GEOMETRIC PROGRESSION
CONTENT
- Definition of Geometric Progression
- Denotations of Geometric progression
- The nth term of a G. P.
- The sum of Geometric series
- Sum of G. P. to infinity
- Geometric mean
Definition of G. P
The sequence 5, 10, 20, 40 has a first term of 5 and the common ratio
Between the term is 2 e.g. (10/5 or 40/2o = 2).
A sequence in which the terms either increase or decrease in a common ratio is called a Geometric Progression
(G. P)
- P: a, ar, ar2, ar3 ………………
Denotations in G. P
a = 1st term
r = common ratio
Un = nth term
Sn = sum
The nth term of a G. P
The nth term = Un
Un = arn-1
1st term = a
2nd term = a x r =ar
3rd term = a x r x r = ar2
4th term = a x r x r x r = ar3
8thterm = a x r x r x r x r x r x r x r = ar7
nth term = a x r x r x r x ……….. arn-1
Example
Given the GP 5, 10, 20, 40. Find its (a) 9th term (b) nth term
Solution
a = 5 r = 10/5 = 2
U9 = arn-1
U9 = 5 (2) 9-1
= 5 (2)8
= 5 x 256 = 1,280
(b) Un = arn-1
= 5(2) n-1
Example 2
The 8th term of a G.P is -7/32. Find its common ratio if it first term is 28.
U8 = -7/32 Un = arn-1
-7/32 = 28 (r)8-1
-7/32 = 28r7
-7/32 x 1/28 =
| -7
896 |
–7/32 x 1/28 = r7
| – 7
32 x 28 |
7 7
r = =
r = – 0.5
Evaluation
- The 6th term of a G.P is 2000. Find its first term if its common ratio is 10.
- Find the 7th term and the nth term of the progression 27, 9 , 3, …
THE SUM OF A GEOMETRIC SERIES
a + ar + ar2 + ar3 + ………………. arn-1
represent a general geometric series where the terms are added.
S = a + ar + ar2 ………… arn-1 eqn 1
Multiply through r
rs = ar + ar2 + ar3 ………. arn ……… eqn 2
subtracteqn 2 from 1
S – rs = a – arn
S (1 – r) =a(1-rn)
1 – r 1-r
S.=a ( 1 – rn) r < 1
1 – r
Multiply through by -1 or subs. eqn. 1 from e.g. 2
rs – s = arn – a
S (r – 1) =a(rn – 1)
r – 1 r – 1
S = a(rn-1)
r -1 for r > 1
Example:
Find the sum of the series.
- ½ + ¼ + 1/8 + …………………… as far as 6th term
- 1 + 3 + 9 + 27 + …………………. 729
Solution
a = ½
r = ½ (r = ¼ ¸ ½ = ½)
\r< 1
S = a (1-rn)
1 – r
S6 = [½ (1 – (½)6]
1 – ½
S6= ½ (1 – 1/64)
½
S6 = 1 – 1 = 64 – 1 = 63
64 64 64
- a = 1, r = 3, n = ? Un = 729
Un = arn-1
729 = 1 x 3n-1 (3n-1 = 3n x 3-1)
729 = 3n
3
3n = 3 x 729
3n = 31 x 36
3n = 37
\ n = 7
S = a(rn-1)
r – 1
S = a(37 – 1) = 2187 – 1
3 – 1 2
2186 = 1093
2
Evaluation: Find the sum of the series 40, -4, 0.4 as far as the 7th term.
SUM OF G. P TO INFINITY
Sum of G. P to infinity is only possible where r is < 1.
Where r is > 1 there is no sum to infinity.
Example:
- Find the sum of G. P. 1 + ½ + ¼ + …………………… (a) to 10 terms (b) to 100 terms. Hence deduce the sum of the series (formula) if it has a very large no. of term or infinity.
(a) a = 1 r = ½
n = 10
S = a (1-rn)
1-r
S = 1(1-(1/2)10) = 1(1-0.0001)
1- ½ 1/2
2 (1 – 0.001)
2 – 0.002 = 1.998.
- n = 100.
S = a (1 – rn)
1 – r
S = 1 (1-(1/2)100) = 1(1- (1/2)10)10
1 – ½ ½
1 (1-(0.001)10
½
1 (1)
½ = 2
Therefore (1/2)100 tend to 0 (infinity).
