A) - 2, - 32
B) - 2, 3
C) - 6, 3
D) - 6, - 32
Correct Answer: A
Solution :
Let r be the common ratio of the G.P. a, b, g, d then \[\beta =\alpha \,r,\] \[\gamma =\alpha \,{{r}^{2}}\] and \[\delta =\alpha \,{{r}^{3}}\] \[\therefore \alpha +\beta =1\] \[\Rightarrow \alpha +\alpha \,r=1\] \[\Rightarrow \alpha (1+r)=1\] ?..(i) \[\alpha \beta =p\Rightarrow \alpha \,(\alpha r)=p\Rightarrow {{\alpha }^{2}}r=p\] ?..(ii) \[\gamma +\delta =4\Rightarrow \alpha {{r}^{2}}+\alpha {{r}^{3}}=4\] \[\gamma +\delta =q\Rightarrow a{{r}^{2}}a{{r}^{3}}\]\[\alpha \,{{r}^{2}}(1+r)=4\] ?..(iii) and \[\gamma \delta =q\Rightarrow \alpha {{r}^{2}}.\alpha {{r}^{3}}=q\] \[\Rightarrow {{\alpha }^{2}}{{r}^{5}}=q\] ?..(iv) Dividing (iii) by (i), we get, \[{{r}^{2}}=4\Rightarrow r=\pm \,2\] If we take \[r=2\], then \[\alpha \]is not integral, so we take \[r=-2,\] Substituting \[r=-2\] in (i), we get \[\alpha =-1\] Now, from (ii), we have \[p={{\alpha }^{2}}r\,=\,{{(-1)}^{2}}(-2)=-2\] and from (iv), we have \[q={{\alpha }^{2}}{{r}^{5}}={{(-1)}^{2}}\,{{(-2)}^{5}}=-32\] Þ (p, q) = (- 2, - 32).
You need to login to perform this action.
You will be redirected in
3 sec
