Arithmetic Series
FURTHER MATHEMATICS SS 1 SECOND TERM
WEEK 2 DATE(S)……………………………………….
SUBJECT: FURTHER MATHEMATICS
CLASS: SS1
TOPIC: SERIES
CONTENT:
- Definition of series.
- Arithmetic sequence
- Geometric series
- Sum to infinity
Sub – Topic: DEFINITION OF SERIES/ARITHMETIC SERIES
Series is the summation (addition) of the phrases of a sequence. For example, the collection equal of the sequence 1,2,3,4,… is 1+2+3+4+… . If the sequence incorporates finite numbers of phrases, then the corresponding collection is finite. If the sequence contains infinite numbers of phrases, then the corresponding collection is infinite.
Arithmetic Series:
These are collection shaped from an arithmetic development. e.g.
1 + 4 + 7 + 10 + …
Generally, if Sn is the sum of n phrases of an arithmetic collection then
Sn = a + (a + d) + (a + 2d) + … + (l – d) +l …………………….. (1)
The place l is the nth time period, a is the primary time period and d is the widespread distinction.
Rewriting the collection above beginning with the nth/final time period, we’ve got.
Sn = l +(l – d) + (l – 2d) +… + (a + d) + a ………………………….(2)
Including equation (1) and (2) we’ve got
2Sn = (a + l) + (a + l) + … + (a + l) + (a + l) in n locations
2Sn = n(a + l)
Sn = n/2(a + l)
However l is the nth time period i.e a + (n – 1) d
Sn= n/2{a +a + (n – 1)d}
Sn = n/2 {2a + (n – 1)d}
Example1 :
Discover the sum of the primary 20 phrases of the collection 3+5+7+9+ …
Answer:
a = 3d = 2n = 20
Sn = n/2 {2a +(n – 1)d}
S20 = 20/2 {2×3 +(20 – 1)2}
S20 = 10{6 + 19 x 2}
= 10{6 + 38}
= 10{44}
S20 = 440
Instance 2:
The sum of the primary 9 phrases of an A.P is 117 and the sum of the subsequent 4 phrases is 104.
Discover the (i) Widespread distinction
(ii) First time period
(iii)twenty fifth time period of the A.P(WAEC)
Answer:
T1, T2, T3, T4 … T8, T9, T10, T11, T12, T13
117 104
S9 = 117 ————————- (*)
n = 9
since a sequence is generally summed from the primary time period
S9 + 4 = 117 + 104
S13 = 221 ————– (**)
n = 13
Sn = n/2 {2a + (n – 1) d}
From (*) above;
117 = 9/2 {2a + (9 – 1) d}
117 = 9/2 x 2a + 9/2 x 8d
117 = 9a + 36d
Divide by by 9 to have;
13 = a + 4d ———————- (1)
From (**) above;
221 = 13/2 {2a + (13 – 1) d}
221 = 13/2 x 2a + 13/2 x 12d
221 = 13a + 78d
Divide by by 13 to have;
17 = a + 6d ———————- (2)
From equation (1) and (2)
Eqn. (1): 13 = a + 4d
Eqn. (2): 17 = a + 6d
-4 = -2d
d = -4
-2
d = 2.
(ii) From equation (1) we’ve got
13 = a + 4 x 2
13 = a + 8
a = 13 – 8
a = 5
(iii) a = 5
d = 2
n = 25
Tn = a + (n – 1) d
T25 = 5 + (25 – 1) 2
= 5 + 24 x 2
= 5 + 48
T25 = 53
CLASS ACTIVITY
(1) Discover the sum of all of the multiples of three between3 to 301. Therefore discover the sum of all of the numbers between 100 and 301 (inclusive), which aren’t multiples of three.
(2) What’s the least variety of phrases of the collection 5 + 11 + 17 + … , which should be added to present a sum better than 1000 ?
Sub – Topic: Geometric collection
The final expression for a geometrical collection is given as
Sn = a + ar + ar2 + ar3 + … + arn-1 —– (1)
The place Sn characterize the sum of n phrases of the collection[mediator_tech]
Multiply each side of equation (1) by r to have
rSn = ar + ar2 + ar3 + ar4 + — + arn —- (2)
Subtract (2) from (1) to have
Sn– rSn = a – arn
Sn (1 – r) = a(1 – rn)
…………………… (3)
If the numerator and denominator of equation (3) is multiplied by –1.
——————– (4)
If r < 1, system (3) is extra convenient
If r >1, formula (4) is more convenient
Instance 1:
Discover the sum of the primary 8 phrases of the G.P 3, 6, 12, 24, …
Answer:
a = 3n = 8r = 2
r > 1
= 3 (255)
S8 = 765
Instance 2: A G.P. has a typical ratio of two. Discover the worth of n for which the sum of 2n phrases is 33 instances the sum of n phrases.
Answer:
, due to this fact
The worth of n for which the sum of 2n phrases is 33 instances the sum of n phrases is gotten by equating to
i.e. = ; simplifying the equation we get hold of
= since r was given as 2, then by substituting we get
; re-arranging, we get hold of
Let , it implies that turns into
; fixing for y we get hold of
i.e.
Therefore n = 0 or n = 5
Sum of a Geometric collection to infinity
Instance 3: The third and sixth phrases of a G.P. are respectively. Discover its sum to infinity.
Answer: ………………….. (1)
…………………… (2)
Dividing equation 2 by equation 1 we get hold of
; simplifying we get
, due to this fact
Substituting in equation 1 we get
Sum to infinity of a G.P. () is given by supplied – 1 ≤ r ≤ 1
Therefore
= 36
CLASS ACTIVITY
- The sum to infinity of a G.P. is , and . Discover r.
- How many terms of the collection 3, – 6, +12, – 24,… are wanted to make a complete of 1 – ?
- If 7 and 189 are the primary and fourth phrases of a G.P. respectively, discover the sum of the primary three phrases of the development.
PRACTICE EXERCISE
- What’s the distinction between the sum of the primary 10 phrases of the collection 2 + 1 + + +…, and its sum to infinity?
- Discover the of sum of first 20 phrases of the collection 8, 12, 16… 96.
- The sum of the primary 2 phrases of an A.P. is 24. The sum of the 4th and fifth phrases is 36. Discover its widespread distinction.
- The sum of the primary three phrases of a geometrical development is half its sum to infinity. Discover the constructive widespread ratio of the development.
- The sum of the primary n phrases of a geometrical development is given by
ASSIGNMENT
- is the sum of the primary n phrases of a collection. Discover the primary 4 phrases, and the nth time period.
- If the sum to infinity of the collection is , discover the worth of
The nth time period of a collection is given by , the place a and b are constants. Use this data to reply questions 3 and 4
- Discover an expression for , the sum of the primary n phrases
- State the limiting worth of as n will increase indefinitely.
- A sequence is shaped by including corresponding phrases of an A.P. and a G.P. , with r=2. The primary, second and third phrases of the sequence are 3,7 and 12 respectively. Discover the A.P. and the G.P., and provides an expression for the sum of the primary n phrases of the sequence.
KEY WORDS
- SERIES (GEOMETRIC AND ARITHMETIC)
- SUM TO INFINITY
- LIMITING VALUE
