2. ARITHMETIC SEQUENCES
In the sequence 2, 5, 8, 11, 14, …, each
term (after the first) can be obtained by
adding three to the term immediately
preceding it. That is,
the second term = the first term + 3
the third term = the second term + 3
and so forth. A sequence like this is given
a special name
3. An arithmetic sequence is a sequence in which
every term after the first term is the sum of the
preceding term and a fixed number called the
common difference of the sequence.
The following notations we will be used are:
𝑎1 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚
𝑎 𝑛 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑡𝑒𝑟𝑚
𝑑 = 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑛 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚 𝑓𝑟𝑜𝑚 𝑎1 𝑡𝑜 𝑎 𝑛
𝑆 𝑛 = 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛 𝑡𝑒𝑟𝑚𝑠
4. For example, the arithmetic sequence is given by
1, 6, 11, 16, …
we can say that on the sequence, 𝑎1 = 1 and
𝑑 = 5 (each term is found by adding 5 to the
preceding term). Thus,
𝑎2 = 𝑎1 + 5 = 1 + 5 = 6
𝑎3 = 𝑎2 + 5 = 6 + 5 = 11
𝑎4 = 𝑎3 + 5 = 11 + 5 = 16
If we take any term and subtract the
preceding term, the difference is always 5. This is why
𝑑 is called the 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 of the sequence
5. FORMULA FOR THE nth TERM
A general formula for calculating any particular term
of an arithmetic sequence is a useful tool. Suppose
we calculate several terms of any arithmetic
sequence.
1𝑠𝑡 𝑡𝑒𝑟𝑚 = 𝑎1 = 𝑎1 + 0𝑑
2𝑛𝑑 𝑡𝑒𝑟𝑚 = 𝑎2 = 𝑎1 + 1𝑑
3𝑟𝑑 𝑡𝑒𝑟𝑚 = 𝑎3 = 𝑎2 + 2𝑑
4𝑡ℎ 𝑡𝑒𝑟𝑚 = 𝑎4 = 𝑎3 + 3𝑑 𝑎𝑛𝑑 𝑠𝑜 𝑓𝑜𝑟𝑡ℎ
In each case, the nth term, is the first term plus (n – 1)
times the common difference d. Thus, we have the
general formula or the nth term formula for arithmetic
sequence
𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 = 𝑎 𝑛 = 𝑎1 + 𝑛 − 1 𝑑
6. Sample Problem
1. Find the 5th term and 11th terms of the
arithmetic sequence with the first term 3 and
the common difference 4.
Answer:
𝑎1 = 3, 𝑑 = 4
𝑎 𝑛 = 𝑎1 + 𝑛 − 1 𝑑
𝑎5 = 3 + 5 − 1 4 = 3 + 16 = 19
𝑎11 = 3 + 11 − 1 4 = 3 + 40 = 43
Therefore, 19 and 43 are the 5th and the 11th
terms of the sequence, respectively.
7. SUM OF THE FIRST n TERMS
The sum of first 𝑛 terms in an arithmetic
sequence can also be obtained by a
formula. Let 𝑆 𝑛 denote the sum of the first 𝑛
terms of an arithmetic sequence. Then,
𝑆 𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + ⋯ + 𝑎 𝑛
𝑆 𝑛 = 𝑎1 + 𝑎1 + 𝑑 + 𝑎1 + 2𝑑 + 𝑎1 + 3𝑑 + ⋯ + 𝑎1 + 𝑛 − 1 𝑑 eq. 1
Reversing the order of addition,
𝑆 𝑛 = 𝑎 𝑛 + 𝑎 𝑛−1 + 𝑎 𝑛−2 + ⋯ + 𝑎1
𝑆 𝑛 = 𝑎1 + 𝑛 − 1 𝑑 + 𝑎1 + 𝑛 − 2 𝑑 + 𝑎1 + 𝑛 − 3 𝑑 + ⋯ + 𝑎1 eq. 2
9. Sample Problem
2. Find the 9th term and the sum of the first nine
terms of the arithmetic sequence with 𝑎1 = −2
and 𝑑 = 5.
Answer:
𝑎 𝑛 = 𝑎1 + 𝑛 − 1 𝑑
𝑎9 = −2 + 9 − 1 5 = −2 + 40 = 38
𝑆 𝑛 =
𝑛
2
2𝑎1 + 𝑛 − 1 𝑑
𝑆9 =
9
2
2 −2 + 9 − 1 5 = 162
Therefore, 38 and 162 are the 9th term and the
sum of the first nine terms, respectively.
10. 3. Find the 20th term and the sum of the first 20 terms
of the arithmetic sequence -7, - 4, -1, 2,…….
Answer:
𝑎1 = −7 and 𝑑 = 3
𝑎 𝑛 = 𝑎1 + 𝑛 − 1 𝑑
𝑎20 = −7 + 20 − 1 3 = 50
𝑆 𝑛 =
𝑛
2
2𝑎1 + 𝑛 − 1 𝑑
𝑆20 =
20
2
2 −7 + 20 − 1 3 = 430
Therefore, 50 and 430 are the 20th term and the sum
of the first 20 terms, respectively.
11. ARITHMETIC MEANS
The terms between 𝑎1 and 𝑎 𝑛 of an
arithmetic sequence are called arithmetic
means of 𝑎1 and 𝑎 𝑛. Thus, the arithmetic
means between 𝑎1 and 𝑎5 are 𝑎2, 𝑎3 and
𝑎4
12. Sample Problem
1. Find four arithmetic means between 8 and -7.
Answer: Since we must insert four numbers between 8 and -7,
there are six numbers in the arithmetic sequence. Thus, 𝑎1 = 8
and 𝑎6 = −7, we can solve for 𝑑 using the formula 𝑎 𝑛 = 𝑎1 +
𝑛 − 1 𝑑.
−7 = 8 + 6 − 1 𝑑
𝑑 = −3
Hence, 𝑎2 = 𝑎1 + 𝑑 = 8 − 3 = 5
𝑎3 = 𝑎2 + 𝑑 = 5 − 3 = 2
𝑎4 = 𝑎3 + 𝑑 = 2 − 3 = −1
𝑎5 = 𝑎4 + 𝑑 = −1 − 3 = −4
Therefore, the four arithmetic means between 8 and -7 are 5,
2, -1, and -4.
13. Break a leg!
1. Write the first five terms of arithmetic
sequence when 𝑎1 = 8 and 𝑑 = −3.
2. Find the 5th term and the sum of the first
five terms if 𝑎1 = 6 and 𝑑 = 3.