The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is \[87\frac{1}{2}\] . Find them.
Solution
Let the required numbers be \[\frac{a}{r},\text { a and ar } .\] Product of the G.P. = 125
\[\Rightarrow a^3 = 125 \]
\[ \Rightarrow a = 5\]
Sum of the products in pairs = \[87\frac{1}{2} = \frac{175}{2}\]
\[\Rightarrow \frac{a}{r} \times a + a \times ar + ar \times \frac{a}{r} = \frac{175}{2}\]
\[ \Rightarrow \frac{a^2}{r} + a^2 r + a^2 = \frac{175}{2}\]
\[\text {Substituting the value of a }\]
\[ \Rightarrow \frac{25}{r} + 25r + 25 = \frac{175}{2}\]
\[ \Rightarrow 50 r^2 + 50r + 50 = 175r\]
\[ \Rightarrow 50 r^2 - 125r + 50 = 0\]
\[ \Rightarrow 25(2 r^2 - 5r + 2) = 0\]
\[ \Rightarrow 2 r^2 - 4r - r + 2 = 0\]
\[ \Rightarrow 2r(r - 2) - 1(r - 2) = 0\]
\[ \Rightarrow (2r - 1)(r - 2) = 0\]
\[ \therefore r = \frac{1}{2}, 2\]
Hence, the G.P. for a = 5 and r = \[\frac{1}{2}\] is 10, 5 and \[\frac{5}{2}\] .
And, the G.P. for a = 5 and r = 2 is \[\frac{5}{2}\] , 5 and 10.
