1.4.2 The nth term of a geometric progression

1.4.2 The nth Term of Geometric Progressions

(C) The nth Term of Geometric Progressions

T n = a r n 1

a = first term
r = common ratio
n = the number of term
Tn = the nth term

Example 1:
Find the given term for each of the following geometric progressions.
(a) 8 ,4 ,2 ,…… T8
(b) 16 27 , 8 9 , 4 3 , ….. ,  T6

Solution:
T n = a r n 1 T 1 = a r 1 1 = a r 0 = a ( First term ) T 2 = a r 2 1 = a r 1 = a r ( S e c o n d term ) T 3 = a r 3 1 = a r 2 ( T h i r d term ) T 4 = a r 4 1 = a r 3 ( Fourth term )

(a)
8 , 4 , 2 , ….. a = 8 , r = 4 8 = 1 2 T 8 = a r 7 T 8 = 8 ( 1 2 ) 7 = 1 16

(b)
16 27 , 8 9 , 4 3 , ….. a = 16 27 r = T 2 T 1 = 16 27 8 9 = 2 3 T 6 = a r 5 = 16 27 ( 2 3 ) 5 = 512 6561


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

Example 2:
Find the number of terms for each of the following geometric progressions.
(a) 2, 4, 8, ….., 8192
(b) 1 4 , 1 6 , 1 9 , ….. , 16 729   
(c) 1 2 , 1 , 2 , ..... , 64  

Solution:
(a)
2 , 4 , 8 , ..... , 8192 ( Last term is given) a = 2 r = T 2 T 1 = 4 2 = 2 T n = 8192 a r n 1 = 8192 ( T h e n th term of GP, T n = a r n 1 ) ( 2 ) ( 2 ) n 1 = 8192 2 n 1 = 4096 2 n 1 = 2 12 n 1 = 12 n = 13

(b)
1 4 , 1 6 , 1 9 , ..... , 16 729 a = 1 4 , r = 1 6 1 4 = 2 3 T n = 16 729 a r n 1 = 16 729 ( 1 4 ) ( 2 3 ) n 1 = 16 729 ( 2 3 ) n 1 = 16 729 × 4 ( 2 3 ) n 1 = 64 729 ( 2 3 ) n 1 = ( 2 3 ) 6 n 1 = 6 n = 7

(c)
1 2 , 1 , 2 , ..... , 64 a = 1 2 , r = 2 1 = 2 T n = 64 a r n 1 = 64 ( 1 2 ) ( 2 ) n 1 = 64 ( 2 ) n 1 = 64 × 2 ( 2 ) n 1 = 128 ( 2 ) n 1 = ( 2 ) 7 n 1 = 7 n = 8


(E) Three consecutive terms of a geometric progression
If e, f and g are 3 consecutive terms of GP, then
g f = f e
Example 3:
If p + 20,   p − 4, p −20 are three consecutive terms of a geometric progression, find the value of p.

Solution:
p 20 p 4 = p 4 p + 20 ( p + 20 ) ( p 20 ) = ( p 4 ) ( p 4 ) p 2 400 = p 2 8 p + 16 8 p = 416 p = 52

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