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GEOMETRIC SEQUENCES

GEOMETRIC SEQUENCES. Consider this sequence. Notice that to go from one term to the next multiplication is required. 10, 20, 40, 80, …….. . ÷2. ÷2. This is called a common ratio . (Term 2 is divided by term 1. Term 3 is divided by term 2 and so on.)

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GEOMETRIC SEQUENCES

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  1. GEOMETRIC SEQUENCES

  2. Consider this sequence. Notice that to go from one term to the next multiplication is required. 10, 20, 40, 80, …….. ÷2 ÷2 This is called a common ratio. (Term 2 is divided by term 1. Term 3 is divided by term 2 and so on.) This makes it a geometric sequence. This could be called a “exponential sequence”. Defined by the formula: an= a1 (r)n-1

  3. WARNING: Don’t confuse geometric with sequences that increase through squaring or cubing. You won’t have a problem as long as you check several quotients when looking for the common ratio.

  4. Find the 15th term of the geometric sequence whose first term is 20 and whose common ratio is 1.05. You could write out the first 15 terms or use the formula. an= a1 (r)n-1 a15= 20 (1.05)15-1

  5. Find the 12th term of the geometric sequence 5, 15, 45,…… After we find the common ratio we can just use the formula. an= a1 (r)n-1 a12= 5 (3)12-1

  6. The fourth term of a geometric sequence is 125 and the 10th term is 125/64. Find the 14th term First we have to play with this formula. an= a1 (r)n-1 Isn’t it true that for example if we had. a10 = a1 (r)10-1 a10= a1 (r)9

  7. And isn’t it true that we COULD rewrite this as…. a10= a1 (r)9 a10= a1 r·r·r·r6 And isn’t it also true that we COULD rewrite this as…. a10 = a1 r·r·r·r6 a10 = a1····r6

  8. And doesn’t everything pretty much cancel? a10 = a1 r·r·r·r6 a10 = a1····r6 X X X And that leaves us with a relationship that we can use on this question. a10 = a4 r6 X X X

  9. The fourth term of a geometric sequence is 125 and the 10th term is 125/64. Find the 14th term And that leaves us with a relationship that we can use on this question. a10 = a4 r6 Notice that the subscript and the exponent on the right add up to the subscript on the left.

  10. The fourth term of a geometric sequence is 125 and the 10th term is 125/64. Find the 14th term Substituting and solving for r we find…. 125/64 = 125 r6 1/64 = r6 ½ = r

  11. The fourth term of a geometric sequence is 125 and the 10th term is 125/64. Find the 14th term Substituting and solving for r we find…. 125/64 = 125 r6 1/64 = r6 ½ = r

  12. The fourth term of a geometric sequence is 125 and the 10th term is 125/64. Find the 14th term NOW we can answer the question a14= a10 r4 a14 = 125/64( ½)4 a14 = 125/1024

  13. Finding the SUM of a FINITE GEOMETRIC SEQUENCE

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