The sum of three terms (numbers) of a G.P. is $3\dfrac{1}{2}$ and their product is 1; the numbers are :

1. $\dfrac{1}{2}, 1$ and 2

2. $\dfrac{1}{3}, 3$ and 9

3. $1, \dfrac{1}{2}$ and 2

4. $2, \dfrac{1}{2}$ and 1

Let three terms of G.P. be $\dfrac{a}{r}, a, ar$.

Given,

Product of three terms of G.P. = 1

$\therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a^3 = 1^3 \\[1em] \Rightarrow a = 1.$

Given,

Sum of three terms of G.P. = $3\dfrac{1}{2}$

$\Rightarrow \dfrac{a}{r} + a + ar = 3\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{r} + 1 + 1(r) = \dfrac{7}{2} \quad [\because a = 1] \\[1em] \Rightarrow \dfrac{1}{r} + 1 + r = \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{1 + r + r^2}{r} = \dfrac{7}{2} \\[1em] \Rightarrow 2(r^2 + r + 1) = 7r \\[1em] \Rightarrow 2r^2 + 2r + 2 = 7r \\[1em] \Rightarrow 2r^2 + 2r - 7r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 5r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 4r - r + 2 = 0 \\[1em] \Rightarrow 2r(r - 2) - 1(r - 2) = 0 \\[1em] \Rightarrow (2r - 1)(r - 2) = 0 \\[1em] \Rightarrow 2r - 1 = 0 \text{ or } r - 2 = 0 \\[1em] \Rightarrow 2r = 1 \text{ or } r = 2 \\[1em] \Rightarrow r = \dfrac{1}{2} \text{ or } r = 2.$

Let r = $\dfrac{1}{2}$

Terms :

⇒ $\dfrac{a}{r}$, a, ar

⇒ $\dfrac{1}{\dfrac{1}{2}}, 1, 1 \times \dfrac{1}{2}$

⇒ 2, 1, $\dfrac{1}{2}$.

Let r = 2

Terms :

⇒ $\dfrac{a}{r}$, a, ar

⇒ $\dfrac{1}{2}, 1, 1 \times 2$

⇒ $\dfrac{1}{2}$, 1, 2.

Hence, Option 1 is the correct option.