Question

The sum of first three terms of a G.P. is \(\frac{39}{10}\) and their product is 1. Find the common ratio and the terms.

Answer 1

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

\(\frac{a}{r}+a+ar=\frac{39}{10}\) ….(1)

\((\frac{a}{r})(a)(ar)=1\) ...(2)

From (2), we obtain

\(a^3=1\)

⇒ a = 1 (Considering real roots only)

Substituting a = 1 in equation (1), we obtain

\(\frac{1}{r}+1+r=\frac{39}{10}\)

⇒ 1 + r + \(r^2=\frac{39}{10}{r}\)

⇒ 10 + 10r + \(10r^2-39r=0\)

⇒ 10\(r^2-29\)r + 10 = 0

⇒ 10\(r^2\) - 25r - 4r + 10 = 0

⇒ 5r (2r - 5) - 2 (2r - 5) = 0

⇒ (5 - r)(2r - 5) = 0

⇒ r = \(\frac {2} {5} or \frac {5} {2}\)

Thus, the three terms of G.P. are \(\frac{5} {2},1,and \frac{2} {5}.\)