11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer If the sum of the \[n\]terms of G.P. is \[S\] product is \[P\] and sum of their inverse is \[R\], than \[{{P}^{2}}\] is equal to  [IIT 1966; Roorkee 1981]

    A) \[\frac{R}{S}\]

    B) \[\frac{S}{R}\]

    C) \[{{\left( \frac{R}{S} \right)}^{n}}\]

    D) \[{{\left( \frac{S}{R} \right)}^{n}}\]

    Correct Answer: D

    Solution :

    Given that sum \[S=\frac{a({{r}^{n}}-1)}{r-1}=\frac{a\,(1-{{r}^{n}})}{1-r}\] ......(i) \[P=a(ar)(a{{r}^{2}})..........(a{{r}^{n-1}})={{a}^{n}}{{r}^{1+2+.........+(n-1)}}\] \[={{a}^{n}}{{r}^{(n-1)n/2}}\ i.e.,\ {{P}^{2}}={{a}^{2n}}{{r}^{n(n-1)}}\]     ......(ii) and \[R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{{r}^{2}}}+..........\]upto \[\frac{1}{49}\] terms            \[=\frac{1}{a}\left( 1+\frac{1}{r}+\frac{1}{{{r}^{2}}}+.........\text{upto}\ n\ \text{terms} \right)\]         \[=\frac{\frac{1}{a}\left[ {{\left( \frac{1}{r} \right)}^{n}}-1 \right]}{\left( \frac{1}{r}-1 \right)}\left( \because \ \frac{1}{r}>1 \right)\] if \[r<1\]        \[=\frac{(1-{{r}^{n}})}{a{{r}^{n-1}}(1-r)}\] ....... (iii) Therefore , \[\frac{S}{R}=\frac{a(1-{{r}^{n}})}{1-r}\times \frac{a{{r}^{n-1}}(1-r)}{(1-{{r}^{n}})}={{a}^{2}}{{r}^{n-1}}\] or \[{{\left( \frac{S}{R} \right)}^{n}}={{({{a}^{2}}{{r}^{n-1}})}^{n}}={{a}^{2n}}{{r}^{n(n-1)}}={{P}^{2}}\].


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