 5.2.2 The nth Term of an Arithmetic Progression

5.2.2 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) = $\frac{116}{4}$  = 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 c – b = 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
$\begin{array}{l}x=\frac{1}{3}\\ \frac{1}{3}+1,\text{2}\left(\frac{1}{3}\right)+3,\text{}6\\ \frac{\text{4}}{\text{3}}\text{, 3}\frac{2}{3}\text{, 6}\\ d=\text{3}\frac{2}{3}-\frac{\text{4}}{\text{3}}=2\frac{1}{3}\end{array}$