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Arithmetic Progression used to find the nth term
Arithmetic Progression as requested by baig772
Arithmetic Progression
An arithmetic progression or arithmetic sequence are a sequence of numbers that increase or decrease by a common difference.
Solving arithmetic progression is interesting since we can find the n^{th} term of a particular sequence in a much easier way.
Within this hub we shall cover the steps involved in progression solving for;
- Finding the n^{th} term
- Find a chosen term as defined by the question
- Finding the A.P.(arithmetic progression ) itself
The general form of an arithmetic sequence is;
a, a + d, a + 2d, a + 3d, a + 4d...
Where the a = first term (number)
d = common difference
To find the n^{th} term of an arithmetic sequence we use the following formula;
t_{n} = a + (n - 1) d
When we add or subtract any constant number with all the terms of the sequence, the arithmetic sequence remains an arithmetic sequence.
Example:
6, 8, 10, 12, 14, 16, 18….. is an A.P with a common difference 2.
Add 4 with all the terms, and we get,
10, 12, 14, 16, 18, 20, 22…. this is also an A.P with a common difference 2.
When we multiply or divide by a non-zero constant with the terms of a sequence, the arithmetic sequence remains an arithmetic sequence.
Example:
10, 20, 30, 40, 50, 60, 70…. Is an A.P with a common difference 10.
Multiply all the terms by 2 , and we get,
20, 40, 60, 80, 100, 120, 140…. this is also an A.P with a common difference 20.
Finding the n^{th }term
( i ) The sequence terms of an A.P. are 1, 12, 23, 34 find the n^{th term}
Therefore;
t_{n }= a + (n - 1) d
a = 1 ( where a = the first term)
d = 12 - 1 = 11 (where d = common difference)
t_{n }= 1 + (n - 1) d
t_{n }= 1 + (n - 1) 11
t_{n }= 1 + 11n - 11
t_{n }= 11n - 10 (n^{th }term)
_{ }
( ii ) Find the 11^{th }term of the sequence 1, 12, 23, 34
Therefore;
a = 1 (first term)
d = 12 - 1 = 11 (common difference)
n = 11 (number of term required)
t_{n} = a + ( n - 1) d
t_{11} = 1 + (11 - 1) x 11
= 1 + (10 x 11)
= 1 + 110
t_{11 }= 111
(iii) Find the A.P. of a sequence when the 7^{th }term is 67
and the 16^{th }term is 166
Consider the A.P. in the form a, a + d, a + 2d, a + 3d, a + 4d......
Therefore;
t_{7 }= a + 6d = 67
t_{16}_{ }= a + 15d = 166
t_{7} - t_{16} = a + 6d = 67
= - a - 15d = -166
= - 9d = -99
-9d = -99 = 11
-9 -9
d = 11
Put "d = 11" into the t_{7 }equation
a + 6d = 67
a = (6 x 11) = 67
a = 66 = 67
a = 67 - 66 = 1
a = 1
t_{1 }= a = 1
t_{2} = a + d = 1 + 11 = 12
t_{3 }= a + 2d = 1 + (2 x 11) = 23
t_{4} = a + 3d = 1 = (3 x 11) = 34
So terms for this A.P. sequence are; 1, 12, 23, 34
Comments
Thank You!!!, sa mga aNsWer... heheheh
hi guys you made my day, i did this 8 yrs ago and now you have refreshad my memory
Thanks- this really helped me in my exam studies
You have some great meths hubs. I do a lot of research with fractals, fibonacci and other works like Gann, Elliott and Murrey in the financial markets. Maths and nature are what makes those studies work and viable. There is nothing as clinical as maths.
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