Let the three numbers be \(\frac{a}{r},a, ar\).
∴ According to the question
\(\Rightarrow\) \(\frac{a}{r}\) + a+ ar = \(\frac{39}{10}\)...... (1)
\(\Rightarrow\) \(\frac{a}{r}\) \(\times a \times ar = 1\) ...... (2)
From 2 we get
⇒ a3 = 1
∴ a = 1. From 1 we get
\(\frac{a + ar + ar^2}{r} = \frac{39}{10}\)
⇒ 10a + 10ar + 10ar2 = 39r …(3)
Substituting a = 1 in 3 we get
⇒ 10(1) + 10(1)r + 10(1)r2 = 39r
⇒10r2 – 29r + 10 = 0
⇒ 10r2 – 25r – 4r + 10 = 0…(4)
⇒ 5r(2r – 5) – 2(2r – 5) = 0
\(\therefore\) r = \(\frac{2}{5}\) or r = \(\frac{5}{2}\)
\(\therefore\) Now the equation will be
⇒ \(\cfrac{1}{\frac{2}{5}}\), 1,1\(\times\) \(\frac{2}{5}\) or \(\frac{1}{\frac{5}{2}}\),1,1 \(\times\) \(\frac{5}{2}\)
⇒ \(\frac{5}{2}, 1, \frac{2}{5}\) or \(\frac{5}{2}, 1, \frac{2}{5}\)
\(\therefore\) The three numbers are \(\frac{5}{2}, 1, \frac{2}{5}\)