Let the first three terms of G.P. be \(\frac{a}{r}\), a, ar

It is  given that \(\frac{a}{r} \times a \times ar = 1\)

\(\Rightarrow\) a³ = 1

⇒ a = 1

And

\(\frac{a}{r} + a + ar = \frac{39}{10}\)

\(\Rightarrow\) a\(\left(\frac{1}{r} + 1 + r\right)\) = \(\frac{39}{10}\)

\(\Rightarrow\)\(\left(\frac{1}{r} + 1 + r\right)\) = \(\frac{39}{10}\) .....(a=1)

\(\Rightarrow\) \(\left(\frac{1}{r} + r\right)\) = \(\frac{39}{10}\) - 1 = \(\frac{29}{10}\)

⇒ 10(1 + r² ) = 29r 

⇒ 10r² - 29r + 10 = 0 

⇒ 10r² - 25r - 4r + 10 = 0 

⇒ 5r(2r - 5) - 2(2r - 5) = 0 

⇒ (2r - 5)(5r - 2) = 0

⇒ r = \(\frac{5}{2}, \frac{2}{5}\)

Therefore the first three terms are:

(i) if r = \(\frac{5}{2}\) then

\(\frac{2}{5}, 1 , \frac{5}{2}\)

(ii) if r = \(\frac{2}{5}\)then

\(\frac{5}{2},1,\frac{2}{5}\)

Hence, the Common ratio r = \(\frac{5}{2},\frac{2}{5}\) and the first three terms are:

(i) if r = \(\frac{5}{2}\) then

\(\frac{2}{5},1,\frac{5}{2}\)

(ii) If r = \(\frac{2}{5}\) then

\(\frac{5}{2},1,\frac{2}{5}\)