Question

The sum of first three terms of a G.P. is 13/12 and their product is − 1. Find the G.P.

Answer 1

Let the first three numbers of the given G.P. be \[\frac{a}{r}, \text { a and ar }\]

∴ Product of the G.P. = −1

\[\Rightarrow\] a³ = −1

\[\Rightarrow\] a = −1
Similarly, Sum of the G.P. =  \[\frac{13}{12}\]

\[\Rightarrow \frac{a}{r} + a + ar = \frac{13}{12}\]

Substituting the value of a = −1

\[\frac{- 1}{r} - 1 - r = \frac{13}{12}\]

\[ \Rightarrow 12 r^2 + 25r + 12 = 0\]

\[ \Rightarrow 12 r^2 + 16r + 9r + 12 = 0\]

\[ \Rightarrow 4r\left( 3r + 4 \right) + 3\left( 3r + 4 \right) = 0\]

\[ \Rightarrow \left( 4r + 3 \right)\left( 3r + 4 \right) = 0\]

\[ \Rightarrow r = - \frac{3}{4}, - \frac{4}{3}\]

Hence, the G.P. for a = −1 and r = \[- \frac{3}{4}\] is \[\frac{4}{3}, - 1 \text { and } \frac{3}{4}\].

And, the G.P. for a = −1 and r =\[- \frac{4}{3}\] is \[\frac{3}{4}, - 1 \text { and } \frac{4}{3}\]