Show that the sequence defined by a n = 2 3 n for all n ∈ N is G.P.

Given that a sequence is defined by a n = 2 3 n a 1 = 2 3 a 2 = 2 3 2 = 2 9 a 3 = 2 3 3 = 2 27 Sequence: 2 3 , 2 9 , 2 27 The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = a n a n - 1 a 2 a 1 = 2 9 2 3 = 1 3 a 3 a 2 = 2 27 2 9 = 1 3 a 2 a 1 = a 3 a 2 = . . . = a n a n - 1 = 1 3 So given sequence is G.P.

Find: the 10 th term of the G.P. - 3 4 , 1 2 , - 1 3 , 2 9 …

Given G.P. is - 3 4 , 1 2 , - 1 3 , 2 9 , … . We have to find the 10 t h term of it The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = a n a n - 1 Common Ratio r = a 2 a 1 = 1 2 - 3 4 = - 2 3 a n = a 1 r n - 1 a 10 = - 3 4 - 2 3 9 = 1 2 2 3 8

Find: the 12 th term of the G.P. 1 a 3 x 3 , a x , a 5 x 5 , …

Given G.P. is 1 a 3 x 3 , a x , a 5 x 5 , … . . The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = a n a n - 1 Common Ratio r = a 2 a 1 = a x 1 a 3 x 3 = a 4 x 4 a n = a 1 r n - 1 a 12 = 1 a 3 x 3 a 4 x 4 11 = a 41 x 41

Find: n th term of the G.P. 3 , 1 3 , 1 3 3 , …

Given G.P. is 3 , 1 3 , 1 3 3 The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = a n a n - 1 Common Ratio r = a 2 a 1 = 1 3 3 = 1 3 a n = a 1 ( r ) n - 1 a n = 3 1 3 n - 1 = 3 3 2 - n