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**Hint:**Using the concept of G.P as let the first term be a and general ratio be r, then nth term can be given as \[{{\text{T}}_{\text{n}}}{\text{ = a}}{{\text{r}}^{{\text{n - 1}}}}\]. So from the given concept form all the terms and then solve it using the condition given in the question.

**Complete step by step answer:**

Let the four terms of a G.P be ,

\[{\text{a,ar,a}}{{\text{r}}^{\text{2}}}{\text{,a}}{{\text{r}}^{\text{3}}}\]

Now, as per the given that sum of all the terms is \[85\]

\[

\Rightarrow {\text{a + ar + a}}{{\text{r}}^{\text{2}}}{\text{ + a}}{{\text{r}}^{\text{3}}} = 85 \\

\Rightarrow {\text{a(1 + r + }}{{\text{r}}^{\text{2}}}{\text{ + }}{{\text{r}}^{\text{3}}}{\text{) = 85}} \\

\]

And the product of the four terms is \[4096\]

\[

\Rightarrow {\text{a(ar)(a}}{{\text{r}}^{\text{2}}}{\text{)(a}}{{\text{r}}^{\text{3}}}) = 4096 \\

\Rightarrow {\text{(}}{{\text{a}}^4}{{\text{r}}^6}{\text{) = 4096}} \\

\]

Calculating the factors of

\[

\Rightarrow {\text{4096 = }}{{\text{a}}^{\text{4}}}{{\text{r}}^{\text{6}}}{\text{ = }}{{\text{2}}^{{\text{12}}}} \\

\Rightarrow {{\text{a}}^{\text{4}}}{{\text{r}}^{\text{6}}}{\text{ = 1}}{\text{.}}{{\text{4}}^{\text{6}}} \\

\Rightarrow {\text{a = 1,r = 4}} \\

\]

And so from this we can conclude that the terms are,

\[

{\text{a = 1,r = 4}} \\

{\text{1,4,16,64}} \\

\]

Hence,

**option (a) is the correct answer.**

**Note:**In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Properties :

If all the terms of G.P are multiplied or divided by the same non-zero constant then the sequence remains in G.P with the same common ratio.

The reciprocals of the terms of a given G.P. form a G.P

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