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Geometric Sequence

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TechMathII - 1.2 - Sequences
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Geometric Sequence

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Geometric Sequence

  1. 1. Grade 10 – Mathematics Quarter I GEOMETRIC SEQUENCE
  2. 2. Objectives: • define geometric sequence; • identify the common ratio and the next term of the sequence; • find the nth term of the geometric sequence; and • solve problems involving geometric sequence.
  3. 3. Find the ratio of the second number to the first number. 2, 81 𝑟 = 8 2 = 𝟒 -3, 92 𝑟 = 9 −3 = −𝟑 1, 1 2 3 𝑟 = 1 2 1 = 𝟏 𝟐
  4. 4. A sequence is geometric if there exist a number r, called the common ratio. The common ratio, r can be determined by dividing any term in the sequence by the term that precedes it.
  5. 5. Identify the common ratio and the next term in the following sequences. 1, 2, 4, 8, …1 𝑟 = 2 1 = 2 The next term is 16 since 8(2) = 16. 80, 20, 5, …2 𝑟 = 20 80 = 1 4 The next term is 5 4 since 5( 1 4 ) = 5 4 . 2, -8, 32, -1283 𝑟 = −8 2 = −4 The next term is 512 since -128(-4) = 512.
  6. 6. State whether each of the following sequences is geometric or not. 1 2 3 5, 20, 80, 320, … 7 2, 5 2, 3 2, 2, … 5, -10, 20, -40, … 4 10 3 , 10 6 , 10 9 , 10 15 20 5 = 4 80 20 = 4 320 80 = 4 5 2 7 2 = 5 7 3 2 5 2 = 3 5 −10 5 = −2 20 −10 = −2 ൗ10 6 ൗ10 3 = 1 2 ൗ10 9 ൗ10 6 = 2 3 ൗ10 15 ൗ10 9 = 3 5
  7. 7. The nth term of a geometric sequence is given by 𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏 , 𝑟 ≠ 0. where 𝑎1 is the first term and r is the common ratio.
  8. 8. What is the 10th term of the geometric sequence 8, 4, 2, 1, …? 𝑟 = 4 8 = 1 2 𝑎 𝑛 = 8 𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏 𝑎10 = 8 1 2 10−1 = 8 1 2 9 𝑎10 = 8 512 = 8 1 512 = 𝟏 𝟔𝟒
  9. 9. Find the missing term in 3, 12, 48, ____, _____ 𝑟 = 12 3 = 4 48(4) = 192 192 192(4) = 768 768 Find the missing term in __, __, 32, 64, 128, … 𝑟 = 64 32 = 2 32 2 = 16 16 16 2 = 8 8
  10. 10. During the initial phase of an outbreak of measles, the number of infections can grow geometrically. If there were 4, 8, 16, … on the first three days of an outbreak of measles, how many will be infected on the 6th day? 𝑟 = 8 4 = 2 𝑎 𝑛 = 4 𝑛 = 6 𝒂 𝒏 = 𝒂 𝟏 𝒓 𝒏−𝟏 𝑎6 = 4 2 6−1 𝑎6 = 4 2 5 𝑎6 = 4(32)= 128 There will be 128 people infected with measles on the 6th day.

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