Correct Answer - Option 2 : 512

__Concepts:__

Let us consider the sequence a1, a2, a3 …. an is a G.P.

- Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
- nth term of the G.P. is an = arn−1
- Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - 1}}\); where r >1
- Sum of n terms = s = \(\frac{{a\;\left( {1 - {r^n}} \right)}}{{1 - r}}\); where r <1
- Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{}}\frac{{\rm{a}}}{{1{\rm{}} - {\rm{r}}}}{\rm{}}\); |r| < 1

Where a is 1st term and r is common ratio.

__Calculation:__

Let 'a' be the first term and 'r' be the common ratio.

We know that Tn = a rn-1

Given: ar^{4} = 2

Now, Product of 9 terms = a × ar × ar^{2} × ar^{3} × ar^{4} × ar^{5} × ar^{6} × ar^{7} × ar^{8}

= a

^{9 }r

^{36} = (ar

^{4})

^{9} = 2

^{9} = 512