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SEQUENCES AND SERIES
GRADE 12
OUTCOMES OF LESSON:
• Revision of linear and quadratic sequences.
• An introduction to the arithmetic sequence and
formula.
• An introduction to the geometric sequence and
formula.
• Lastly , you will learn about Series for example both
arithmetic and geometric series.
REVISION OF QUADRATIC PATTERNS
• In grades 10 and 11 we dealt with quadratic number patterns
of the general term 𝑇𝑛 = 𝑎𝑛2
+bn+c and linear or (arithmetic
sequence) with the general term 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑
• Remember a quadratic pattern has a second constant
difference among consecutive terms and linear patterns a 1st
constant difference.
EXAMPLE
CONSIDER THE FOLLOWING NUMBER PATTERN
2 ; 3 ; 6 ; 11 ; …
1. Determine the 𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 (𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑡𝑒𝑟𝑚) and
hence the value of the 42nd term.
2. Determine which term will equal 1091.
REVISION EXERCISE: DETERMINE THE GENERAL TERM
FOR EACH NUMBER PATTERN.
A.2 ; 6 ; 14 ; 26 ; …
B.4 ; 9 ; 16 ; 25 ; …
C.1; 3 ; 6 ; 10 ; …
D.-1 ; 0 ; 3 ; 8 ; …
E.-3 ; -6 ; -11 ; -18 ; …
ARITHMETIC SEQUENCES
• Note we can rewrite this pattern using only the first (1st) term
and the Constant difference
• Consider the following linear number pattern: 7 ; 10 ; 13 ; 16
; 19 ; …
• Term 1 T1=7 and constant difference = T2-T1=T3-T2=d=3
ARITHMETIC SEQUENCES
T1=7
T2=7+3=10
T3=7+3+3=7+2(3)=13
T4=7+3+3+3=7+3(3)=16
THEREFORE, WE CAN CONCLUDE THAT T10=7+9(3)=34
ARITHMETIC SEQUENCES
• Therefore, the general term of the pattern is 𝑇𝑛 = 7 + 𝑛 − 1 (3) and we can therefore simplify this
expression into 𝑇𝑛 = 4 + 3𝑛.
• In most cases 𝑎 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝑎𝑛𝑑 𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 and we can use
these variables to write an expression for the general term that is 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 .
𝑎 = 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚
𝑑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑛 = 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓
𝑡𝑒𝑟𝑚
𝑡𝑒𝑟𝑚
𝑛𝑢𝑚𝑏𝑒𝑟
𝑇𝑛 = 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑟 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑡𝑒𝑟𝑚.
EXAMPLE 1: CONSIDER THE FOLLOWING ARITHMETIC
SEQUENCE 3 ; 5 ; 7 ; 9 ; …
a)Determine a formula for the general
term of the above sequence.
b)Find the value of the 50th term.
EXAMPLE 2.CONSIDER THE FOLLOWING SEQUENCE:
-5 ; -9 ; -13 ; -17 ; …
I. Show that the following sequence is
arithmetic.
II.Find the value of the 25th term of the
sequence.
EXERCISE 1
a. 𝑥 ; 4𝑥 + 5 ; 10𝑥 − 5 ; are the first three terms of an
arithmetic sequence. Determine the value of x and hence
the sequence.
b.Determine the general term of the following linear
sequence 4 ; -2 ; -8 ; …
c. 𝑝 ; 2𝑝 + 2 ; 5𝑝 + 3 ; … are the first three terms of an
arithmetic sequence. Calculate the value of p. Moreover,
determine the sequence and calculate the 49th term.
Which term is equal to 100,5?
GEOMETRIC SEQUENCES: LET US TAKE A LOOK AT THE
FOLLOWING NUMBER PATTERN 6 ; 12 ; 24 ; 48 ; 96 ; ….
• It is clear that this pattern is not linear (arithmetic) since there is
no constant difference between consecutive terms. Neither , is
there a second constant difference. However, this sequence is
obtained by multiplying the previous term by 2 .
• Notice ,that we can calculate a ratio among consecutive terms in
the following way
𝑇2
𝑇1
=
𝑇3
𝑇2
=
𝑇4
𝑇3
=r
• This ratio we calculate is called the “constant ratio” and therefore
is used to determine the next term by either multiplying or
dividing by this constant ratio.
GEOMETRIC SEQUENCES: 6 ; 12 ; 24 ; 48 ; 96 ; …
• This kind of number pattern is called a geometric or exponential pattern.
Moreover, we can rewrite this pattern using only the first term and the
constant ratio.
• T1=6
• T2=6×2=12
• T3=6×2×2=6×22 =24
• T4=6×2×2×2=6×23
=48
• Therefore, T100=6×299 and therefore in algebraic terms 𝑇𝑛 = 6 × 2 𝑛−1 and
is known as the general pattern of the number pattern.
