Sequence
Arithmetic Progression
In an Arithmetic Progression the difference between one term and the next is a constant.In other words, we just add the same value each time ... infinitely.
Example:
1, 4, 7, 10, 13, 16, 19, 22, 25, ... |
This sequence has a difference of 3 between each number.
In General we could write an arithmetic sequence like this:
{a, a+d, a+2d, a+3d, ... }
where:- a is the first term, and
- d is the difference between the terms (called the "common difference")
Example: (continued)
1, 4, 7, 10, 13, 16, 19, 22, 25, ... |
- a = 1 (the first term)
- d = 3 (the "common difference" between terms)
{a, a+d, a+2d, a+3d, ... }
{1, 1+3, 1+2×3, 1+3×3, ... }
{1, 4, 7, 10, ... }
Rule
We can write an Arithmetic Sequence as a rule:
Tn = a + (n-1)d
Example 1: Calculate the 4th term for
3, 8, 13, 18, 23, 28, 33, 38, ... |
This sequence has a difference of 5 between each number.
The values of a and d are:- a = 3
- d = 5
Tn = a + (n-1)d
T4 = a + (4-1)d
= a + 3d
= 3 + 3(5)
=18
Example 2: If T3 = 8 and T6 = 17, find T14.
Solution:
T3 = a + 2d
8 = a + 2d -------①
T6 = a +5d
17 = a + 5d --------②
② - ①
17-8 = a-a + (5d-2d)
9 = 3d
d=3
substitute d=3 into ①
8 = a + 2(3)
a = 2
T14 = a +13d
= 2 + 13(3)
T14 = 41
Summing an Arithmetic Series
To sum up the terms of this arithmetic sequence:
Solution:
a = 0.6, d = 1.1 and n = 100
Sum of n terms can be given as follows:
Thus, sum of the 100 term,
S100= 5505
T20 = 200 +(20 - 1 ) (-10)
a + (a+d) + (a+2d) + (a+3d) + ...
use this formula:Example1:
Add up the first 10 terms of the arithmetic sequence:
{ 1, 4, 7, 10, 13, ... }
The values of a, d and n are:- a = 1
- d = 3
- n = 10 (how many terms to add up)
- S10 = 10/2 [2(1) + ( 10-1) 3]
- = 5 [ 2+9(3)]
- = 5[29]
- = 145
Example2:
0.6, 1.7, 2.8, ……………… to 100 termsSolution:
a = 0.6, d = 1.1 and n = 100
Sum of n terms can be given as follows:
Thus, sum of the 100 term,
S100= 5505
Problem 1:
The first term of an arithmetic sequence is equal to 200 and the common difference is equal to
-10. Find the value of the 20 th term
The first term of an arithmetic sequence is equal to 200 and the common difference is equal to
-10. Find the value of the 20 th term
Solution to Problem 1:
Use the value of the common difference d = -10 and the first term a1 = 200 in the formula for the n th term given above and then apply it to the 20 th term
T20 = 200 +(20 - 1 ) (-10)
= 10
Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.
Problem 2:
An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.
An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.
Solution to Problem 2:
-
We use the n th term formula for the 6 th term, which is known, to write
T6 = 52 - = a +(6 - 1 ) (10)
- 52 = a +50
- a = 2
-
The above equation allows us to calculate a.
a = 2
-
Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows.
T15 = 2 + (15 - 1)(10) = 142
Problem 3:
Find the sum of all the integers from 1 to 1000.
Find the sum of all the integers from 1 to 1000.
Solution to Problem 3:
-
The sequence of integers starting from 1 to 1000 is given by
1 , 2 , 3 , 4 , ... , 1000
-
The
above sequence has 1000 terms. The first term is 1 and the last term is
1000 and the common difference is equal to 1. We have the formula that
gives the sum of the first n terms of an arithmetic sequence knowing the
first and last term of the sequence and the number of terms (see
formula above).
s1000 = 1000 (1 + 1000) / 2 = 500500
Problem 4:
Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.
Solution to Problem 4:
-
The first few terms of a sequence of positive integers divisible by 5 is given by
5 , 10 , 15 , ...
-
The
above sequence has a first term equal to 5 and a common difference d =
5. We need to know the rank of the term 1555. We use the formula for the
n th term as follows
1555 = a1 + (n - 1 )d
-
Substitute a1 and d by their values
1555 = 5 + 5(n - 1 )
-
Solve for n to obtain
n = 311
-
We now know that 1555 is the 311 th term, we can use the formula for the sum as follows
s311 = 311 (5 + 1555) / 2 = 242580
We can use this fomula to calculate the sum of the AP :
where a = first term, l = last term
Problem 5:
Find the sum of the first 50 even positive integers.
Solution to Problem 5:
-
The sequence of the first 50 even positive integers is given by
2 , 4 , 6 , ...
-
The
above sequence has a first term equal to 2 and a common difference d =
2. We use the n th term formula to find the 50 th term
T50 = 2 + 49(2) = 100
-
We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms
s50 = 50 (2 + 100) / 2 = 2550
Thank you teacher
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