝒅
𝟏𝟎𝟎 − 𝟖 = 𝟒𝒅
𝟗𝟐 = 𝟒𝒅
𝟗𝟐
𝟒
=
𝟒𝒅
𝟒
𝟐𝟑 = 𝒅

26. Find the common difference
of the arithmetic sequence
with 𝒂𝟒 = 𝟏𝟎 𝒂𝒏𝒅 𝒂𝟏𝟏 = 𝟒𝟓

27. 1. 55 is the __th term of 4, 7, 10, …

28. 2. 163 is the ___th term of -5, 2, 9, …

29. Solve the following questions.
1. Given the sequence
3, 1, –1, –3, …, find 𝒂𝟏𝟐.

30. 2. Find the 9th term of the
arithmetic sequence
12, 24, 36, …

31. 3. If 𝒂𝟏 = –17 and d = 4,
find 𝒂𝟐𝟐 of the arithmetic
sequence.

32. 4. Find the 16th term of the
arithmetic sequence whose
first term is 6 and the common
difference is 0.25.

33. 5. Which term is 27 in the
arithmetic sequence
54, 51, 48, …?

34. In Arithmetic Sequence
the terms between any
two terms are called
Arithmetic Means.

35. Insert three arithmetic means between 18 and 30.
18, __, __, __, 30
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= ?
𝒂𝟏 = 𝟏𝟖
𝒂𝟓 = 𝟑𝟎
n= 𝟓
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟑𝟎 = 𝟏𝟖 + 𝟓 − 𝟏 𝒅
𝟑𝟎 = 𝟏𝟖 + 𝟒 𝒅
𝟑𝟎 = 𝟏𝟖 + 𝟒𝒅
𝟑𝟎 − 𝟏𝟖 = 𝟒𝒅
𝟏𝟐 = 𝟒𝒅
d= 3

36. Insert four arithmetic means between 7 and 37.
7, __, __, __, __, 37
Solution:
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
d= ?
𝒂𝟏 = 𝟕
𝒂𝟔 = 𝟑𝟕
n= 𝟔
𝒂𝒏
= 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟑𝟕 = 𝟕 + 𝟔 − 𝟏 𝒅
𝟑𝟕 = 𝟕 + 𝟓 𝒅
𝟑𝟎 = 𝟓𝒅
6=d or d=6

37. The arithmetic mean or
average of the number x and
y is given by
m =
𝑥 + 𝑦
2

38. Find the arithmetic mean between 8 and 20
m =
𝑥 + 𝑦
2
8, ___, 20
m =
8 + 20
2
m =
28
2
m =14

39. Find the arithmetic mean between -5 and 7
m =
𝑥 + 𝑦
2
-5, ___, 7
m =
−5 + 7
2
m =
2
2
m =1

40. . Flower farms in Dujali grew different variety of
flowers including anthurium. Monica, a flower
arranger, went to Dujali to buy anthurium. She
plans to arrange the flowers following an arithmetic
sequence with four (4) layers. If she put one (1)
anthurium on the first layer and seven (7) on the
fourth layer, how many anthurium should be placed
on the second and third layer of the flower
arrangement?

41. _1_, ___, ___, _7_
Solution:
d= ?
𝒂𝟏 = 𝟏
𝒂𝟒 = 𝟕
n= 𝟒
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟕 = 𝟏 + 𝟒 − 𝟏 𝒅
𝟕 = 𝟏 + 𝟑 𝒅
𝟕 − 𝟏 = 𝟑𝒅
𝟔 = 𝟑𝒅
2=d or d=2

42. •St. Mary Magdalene Parish Church in Kawit, one
of the oldest churches in Cavite, established in
1624 by Jesuit Missionaries. The church is made of
red bricks preserved for more than a hundred years.
Suppose that the lowest part of the church wall
contains five (5) layers of red bricks, 4bricks on the
top and 16 bricks on the bottom layer. Assuming an
arithmetic sequence, how many bricks are there in
the 2nd, 3rd and 4th layer of the wall?

43. _4_, ___, ___, ___, _16_
Solution:
d= ?
𝒂𝟏 = 𝟒
𝒂𝟓 = 𝟏𝟔
n= 𝟓
𝒂𝒏 = 𝒂𝟏 + 𝒏 − 𝟏 𝒅
𝟏𝟔 = 𝟒 + 𝟓 − 𝟏 𝒅
𝟏𝟔 = 𝟒 + 𝟒 𝒅
𝟏𝟔 − 𝟒 = 𝟒𝒅
𝟏𝟐 = 𝟒𝒅
3 = d

44. Answer the following.
1. Insert two arithmetic means between 20 and 38.
2. Insert three arithmetic means between 52 and 40.
3. Find the missing terms of the arithmetic sequence 5, __, __, __, __, 25.
4. Find the missing terms of the arithmetic sequence 0, __, __, __, __, __, 15.
5. The fifteenth term of an arithmetic sequence is –3 and the first term is 25.
Find the common difference and the tenth term.

45. Use the following numbers inside the box to complete the
arithmetic sequence below. You may use a number more than
once.
1) 2, ___, ___, 14
2) 4, ___, ___, ___, 10
3) 6, ___, ___, ___, 16
4) 9, ___, ___, ___, __, 24
5) ___, 17, ___, ___, 11