A) \[ax{{n}^{-1}}\]
B) \[a{{x}^{\frac{n}{2}-1}}\]
C) \[a{{x}^{\frac{n}{2}}}\]
D) \[a{{x}^{\frac{n}{2}+1}}\]
Correct Answer: C
Solution :
(c): \[a,ax,a{{x}^{2}},a{{x}^{3}},....a{{x}^{n}}\] Such questions are solved by two methods: (a) RIDOROUS METHOD (b) COMMON SENSE METHOD RIGORDUS METHOD Since no of terms is odd, n can be written as (2k) since first term has \[x{}^\circ (a=ax{}^\circ )\] \[\therefore \] terms are \[\begin{align} & ax{}^\circ ,ax{}^\circ ,a{{x}^{1}},a{{x}^{2}}.......a{{x}^{k}},a{{x}^{k+1}},......a{{x}^{2k}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\uparrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\uparrow \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(k+1)th\,\,term\,\,\,\,\,\,\,\,(2k+1)th\,\,term \\ \end{align}\] Median \[=\frac{(n+1)}{2}th\,\,term=\left( \frac{2k+1+1}{2} \right)th\,\,term\] \[=(k+1)th\,\,term=a{{x}^{k}}=a{{x}^{n/2}}\] COMMON SENSE METHOD No. of terms = odd Trivial values for 3 terms series, 5 terms series, 7 terms series. For 3 terms series;\[a,ax,a{{x}^{2}}\]: Median\[=ax=a{{x}^{(2/2)}}\] For 5 terms series; \[a,ax,a{{x}^{2}},a{{x}^{3}},a{{x}^{4}}\] Median \[=a{{x}^{2}}=a{{x}^{(4/2)}}\] For 7 term series, \[a,ax,a{{x}^{2}},a{{x}^{3}},a{{x}^{4}},a{{x}^{5}},a{{x}^{6}}\] \[\therefore \]Median \[=a{{x}^{3}}=a{{x}^{(6/2)}}\] From observation of above pattern, answer should be \[a{{x}^{(n/2)}}\]
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