1.6 The nth term of a geometric progression

1.6 The nth Term of Geometric Progressions

(C) The nth Term of Geometric Progressions
   T n =a r n1   
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 n1 T 1 =a r 11 =a r 0 =a (First term) T 2 =a r 21 =a r 1 =ar (Second term) T 3 =a r 31 =a r 2  (Third term) T 4 =a r 41 =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 n1 =8192(Thenth term of GP,  T n =a r n1 ) ( 2 ) ( 2 ) n1 =8192 2 n1 =4096 2 n1 = 2 12 n1=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 n1 = 16 729 ( 1 4 ) ( 2 3 ) n1 = 16 729 ( 2 3 ) n1 = 16 729 ×4 ( 2 3 ) n1 = 64 729 ( 2 3 ) n1 = ( 2 3 ) 6 n1=6 n=7
(c)
1 2 , 1, 2,....., 64 a= 1 2 , r= 2 1 =2 T n =64 a r n1 =64 ( 1 2 ) ( 2 ) n1 =64 ( 2 ) n1 =64×2 ( 2 ) n1 =128 ( 2 ) n1 = ( 2 ) 7 n1=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:
p20 p4 = p4 p+20 ( p+20 )( p20 )=( p4 )( p4 ) p 2 400= p 2 8p+16 8p=416 p=52