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