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 nth 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 nth 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 nth term of an arithmetic sequence we use the following formula;
tn = 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 nth term
( i ) The sequence terms of an A.P. are 1, 12, 23, 34 find the nth term
Therefore;
tn = a + (n - 1) d
a = 1 ( where a = the first term)
d = 12 - 1 = 11 (where d = common difference)
tn = 1 + (n - 1) d
tn = 1 + (n - 1) 11
tn = 1 + 11n - 11
tn = 11n - 10 (nth term)
( ii ) Find the 11th 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)
tn = a + ( n - 1) d
t11 = 1 + (11 - 1) x 11
= 1 + (10 x 11)
= 1 + 110
t11 = 111
(iii) Find the A.P. of a sequence when the 7th term is 67
and the 16th term is 166
Consider the A.P. in the form a, a + d, a + 2d, a + 3d, a + 4d......
Therefore;
t7 = a + 6d = 67
t16 = a + 15d = 166
t7 - t16 = a + 6d = 67
= - a - 15d = -166
= - 9d = -99
-9d = -99 = 11
-9 -9
d = 11
Put "d = 11" into the t7 equation
a + 6d = 67
a = (6 x 11) = 67
a = 66 = 67
a = 67 - 66 = 1
a = 1
t1 = a = 1
t2 = a + d = 1 + 11 = 12
t3 = a + 2d = 1 + (2 x 11) = 23
t4 = a + 3d = 1 = (3 x 11) = 34
So terms for this A.P. sequence are; 1, 12, 23, 34