![Example 6: $100 is invested into an account (earning 5% compound
interest per annum), at the start of every year. Find the
amount in the account at the end of the 8th
year.
The amount in the account at the start of each year earns 5% interest.
i.e. The amount is increased by 5%.
To increase an amount by 5%, multiply by 1.05
At the end of the 1st
year
The amount in the account:
= (100×1.05)
At the start of the 2nd
year= 100 + (100×1.05)
At the end of the 2nd
year = [100 + (100×1.05)]×1.05
= (100×1.05) + (100×1.052
)
At the start of the 3rd
year = 100 + (100×1.05) + (100×1.052
)
At the end of the 3rd
year = {100 + (100×1.05) + (100×1.052
)}×1.05
= (100×1.05) + (100×1.052
) + (100×1.053
)](https://image.slidesharecdn.com/geometricprogressions-130813005302-phpapp01/85/Geometric-Progressions-9-320.jpg)
Geometric Progressions
- 2. A Geometric Progression (GP) or Geometric Series is one in which each term is found by multiplying the previous term by a fixed number (common ratio). If the first term is denoted by a, and the common ratio by r, the series can be written as: a + e.g. 5 + 10 + 20 + 40 + … Hence the nth term is given by: 1− = n n aru or 2 – 4 + 8 –16 + … ar2 +ar + ar3 + …
- 3. The sum of the first n terms, Sn is found as follows: Sn = a + ar + ar2 + ar3 +…arn–2 + arn–1 …(1) Multiply throughout by r: r Sn = ar + ar2 + ar3 + ar4 + …arn–1 + arn …(2) Now subtract (2) – (1): r Sn – Sn = arn – a Factorise: Sn (r – 1) = a (rn – 1 ) Hence: ( 1) 1 n n a r S r − = −
- 4. Example 1: For the series 2 + 6 + 18 + 54 + … Find a) The 10th term. b) The sum of the first 8 terms. a) For the series, we have: a = 2, r = 3 Using: un = arn–1 u10 = 2(39 ) = 39366 ( 1) b) Using 1 n n a r S r − = − 8 8 2(3 1) 3 1 S − = − = 6560
- 5. Example 2: For the series 32 – 16 + 8 – 4 + 2 … Find a) The 12th term. b) The sum of the first 7 terms. a) For the series, we have: a = 32, r = Using: un = arn–1 u12 = 32 ( 1) b) Since 1 n n a r S r − = − 71 2 7 1 2 32(1 ( ) ) 1 ( ) S − − = − − = 21.5 1 2 – ( ) 1 2 – 11 = 1 64 – We can write this as: (1 ) 1 n n a r S r − = − (This ensures that the denominator is positive).
- 6. Example 3: ∑∑ == − 12 1 10 0 )203(b))2(6a):Find r r r r a) Firstly, we need to find the first few terms: The series is: 6 We have a = 6, r = 2 The number of terms, n = 11 ( 1) Using 1 n n a r S r − = − 11 11 6(2 1) 2 1 S − ∴ = − = 12282 b) ∑= − 12 1 )203( r r = (31 – 20 ) + (32 – 20 ) + (33 – 20 ) + … = (31 + 32 + 33 + …) – ( 20 + 20 + 20 + …) A Geometric Series 12 of these 13 )13(3 12 − − = – (20 × 12) = 797160 – 240 = 796920 + 12 + 24 + 48 + …
- 7. Example 4: In a Geometric Series, the third term is 36, and the sixth term is 121.5. For the series, find the common ratio, the first term and the twentieth term. The third term is 36 i.e. u3 = 36 The sixth term is 121.5 i.e. u6 = 121.5 Using: un = arn–1 ar2 = 36 ….(1) ar5 = 121.5 ….(2) Now, divide equation (2) by equation (1): 36 5.121 2 5 = ar ar So r3 = 3.375 3 375.3=∴r r = 1.5 Substitute this value into equation (1): a(1.5)2 = 36 a = 16 Now the 20th term, u20 = ar19 = 16 (1.5)19 = 35469 (To the nearest integer)
- 8. Example 5: The first three terms of a geometric progression are x, x + 3, 4x. Find the two possible values for the common ratio. For each value find these first three terms, and the common ratio. The ratio of a G.P. is found by dividing a term by the previous term: 3 4 i.e. + = x x r x x r 3 and + = x x x x 3 3 4 + = + ∴ Now, x (4x) = (x + 3)(x + 3) 3x2 – 6x – 9 = 0 Divide by 3: x2 – 2x – 3 = 0 (x – 3)(x + 1) = 0 So, either x = 3, giving the terms: 3, 6, 12 with ratio r = 2 or x = –1, giving the terms: –1, 2, –4 with ratio r = –2 4x2 = x2 + 6x + 9∴
- 9. Example 6: $100 is invested into an account (earning 5% compound interest per annum), at the start of every year. Find the amount in the account at the end of the 8th year. The amount in the account at the start of each year earns 5% interest. i.e. The amount is increased by 5%. To increase an amount by 5%, multiply by 1.05 At the end of the 1st year The amount in the account: = (100×1.05) At the start of the 2nd year= 100 + (100×1.05) At the end of the 2nd year = [100 + (100×1.05)]×1.05 = (100×1.05) + (100×1.052 ) At the start of the 3rd year = 100 + (100×1.05) + (100×1.052 ) At the end of the 3rd year = {100 + (100×1.05) + (100×1.052 )}×1.05 = (100×1.05) + (100×1.052 ) + (100×1.053 )
- 10. The amount we now have: At the end of the 3rd year = (100×1.05) + (100×1.052 ) + (100×1.053 ) This is a GP With a = 100 × 1.05 = 105 and r = 1.05 We want the sum to 8 terms, i.e. n = 8 ( 1) Using 1 n n a r S r − = − 8 8 105 (1.05 1) 1.05 1 S − ∴ = − = 1002.66 Hence, the amount in the account after 8 years is £1002.66 (To the nearest penny.)
- 11. ( 1) 1 n n a r S r − = − A Geometric Series is one in which the terms are found by multiplying each term by a fixed number (common ratio). The nth term is given by: The sum of the first n terms is given by: 1− = n n aru In problems where r < 1 it is better to write the above as: (1 ) 1 n n a r S r − = − Summary of key points:
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