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: 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

Find: the 10 th term of the G.P. 2 , 1 2 , 1 2 2 , …

Given G.P. is 2 , 1 2 , 1 2 2 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 2 = 1 2 a n = a 1 r n - 1 a 10 = ( 2 ) 1 2 9 = 1 2 1 2 8

Find the 4 th term from the end of the G.P. 2 27 , 2 9 , 2 3 , … , 162

Given G.P. is 2 27 , 2 9 , 2 3 , … . . 162 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 r = 2 9 ( 2 27 ) r = 3 First term will be n t h term from end a n = l 1 r n - 1 l = last term = 162 n = number of term from end a 4 = 162 1 3 4 - 1 = 162 1 3 3 = 6 So, 4 t h term from the end is 6

Which term of the G.P.: 2 , 1 2 , 1 2 2 , 1 4 2 , … is 1 512 2 ?

Given progression is 2 , 1 2 , 1 2 2 , 1 4 2 , … . . 1 512 2 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 Here a 1 = 2 a 2 = 1 2 , T n = 1 512 2 Common Ratio r = a 2 a 1 r = 1 2 2 r = 1 2 a n = 1 512 2 and n = ? a n = a 1 r n - 1 1 512 2 = 2 1 2 n - 1 1 512 2 × 1 2 = 1 2 n - 1 1 2 10 = 1 2 n - 1 ⇒ n - 1 = 10 n = 11 So,it is 11 th term

Which term of the G.P.: 2 , 2 2 , 4 , . . . is 128 ?

Given progression is 2 , 2 2 , 4 , … … … . 128 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 1 = 2 a 2 = 2 2 a n = 128 Common Ratio r = a 2 a 1 = 2 2 2 = 2 a n = a 1 r n - 1 ⇒ 128 = 2 2 n - 1 ⇒ 64 = 2 n - 1 2 ⇒ n - 1 2 = 6 ⇒ n = 13 So,it is 13 th term

Which term of the G.P.: 3 , 3 , 3 3 , … is 729 ?

Given progression series is 3 , 3 3 3 , … … … . 729 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 Here a 1 = 3 a 2 = 3 a n = 729 Common Ratio r = a 2 a 1 = 3 3 = 3 a n = a 1 r n - 1 ⇒ 729 = 3 3 n - 1 ⇒ 3 6 = 3 n ⇒ n 2 = 6 ⇒ n = 12 So,it is 12 th term

Which term of the G.P.: 1 3 , 1 9 , 1 27 … is 1 19683 ?

Given progression series is 1 3 , 1 9 , 1 27 , … … . 1 19683 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 Here a 1 = 1 3 a 2 = 1 9 a n = 1 19683 Common Ratio r = a 2 a 1 = ( 1 9 ) 1 3 = 1 3 a n = a 1 r n - 1 1 19683 = 1 3 1 3 n - 1 1 3 9 = 1 3 n ⇒ n = 9 So,it is 9 th term