A) \[r=-2/3\] or \[-3/2\]
B) \[r=-2/3\] or \[3/2\]
C) \[r=2/3\] or \[-3/2\]
D) \[r=2/3\] or \[3/2\]
Correct Answer: A
Solution :
Let first three terms of a GP are a, ar, \[a{{r}^{2}}.\] Then, \[sum=\frac{7}{9}\] (given) \[\Rightarrow \] \[a+ar+a{{r}^{2}}=\frac{7}{9}\] \[\Rightarrow \] \[a(1+r+{{r}^{2}})=\frac{7}{9}\] ?.(i) and Product \[=-\frac{8}{27}\] \[\Rightarrow \] \[a.ar.ar=-\frac{8}{27}\] \[\Rightarrow \] \[{{a}^{3}}{{r}^{3}}=-\frac{8}{27}={{\left( \frac{-2}{3} \right)}^{3}}\] On comparing the powers, we get \[ar=\frac{-2}{3}\] ?..(ii) On dividing Eq. (i) by (ii), we get \[\frac{1+r+{{r}^{2}}}{r}=\frac{\frac{7}{9}}{\frac{-2}{3}}\] \[\Rightarrow \] \[\frac{1+r+{{r}^{2}}}{r}=-\frac{7}{9}\times \frac{3}{2}=\frac{-7}{6}\] \[\Rightarrow \] \[1+r+{{r}^{2}}=\frac{-7}{6}r\] \[\Rightarrow \] \[{{r}^{2}}+\left( 1+\frac{7}{6} \right)r+1=0\] \[\Rightarrow \] \[{{r}^{2}}+\frac{13}{6}r+1=0\] \[\Rightarrow \] \[\left( r+\frac{3}{2} \right)\left( r+\frac{2}{3} \right)=0\] \[\Rightarrow \] \[r=\frac{-3}{2}\] or \[\frac{-2}{3}\]
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