GEOMETRIC SEQUENCES:
• If a is the first term and r is the constant ratio among consecutive terms of
a geometric sequence then the pattern can be written as follows:
• T1=a
• T2=a×r
• T3=a×𝑟2
• T4=a× 𝑟3
• T100=a×𝑟99
• 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑇𝑛 = 𝑎 × 𝑟 𝑛−1
Chapter 1 sequences and series lesson

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Chapter 1 sequences and series lesson

  • 2. OUTCOMES OF LESSON: • Revision of linear and quadratic sequences. • An introduction to the arithmetic sequence and formula. • An introduction to the geometric sequence and formula. • Lastly , you will learn about Series for example both arithmetic and geometric series.
  • 3. REVISION OF QUADRATIC PATTERNS • In grades 10 and 11 we dealt with quadratic number patterns of the general term 𝑇𝑛 = 𝑎𝑛2 +bn+c and linear or (arithmetic sequence) with the general term 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 • Remember a quadratic pattern has a second constant difference among consecutive terms and linear patterns a 1st constant difference.
  • 4. EXAMPLE CONSIDER THE FOLLOWING NUMBER PATTERN 2 ; 3 ; 6 ; 11 ; … 1. Determine the 𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 (𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑡𝑒𝑟𝑚) and hence the value of the 42nd term. 2. Determine which term will equal 1091.
  • 5. REVISION EXERCISE: DETERMINE THE GENERAL TERM FOR EACH NUMBER PATTERN. A.2 ; 6 ; 14 ; 26 ; … B.4 ; 9 ; 16 ; 25 ; … C.1; 3 ; 6 ; 10 ; … D.-1 ; 0 ; 3 ; 8 ; … E.-3 ; -6 ; -11 ; -18 ; …
  • 6. ARITHMETIC SEQUENCES • Note we can rewrite this pattern using only the first (1st) term and the Constant difference • Consider the following linear number pattern: 7 ; 10 ; 13 ; 16 ; 19 ; … • Term 1 T1=7 and constant difference = T2-T1=T3-T2=d=3
  • 8. ARITHMETIC SEQUENCES • Therefore, the general term of the pattern is 𝑇𝑛 = 7 + 𝑛 − 1 (3) and we can therefore simplify this expression into 𝑇𝑛 = 4 + 3𝑛. • In most cases 𝑎 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝑎𝑛𝑑 𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 and we can use these variables to write an expression for the general term that is 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑 . 𝑎 = 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝑑 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑛 = 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑒𝑟𝑚 𝑡𝑒𝑟𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑇𝑛 = 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑛𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑟 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑡𝑒𝑟𝑚.
  • 9. EXAMPLE 1: CONSIDER THE FOLLOWING ARITHMETIC SEQUENCE 3 ; 5 ; 7 ; 9 ; … a)Determine a formula for the general term of the above sequence. b)Find the value of the 50th term.
  • 10. EXAMPLE 2.CONSIDER THE FOLLOWING SEQUENCE: -5 ; -9 ; -13 ; -17 ; … I. Show that the following sequence is arithmetic. II.Find the value of the 25th term of the sequence.
  • 11. EXERCISE 1 a. 𝑥 ; 4𝑥 + 5 ; 10𝑥 − 5 ; are the first three terms of an arithmetic sequence. Determine the value of x and hence the sequence. b.Determine the general term of the following linear sequence 4 ; -2 ; -8 ; … c. 𝑝 ; 2𝑝 + 2 ; 5𝑝 + 3 ; … are the first three terms of an arithmetic sequence. Calculate the value of p. Moreover, determine the sequence and calculate the 49th term. Which term is equal to 100,5?
  • 12. GEOMETRIC SEQUENCES: LET US TAKE A LOOK AT THE FOLLOWING NUMBER PATTERN 6 ; 12 ; 24 ; 48 ; 96 ; …. • It is clear that this pattern is not linear (arithmetic) since there is no constant difference between consecutive terms. Neither , is there a second constant difference. However, this sequence is obtained by multiplying the previous term by 2 . • Notice ,that we can calculate a ratio among consecutive terms in the following way 𝑇2 𝑇1 = 𝑇3 𝑇2 = 𝑇4 𝑇3 =r • This ratio we calculate is called the “constant ratio” and therefore is used to determine the next term by either multiplying or dividing by this constant ratio.
  • 13. GEOMETRIC SEQUENCES: 6 ; 12 ; 24 ; 48 ; 96 ; … • This kind of number pattern is called a geometric or exponential pattern. Moreover, we can rewrite this pattern using only the first term and the constant ratio. • T1=6 • T2=6×2=12 • T3=6×2×2=6×22 =24 • T4=6×2×2×2=6×23 =48 • Therefore, T100=6×299 and therefore in algebraic terms 𝑇𝑛 = 6 × 2 𝑛−1 and is known as the general pattern of the number pattern.
  • 14. GEOMETRIC SEQUENCES: • If a is the first term and r is the constant ratio among consecutive terms of a geometric sequence then the pattern can be written as follows: • T1=a • T2=a×r • T3=a×𝑟2 • T4=a× 𝑟3 • T100=a×𝑟99 • 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑇𝑛 = 𝑎 × 𝑟 𝑛−1