The document outlines objectives for Grade 10 mathematics focused on arithmetic series, including definitions and methods for calculating the sum of n terms. It provides formulae for the sum of terms in an arithmetic sequence and includes worked examples for various sequences, demonstrating how to find partial sums. Key examples include finding the sums of sequences like 50, 47, 44 and the first 30 natural numbers.

1. Grade 10 – Mathematics
Quarter I
ARITHMETIC SERIES

2. Objectives:
•define arithmetic series; and
•find the sum of n terms of arithmetic
sequence given the first few terms of an
arithmetic sequence.

3. The sum of the terms of an
arithmetic sequence forms an
arithmetic series.
The sum of the first n terms of a
sequence, called a partial sum, is
denoted by 𝑆 𝑛.

4. The sum of n a terms of an arithmetic
sequence is given by:
𝑖𝑓 𝑎 𝑛 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛, 𝑆 𝑛 =
𝑛
2
𝑎1 + 𝑎 𝑛
𝑖𝑓 𝑎 𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛, 𝑆 𝑛 =
𝑛
2
2𝑎 + 𝑛 − 1 𝑑

5. Find the sum of the first 12 terms of the arithmetic
sequence 50, 47, 44, 41, 38, …
Solution: 𝑎 = 50, 𝑑 = −3, 𝑛 = 12
𝑆 𝑛 =
𝑛
2
2𝑎 + 𝑛 − 1 𝑑
𝑆 𝑛 =
12
2
2 50 + 12 − 1 − 3
𝑆12= 6 100 − 33 𝑺 𝟏𝟐= 402

6. Find the sum of the first 18 terms of the arithmetic
sequence 3, 5, 7, 9, 11, …
Solution: 𝑎 = 3, 𝑑 = 2, 𝑛 = 18
𝑆 𝑛 =
𝑛
2
2𝑎 + 𝑛 − 1 𝑑
𝑆 𝑛 =
18
2
2 3 + 18 − 1 2
𝑆18= 9 6 + 34 𝑺 𝟏𝟖= 40

7. Find the sum of the first 30 natural numbers.
Solution: 𝑎1 = 1, 𝑎 𝑛 = 30, 𝑛 = 30
𝑆 𝑛 =
𝑛
2
𝑎1 + 𝑎 𝑛
𝑆 𝑛 =
30
2
1 + 30
𝑆30= 15 31 𝑺 𝟑𝟎= 465

8. Find the sum of the first 50 multiples of 5.
Solution: 𝑎1 = 5, 𝑎 𝑛 = 50, 𝑛 = 50
𝑆 𝑛 =
𝑛
2
𝑎1 + 𝑎 𝑛
𝑆 𝑛 =
50
2
5 + 50
𝑆50= 25 55 𝑺 𝟓𝟎= 1375

9. Find the sum of all multiples of 3 between 1 and 100.
Solution: 𝑎1 = 3, 𝑎 𝑛 = 99, 𝑑 = 3
First, we determine how many multiples
of 3 there are between 1 and 100.
𝑎 𝑛 = 𝑎 + 𝑛 − 1 𝑑
99 = 3 + 𝑛 − 1 3
99 = 3 + 3𝑛 − 3
99 = 3𝑛
𝒏 = 𝟑𝟑
𝑆 𝑛 =
𝑛
2
𝑎1 + 𝑎 𝑛
𝑆33 =
33
2
3 + 99
𝑺 𝟑𝟑 = 1, 683

10. Find the sum of all multiples of 6 between 1 and 100.
Solution: 𝑎1 = 6, 𝑎 𝑛 = 96, 𝑑 = 6
First, we determine how many multiples
of 6 there are between 1 and 100.
𝑎 𝑛 = 6 + 𝑛 − 1 𝑑
96 = 6 + 𝑛 − 1 6
96 = 6 + 6𝑛 − 6
96 = 6𝑛
𝒏 = 𝟏𝟔
𝑆 𝑛 =
𝑛
2
𝑎1 + 𝑎 𝑛
𝑆33 =
16
2
6 + 96
𝑺 𝟑𝟑 = 𝟖𝟏𝟔