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Mastering Geometric Sequences: Solving Problems & Finding Sums

Learn how to calculate sums and solve problems involving geometric sequences. Understand the concept of series and how to find the sum of terms. Includes examples and step-by-step explanations.

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Mastering Geometric Sequences: Solving Problems & Finding Sums

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  1. un = u1 x r n-1 Recap… If I put £400 in a bank account on my 16th birthday and get 5% interest per year. How much money would I have on my 40th birthday? u1 = 400 r = 1.05 n = 25 un = 400 x 1.0524 un = £1290.04

  2. Sequences and Series 2 Learn how to find the sum of a terms in a geometric sequence

  3. A simple problem… A man vows to give his niece 1p on her first birthday and double the previous amount on each of her next birthdays until she is 21. How much money will she have recieved altogether on her 21st birthday? This is n geometric series as we are adding the terms in a geometric sequence. Estimate!!! S = 1 + 2 + 4 + 8 + 16 + 32….1048576 Finding the sum… S = 1 + 2 + 22 + 23 + 24 + 25….219 + 220 Writing as powers of 2… 2S = 2 + 22 + 23 + 24 + 25 + 26….220 + 221 Doubling the series… 2S – S = 221 - 1 Subtracting… S = 221 – 1 = 2097152 – 1 = 2097151 = £20971.51

  4. Let a = u1 = 1st term r = common ratio n = number of terms We found the sum of this bit to be: S = a (rn – 1) S = rn – 1 r – 1 r – 1 Sum of Terms A geometric series is the sum of a geometric sequence S = a + ar + ar2 + ar3 + ar4….arn-2 + arn-1 Writing as powers of r… S = a (1 + r + r2 + r3 + r4….rn-2 + rn-1) Factorise right hand side… We need to multiply by a

  5. Sum of Terms- the important bit The sum of terms in a finite geometric sequence can be calculated using: S = u1 (rn – 1) r – 1 Using sigma notation, we can write…

  6. S = u1 (rn – 1) 108 = 1 (2n – 1) r – 1 2 – 1 The old washing up trick Offer to do the washing up for 1c today, but twice as much every day after. How long until you are a millionaire? Sum = 1000000 x 100 = 108 u1 = 1 r = 2 n = ? 2n = 108 + 1 log 2n = log (108 + 1) n log 2 = log (108 + 1) n = log (108 + 1) / log 2 n = 26.57… In 27 days (less than a month) you will be a millionaire! 108 = 2n – 1

  7. S = u1 (rn – 1) S = 3 (410 – 1) r – 1 4 – 1 Example 1 Find the sum of the first 10 terms of the series 3 + 12 + 48... u1 = 3 r = 4 n = 10 S = 1,048,575

  8. S = u1 (rn – 1) S = 2.5 (310 – 1) r – 1 3 – 1 un = u1 x r n-1 Example 2 Evaluate: 2.5 r = 3 n = 10 u1 = S = 73,810

  9. S = u1 (rn – 1) S = 3 (-0.510 – 1) r – 1 -0.5 – 1 Example 3 Find the sum of the first 10 terms of the series 3 + -1.5 + 0.75... u1 = 3 r = -0.5 n = 10 S = 2.002

  10. Information booklet 2 versions provided. Why? • Either can be used • 2nd version is simpler if r<1

  11. S = 3 (1 -0.510) 1- 0.5 Example 4 Find the sum of the first 10 terms of the series 3 + 1.5 + 0.75... u1 = 3 r = 0.5 n = 10 S = 5.994 What do you think will happen if we increase n?

  12. Activity Turn to page 193-4 of your textbook and answer questions 3,5,6+ in exercise 6D

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