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A Geometric Progression (GP) or Geometric Series isone in which each term is found by multiplying theprevious 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 + …
The sum of the first n terms, Sn is found as follows:
Sn = a + ar + ar2 + ar3 +…ar n–2 + ar
n–1 …(1)
Multiply throughout by r:
r Sn = ar + ar2 + ar3 + ar4 + …ar n–1 + ar
n …(2)
Now subtract (2) – (1): r Sn – Sn = ar n – a
Factorise: Sn (r – 1) = a (r n – 1 )
Hence: ( 1)
1
n
n
a rS
r
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 = ar n–1 u10 = 2(3
9) = 39 366
( 1)b) Using
1
n
n
a rS
r
8
8
2(3 1)3 1
S
= 6560
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 = ar n–1 u12 = 32
( 1)b) Since
1
n
n
a rS
r
712
7 12
32(1 ( ) )1 ( )
S = 21.5
1 2
–
( ) 1 2
–11
= 1 64
–
We can write this as:(1 )1
n
n
a rS
r
(This ensures that the denominator is positive).
Example 3:
12
1
10
0
)203( b) )2(6 a) :Findr
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 rS
r
11
11
6(2 1)2 1
S
= 12 282
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) = 797 160 – 240
= 796 920
+ 12 + 24 + 48 + …
Example 4: In a Geometric Series, the third term is 36, and thesixth term is 121.5. For the series, find the commonratio, 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 = ar n–1 ar
2 = 36 ….(1)ar
5 = 121.5 ….(2)
Now, divide equation (2) by equation (1):36
5.1212
5
ar
ar
So r3 = 3.3753 375.3r 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 = 35 469
(To the nearest integer)
Example 5: The first three terms of a geometric progression arex, x + 3, 4x. Find the two possible values for thecommon ratio. For each value find these first threeterms, 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
xr
x
xr
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
Example 6: $100 is invested into an account (earning 5% compoundinterest per annum), at the start of every year. Find theamount 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)
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.)
( 1)1
n
n
a rS
r
A Geometric Series is one in which the terms are found bymultiplying 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 nn aru
In problems where r < 1 it is better to write the above as:
(1 )1
n
n
a rS
r
Summary of key points:
The End
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