2. ARITHMETIC SEQUENCES
ο In the sequence 2, 5, 8, 11, 14, β¦, 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. ο 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.
ο The following notations we will be used are:
π1 = πππ π‘βπ ππππ π‘ π‘πππ
π π = πππ π‘βπ ππ‘β π‘πππ
π = ππππππ ππππππππππ
π = πππ π‘βπ ππ’ππππ ππ π‘πππ ππππ π1 π‘π π π
π π = πππ π‘βπ π π’π ππ π‘βπ ππππ π‘ π π‘ππππ
4. ο For example, the arithmetic sequence is given by
1, 6, 11, 16, β¦
we can say that on the sequence, π1 = 1 and
π = 5 (each term is found by adding 5 to the
preceding term). Thus,
π2 = π1 + 5 = 1 + 5 = 6
π3 = π2 + 5 = 6 + 5 = 11
π4 = π3 + 5 = 11 + 5 = 16
If we take any term and subtract the
preceding term, the difference is always 5. This is why
π is called the ππππππ ππππππππππ of the sequence
5. FORMULA FOR THE nth TERM
ο 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.
1π π‘ π‘πππ = π1 = π1 + 0π
2ππ π‘πππ = π2 = π1 + 1π
3ππ π‘πππ = π3 = π2 + 2π
4π‘β π‘πππ = π4 = π3 + 3π πππ π π ππππ‘β
In each case, the nth term, is the first term plus (n β 1)
times the common difference d. Thus, we have the
general formula or the nth term formula for arithmetic
sequence
ππ‘β π‘πππ = π π = π1 + π β 1 π
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:
π1 = 3, π = 4
π π = π1 + π β 1 π
π5 = 3 + 5 β 1 4 = 3 + 16 = 19
π11 = 3 + 11 β 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
ο The sum of first π terms in an arithmetic
sequence can also be obtained by a
formula. Let π π denote the sum of the first π
terms of an arithmetic sequence. Then,
π π = π1 + π2 + π3 + π4 + β― + π π
π π = π1 + π1 + π + π1 + 2π + π1 + 3π + β― + π1 + π β 1 π eq. 1
Reversing the order of addition,
π π = π π + π πβ1 + π πβ2 + β― + π1
π π = π1 + π β 1 π + π1 + π β 2 π + π1 + π β 3 π + β― + π1 eq. 2
9. Sample Problem
2. Find the 9th term and the sum of the first nine
terms of the arithmetic sequence with π1 = β2
and π = 5.
Answer:
π π = π1 + π β 1 π
π9 = β2 + 9 β 1 5 = β2 + 40 = 38
π π =
π
2
2π1 + π β 1 π
π9 =
9
2
2 β2 + 9 β 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,β¦β¦.
Answer:
π1 = β7 and π = 3
π π = π1 + π β 1 π
π20 = β7 + 20 β 1 3 = 50
π π =
π
2
2π1 + π β 1 π
π20 =
20
2
2 β7 + 20 β 1 3 = 430
Therefore, 50 and 430 are the 20th term and the sum
of the first 20 terms, respectively.
11. ARITHMETIC MEANS
ο The terms between π1 and π π of an
arithmetic sequence are called arithmetic
means of π1 and π π. Thus, the arithmetic
means between π1 and π5 are π2, π3 and
π4
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, π1 = 8
and π6 = β7, we can solve for π using the formula π π = π1 +
π β 1 π.
β7 = 8 + 6 β 1 π
π = β3
Hence, π2 = π1 + π = 8 β 3 = 5
π3 = π2 + π = 5 β 3 = 2
π4 = π3 + π = 2 β 3 = β1
π5 = π4 + π = β1 β 3 = β4
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 π1 = 8 and π = β3.
2. Find the 5th term and the sum of the first
five terms if π1 = 6 and π = 3.