Correct Answer - Option 4 : 9

__Concept:__

Let us consider sequence a

1, a

2, a

3 …. a

n is a G.P.

- Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}\)
- nth term of the G.P. is an = arn−1
- Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1
- Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {1 - {\rm{\;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{\;r}}}}\); where r <1
- Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{\;}}\frac{{\rm{a}}}{{1{\rm{\;}} - {\rm{\;r}}}}{\rm{\;}}\) ; |r| < 1

__Calculation__**:**

Given series is 4, 8, 16, ...

Here, a = 4, r = 2

Sum of n numbers = s_{n} = 2044

To Find: nAs we know that, Sum of n terms of GP = sn = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1

∴ sn = \(\frac{{{\rm{4\;}}\left( {{{\rm{2}}^{\rm{n}}} - 1} \right)}}{{{\rm{2}}\; - {\rm{\;}}1}}\)

2044 = 4 × (2^{n} - 1)

⇒ 511 = (2n - 1)

⇒ 2^{n} = 512

⇒ 2n = 2^{9}

∴ n = 9