(a) Let 128 be the nth term of the given G.P. Here a = 2 and r = √2
Therefore, arn–1 = 128
⇒ 2(√2)n−1 = 128 ⇒ (√2)n−1 = 64 ⇒ (2)

n−1

2

= (2)6
∴

n−1

2

= 6
⇒ n – 1 = 12 ⇒ n = 13.
Hence, 128 is the 13th term of the given G.P.
(b) Let 729 be the nth term of the given G.P.
Here a = √3 and r = √3
Therefore, arn−1 = 729
⇒ (√3)(√3)n−1 = 729 ⇒ (√3)(3)

n−1

2

= 729
⇒ 3

n−1

2

+

1

2

= 729 ⇒ 3

n

2

= (3)6
∴

n

2

= 6 ⇒ n = 12
Hence 729 is the 12th term of the given G.P.
(c) Let

1

19683

be the nth term of the given G.P. Here a =

1

3

and r =

1

3

∴ arn−1 =

1

19683

⇒ (

1

3

)(

1

3

)n−1 =

1

19683

⇒ (

1

3

)n = (

1

3

)9
∴ n = 9
Hence,

1

19683

is the 9th term of the given G.P.