11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer If \[n\] geometric means between \[a\] and \[b\]be \[{{G}_{1}},\ {{G}_{2}},\ .....\]\[{{G}_{n}}\]  and a geometric mean be \[G\], then the true relation is

    A) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}=G\]

    B) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}={{G}^{1/n}}\]

    C) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}={{G}^{n}}\]

    D) \[{{G}_{1}}.{{G}_{2}}........{{G}_{n}}={{G}^{2/n}}\]

    Correct Answer: C

    Solution :

    Here \[G={{(ab)}^{1/2}}\]and \[{{G}_{1}}=a{{r}^{1}},\ {{G}_{2}}=a{{r}^{2}},........{{G}_{n}}=a{{r}^{n}}\] Therefore \[{{G}_{1}}.\ {{G}_{2}}.\ {{G}_{3}}.....{{G}_{n}}={{a}^{n}}{{r}^{1+2+...+n}}={{a}^{n}}{{r}^{n(n+1)/2}}\] But \[a{{r}^{n+1}}=b\Rightarrow r={{\left( \frac{b}{a} \right)}^{1/(n+1)}}\] Therefore, the required product is \[{{a}^{n}}{{\left( \frac{b}{a} \right)}^{1/(n+1)\ .\ n(n+1)/2}}={{(ab)}^{n/2}}={{\left\{ {{(ab)}^{1/2}} \right\}}^{n}}={{G}^{n}}\]. Note: It is a well-known fact.


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