Which term of the following sequences:2, 2√2, 4,… is 128 ?

(a) The given sequence, 2,\(2 \sqrt{2}\), 4,…

We have,

a = 2 and r =\( \dfrac{2\sqrt{2}}{2}\) = \( \sqrt{2}\)

Taking the n^th term of this sequence as 128, we have

\( a_n=ar^{n-1}\)

\( (2)(\sqrt{2})^{n-1}=128\)

\( (2)(2)^\dfrac{n-1}{2}=(2)^7\)

\( (2)^\dfrac{n-1}{2}+1=7\)

\( \dfrac{n-1}{2}=6\)

n-1=12

n = 13 

Therefore, the 13^th term of the given sequence is 128.

(ii) Given sequence, \( \sqrt{3}\), 3, \( 3\sqrt{3}\),…

We have,

a = \( \sqrt{3}\) and r = \( 3\sqrt{3}\) = \( \sqrt{3}\)

Taking the n^th term of this sequence to be 729, we have

\( a_n=ar^{n-1}\)
\( ar^{n-1}=729\)

\((3)^{\dfrac{1}{2}}\)\( (3)^{\dfrac{n-1}{2}}\)= (3)⁶

^{\( (3)^{\dfrac{1}{2}+{\dfrac{n-1}{2}}}\) =}(3)⁶

Equating the exponents, we have

\( \dfrac{1}{2}+{\dfrac{n-1}{2}}\) =6

\( \dfrac{1+n-1}{2}\) = 6

n = 12

(iii) Given sequence, \( \dfrac{1}{3}\), \( \dfrac{1}{9}\), \( \dfrac{1}{27}\), …

a = \( \dfrac{1}{3}\) and r = (1/9)/(1/3) = \( \dfrac{1}{3}\)

Taking the n^th term of this sequence to be \( \dfrac{1}{19683}\), we have

\( a_n=ar^{n-1}=\dfrac{1}{19683}\)

\(( \dfrac{1}{3}) (\dfrac{1}{3})^{n-1}=\dfrac{1}{19683}\)

\( ( \dfrac{1}{3}) ^n=( \dfrac{1}{3}) ^9\)

= 9

Therefore, the 9^th term of the given sequence is \( \dfrac{1}{19683}\).