An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. The formula for the nth term of an arithmetic sequence is an = a1 + (n-1)d, where a1 is the first term, n is the term number, and d is the common difference. This document provides examples of using the arithmetic sequence formula to find terms, common differences, and formulas for arithmetic sequences given various terms or conditions.

1. ARITHMETIC
SEQUENCE

2. ARITHMETIC SEQUENCE
•IS A SEQUENCE OF NUMBER WHOSE THE
DIFFERENCE BETWEEN CONSECUTIVE TERM IS
THE SAME

3. FORMULA
•an=a1+(n-1)d
•a1-first term
•n- number of term
•d-common difference

4. Example #1:
•Verify if the following are arithmetic.
•1. 7,11,15,19
•Common difference
•4

5. Example # 2
1,4,7,10
Common difference
_____

6. Example # 3
• Find the formula for the nth term of an arithmetic sequence with
common difference of 3 and whose first term is 12.
• an=a1+(n-1)d
• Given
• a1=12
• d=3

7. Solution
•an=12+(n-1)3
•an=12+3n-3
•an=3n+9

8. Example #4
• Given the first 4 terms of the arithmetic sequence : 10,8,6,4 find the
23rd term of the sequence

9. Solution: an=a1+(n-1)d
•Given
•a1=10
•d=-2
•a23=10+(23-1)(-2)
•a23=10+(22)(-2)
•a23=10-44
• -34

10. Example # 5
• If a sequence has a first term of a1 = 12 and a common difference d=-
7. Write the formula that describes this sequence. Use the formula of
the arithmetic sequence.
• Given
•a1=12
•d=-7

11. Solution
•an=a1+(n-1)d
•an=12+(n-1)(-7)
•12+(-7n+7)
•12-7n+7
•an=-7n+9

12. Example # 6
• Write the formula describing the sequence 6,14,22,30,…
• Given:
•a1=6
• d=8
•an=a1+(n-1)d
•an=6+(n-1)(8)
•an=6+8n-8
•an =8n-2

13. Example # 7
• Find the 45th term of the arithmetic sequence -9,-2,5,7,12,…
•an=a1+(n-1)d
•a45=-9+(45-1)(7)
•-9+308
•a45= 299

14. Example # 8
• Write the formula of a sequence with two given terms
•a5=-32
•a18=85
• Use the information of each term to construct an equation with two
unknowns using the arithmetic sequence formula.

15. For a5=-32
•an=a1+(n-1)d
•-32 =a1+(5-1)d
•-32 =a1+(4)d

16. For a18=85
•an=a1+(n-1)d
•85 =a1+(18-1)d
•85=a1+(17)d

17. Solve the system of equations using
the Elimination Method. Multiply Equation # 1
by −1 and add it to Equation #2 to eliminate a1
•-32 =a1+(4)d (-1)
•32 =-a1-4d
•32 =-a1-4d
•85 = a1+17d
•117=13d
• 9=d

18. After finding the value of the common difference,
it is now easy to find the value of the first term.
Back substitute d=9 to any of the two equations to
find a1​.
•-32 =a1+(4)d
•-32 =a1+(4)(9)
•-32 =a1+36
•-32 – 36 =a1
•– 68 =a1

19. a1=-68
d=9
• The formula were looking for is
• an=a1+(n-1)d
• an=-68+(n-1)d
• an=-68+9n-9
•an= 9n - 77