In general,
S = a (1-rn)= a(1-0) = a__
1-r 1-r 1 – r
\ S¥= a__ = n → ¥
1 – r
Example 2:
Find the sum of the series 45 + 30 + 20 + ……………… to infinity.
a = 45, r = 2/3, n = infinity
S∞ = a S = 45__
1 – r 1- 2/3
S∞ = 45 ÷ 1/3
45 x 3/1
= 135
Evaluation
- The sum to infinity of a Geometric Series is 100. Find the first term if the common ratio is –1/2.
- The 3rd and 6th term of a G. P. are 48 and 142/9 respectively, write down the first four terms of the G. P.
- The sum of a G. P. is 100 find its first term if the common ratio is 0.8.
GEOMETRIC MEAN
If three numbers such as x , y and z are consecutive terms of a G.P then their common ratio will be
y =z
x y
y2 = xz
y = xz
The middle value , y is the geometric mean (GM). We can conclude by saying that the GM of two numbers is the positive square root of their products.
Example
Calculate the geometric mean of I. 3 and 27 II. 49 and 25
4
Solution
- M of 3 and 27 II. G.M of 49 and 25
= √ 3 x 27 4
= √ 81 = 49 x 25
= 9 4
= 7 x 5
2
= 35 = 17 1/2
2
Example
The first three terms of a GP are k + 1, 2k – 1, 3k + 1. Find the possible values of the common ratio.
Solution
The terms are k + 1, 2k – 1, 3k + 1
2k -1 =3k + 1
k + 1 2k – 1
(2k-1)(2k-1) = (k+1)(3k+1)
4k2-2k-2k +1 = 3k2 +k+3k + 1
4k2– 4k +1 = 3k2 +4k + 1
4k2 – 3k2 – 4k – 4k + 1-1 = 0
k2 -8k = 0
k(k-8) = 0
k = 0 or k – 8 = 0
k = 0 or 8
The common ratio will have two values due to the two values of k
When k=0 when k= 8
K+1 = 0+1 =1 k+1 = 8+1 = 9
2k- 1= 2×0 – 1 = -1 2k- 1 = 2×8 – 1 = 15
3k+ 1= 3×0+ 1 = 1 3k+1 = 3×8 +1 = 25
terms are 1 , -1 , 1 terms are 9,15,25
common ratio, r = -1/1 common ratio,r = 15/9
r = -1
EVALUATION
The third term of a G.P. is 1/81. Determine the first term if the common ratio is 1/3.
GENERAL EVALUATION /REVISION QUESTION
- p – 6, 2p and 8p + 20 are three consecutive terms of a GP. Determine the value of (a) p (b) the common ratio
- If 1 , x , 1 , y , ….are in GP , find the product of x and y
16 4
3.The third term of a G.P is 45 and the fifth term 405.Find the G.P. if the common ratio r is positive.
4.Find the 7th term and the nth term of the progression 27,9,3,…
5.In a G.P, the second and fourth terms are 0.04 and 1 respectively. Find the (a) common ratio (b) first term
WEEKEND ASSIGNMENT
- In the 2nd and 4th term of a G.P are 8 and 32 respectively, what is the sum of the first four terms. (a) 28 (b) 40 (c) 48 (d) 60
- The sum of the first five term of the G.P. 2, 6, 18, is (a) 484 (b) 243 (c) 242 (d) 130
- The 4th term of a GP is -2/3 and its first term is 18 what is its common ratio. (a) ½ (b) 1/3
(c) -1/3 (d) -1/2
- If the 2nd and 5th term of a G. P. are -6 and 48 respectively, find the sum of the first four terms: (a) -45 (b) -15 (c) 15 (d) 33
- Find the first term of the G.P. if its common ratio and sum to infinity – 3/3 and respectively (a) 48 (b) 18 (c) 40 (d) -42
THEORY
1.The 3rd term of a GP is 360 and the 6th term is 1215. Find the
(i) Common ratio (ii) First term (iii) Sum of the first four terms
1b. If (3- x) + (6) + (7- 5x) is a geometric series, find two possible values for
(i) x (ii) the common ratio, r (iii) the sum of the G.P
2.The first term of a G. P. is 48. Find the common ratio between its terms if its sum to infinity is 36.
See also