The sum of first three terms of a G.P. is $\dfrac{39}{10}$ and their product is 1, find :

(i) the common ratio

(ii) the first three terms of this G.P.

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

Given,

Product of the numbers = 1

$\therefore \dfrac{a}{r} \times a \times ar$ = 1

⇒ a³ = 1

⇒ a = $\sqrt[3]{1}$ = 1.

Given,

Sum of first three terms = $\dfrac{39}{10}$

$\therefore \dfrac{a}{r} + a + ar = \dfrac{39}{10}\\[1em] \Rightarrow \dfrac{a + ar + ar^2}{r} = \dfrac{39}{10}\\[1em] \Rightarrow 10(a + ar + ar^2) = 39r \\[1em] \Rightarrow 10a(1 + r + r^2) = 39r \\[1em] \Rightarrow 10 \times 1 \times (1 + r + r^2) = 39r \\[1em] \Rightarrow 10 + 10r + 10r^2 = 39r \\[1em] \Rightarrow 10r^2 + 10r - 39r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 29r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \\[1em] \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \\[1em] \Rightarrow (5r - 2)(2r - 5) = 0 \\[1em] \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \\[1em] \Rightarrow 5r = 2 \text{ or } 2r = 5 \\[1em] \Rightarrow r = \dfrac{2}{5} \text{ or } r = \dfrac{5}{2}.$

Hence, common ratio = $\dfrac{2}{5} \text{ or } \dfrac{5}{2}$.

(ii) Let r = $\dfrac{2}{5}$

Substituting values of a and r in $\dfrac{a}{r}, a, ar$, we get :

$\Rightarrow \dfrac{1}{\dfrac{2}{5}}, 1, 1 \times \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{5}{2}, 1, \dfrac{2}{5}.$

Let r = $\dfrac{5}{2}$

Substituting values of a and r in $\dfrac{a}{r}, a, ar$, we get :

$\Rightarrow \dfrac{1}{\dfrac{5}{2}}, 1, 1 \times \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{2}{5}, 1, \dfrac{5}{2}.$

Hence, first three terms of the G.P. are $\dfrac{5}{2}, 1, \dfrac{2}{5} \text{ or } \dfrac{2}{5}, 1, \dfrac{5}{2}$.