1.2.2 The nth Term of an Arithmetic Progression


1.2.2 The nth term of an Arithmetic Progression

(C) The nth term of an Arithmetic Progression
 
Tn = a + (n − 1) d
where
a = first term
d = common difference
n = the number of term
Tn  = the nth term




(D) The Number of Terms in an Arithmetic Progression
Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term

Example 1
:
Find the number of terms for each of the following arithmetic progressions.
(a) 5, 9, 13, 17… , 121
(b) 1, 1.25, 1.5, 1.75,…, 8

Solution:
(a)
5, 9, 13, 17… , 121
AP,
a = 5, d = 9 – 5 = 4
The last term, Tn = 121
a + (n – 1) d = 121
5 + (n – 1) (4) = 121
(n – 1) (4) = 116
(n – 1) = 116 4  = 29
n = 30

(b)
1, 1.25, 1.5, 1.75,..., 8
AP,
a = 1, d = 1.25 – 1 = 0.25
Tn = 8
a + (n – 1) d = 8
1 + (n – 1) (0.25) = 8
(n – 1) (0.25) = 7
(n – 1) = 28
n = 29


(E) The Consecutive Terms of an Arithmetic Progression
 
If a, b, c are three consecutive terms of an arithmetic progression, then
cb = b a

Example 2:
If x + 1, 2x + 3 and 6 are three consecutive terms of an arithmetic progression, find the value of x and its common difference.

Solution:
x + 1, 2x + 3, 6
cb = a
6 – (2x + 3) = (2x + 3) – (x + 1)
6 – 2x – 3 = 2x + 3 – x – 1
3 – 2x = x + 2
x = 1 3 1 3 + 1 , 2 ( 1 3 ) + 3 , 6 4 3 , 3 2 3 , 6 d = 3 2 3 4 3 = 2 1 3

Leave a Comment