A) \[a\,{{\left( \frac{b}{a} \right)}^{\frac{n}{n-1}}}\]
B) \[a\,{{\left( \frac{b}{a} \right)}^{\frac{n-1}{n}}}\]
C) \[a\,{{\left( \frac{b}{a} \right)}^{\frac{n}{n+1}}}\]
D) \[a\,{{\left( \frac{b}{a} \right)}^{\frac{1}{n}}}\]
Correct Answer: C
Solution :
If \[n\] geometric means \[{{g}_{1}},{{g}_{2}}.......{{g}_{n}}\] are to be inserted between two positive real numbers \[a\] and \[b\], then \[a,\ {{g}_{1}},\ {{g}_{2}}......{{g}_{n}},\ b\]are in G.P. Then \[{{g}_{1}}=ar,\ {{g}_{2}}=a{{r}^{2}}........{{g}_{n}}=a{{r}^{n}}\] So \[b=a{{r}^{n+1}}\Rightarrow r={{\left( \frac{b}{a} \right)}^{1/(n+1)}}\] Now \[{{n}^{th}}\] geometric mean \[\]\[({{g}_{n}})=a{{r}^{n}}=a\,\,{{\left( \frac{b}{a} \right)}^{n/(n+1)}}\]. Aliter : As we have the \[{{m}^{th}}\] G.M. is given by \[{{G}_{m}}=a{{\left( \frac{b}{a} \right)}^{\frac{m}{n+1}}}\] Now replace \[m\] by \[n\] we get the required result.
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