A) \[n{{a}^{2}}=bc{{(n+1)}^{2}}\]
B) \[n{{b}^{2}}=ac{{(n+1)}^{2}}\]
C) \[n{{c}^{2}}=ab{{(n+1)}^{2}}\]
D) None of these
Correct Answer: B
Solution :
Let the roots are \[\alpha \]and \[n\alpha \] Sum of roots\[\alpha +n\alpha =-\frac{b}{a}\]Þ\[\alpha =-\frac{b}{a(n+1)}\] .....(i) and product, \[\alpha .n\alpha =\frac{c}{a}\]Þ \[{{\alpha }^{2}}=\frac{c}{na}\] ....(ii) From (i) and (ii), we get Þ \[{{\left[ -\frac{b}{a(n+1)} \right]}^{2}}=\frac{c}{na}\]Þ \[\frac{{{b}^{2}}}{{{a}^{2}}{{(n+1)}^{2}}}=\frac{c}{na}\] \[\Rightarrow \] \[n{{b}^{2}}=ac{{(n+1)}^{2}}\]. Note: Students should remember this question as a fact.
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