A) \[a\ge b\ge c\]
B) \[a+c=b\]
C) \[ac-{{b}^{2}}=0\]
D) (a) and (c) both
Correct Answer: D
Solution :
Let \[\alpha ,\beta \] be the first and \[{{(2n-1)}^{th}}\] terms of the A.P., the G.P. and the H.P. respectively. Then we have For A.P. : \[\beta =\alpha +(2n-2)d\Rightarrow d=\frac{\beta -\alpha }{2n-2}\] \[{{n}^{th}}\]term \[=a=\alpha +(n-1)d=\frac{1}{2}(\alpha +\beta )\] ?..(i) Again for G.P. : \[\beta =\alpha .{{r}^{2n-2}}\Rightarrow r={{\left( \frac{\beta }{\alpha } \right)}^{\frac{1}{2n-2}}}\] \[\therefore \]\[{{n}^{th}}\] term \[=b=\alpha {{r}^{n-1}}=\alpha {{\left( \frac{\beta }{\alpha } \right)}^{\frac{n-1}{2n-2}}}=\alpha {{\left( \frac{\beta }{\alpha } \right)}^{\frac{1}{2}}}\] or \[b={{(\alpha \beta )}^{1/2}}=\sqrt{\alpha \beta }\] ?..(ii) Again for H.P. : \[\frac{1}{\beta }=\frac{1}{\alpha }+(2n-2)d'\] \[\Rightarrow \]\[\frac{1}{c}=\frac{1}{\alpha }+(n-1)d'=\frac{1}{\alpha }+\frac{\alpha -\beta }{2\alpha \beta }=\frac{\alpha +\beta }{2\alpha \beta }\] \[\Rightarrow \]\[c=\frac{2\alpha \beta }{\alpha +\beta }\] ?..(iii) Now, more than one of the alternative answers may be correct. We try for (a): \[a-b=\frac{\alpha +\beta }{2}-\sqrt{\alpha \beta }=\frac{1}{2}{{(\sqrt{\alpha }-\sqrt{\beta })}^{2}}\ge 0\Rightarrow a\ge b\] \[b-c=\sqrt{\alpha \beta }-\frac{2\alpha \beta }{\alpha +\beta }=\frac{\sqrt{\alpha \beta }}{\alpha +\beta }(\alpha +\beta -2\sqrt{\alpha \beta })\] \[=\frac{\sqrt{\alpha \beta }}{(\alpha +\beta )}{{(\sqrt{\alpha }-\sqrt{\beta })}^{2}}\ge 0\Rightarrow b\ge c\] \[\therefore \]\[a\ge b\ge c\] ?..(iv) Now we try for : \[ac=\frac{\alpha +\beta }{2}.\frac{2\alpha \beta }{\alpha +\beta }=\alpha \beta ={{b}^{2}}\] \[\therefore \]\[ac-{{b}^{2}}=0\] ?..(v) Obviously it can be seen that \[a+c\ne b\] ?..(vi) Hence (a) and (b) both hold good.
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