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प्रश्न
If Sn denotes the sum of first n terms of an A.P. < an > such that
विकल्प
\[\frac{2 m + 1}{2 n + 1}\]
\[\frac{2 m - 1}{2 n - 1}\]
\[\frac{m - 1}{n - 1}\]
\[\frac{m + 1}{n + 1}\]
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उत्तर
\[\frac{2 m - 1}{2 n - 1}\]
\[\frac{S_m}{S_n} = \frac{m^2}{n^2}\]
\[ \Rightarrow \frac{\frac{m}{2}\left\{ 2a + \left( m - 1 \right)d \right\}}{\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}} = \frac{m^2}{n^2}\]
\[ \Rightarrow \frac{\left\{ 2a + \left( m - 1 \right)d \right\}}{\left\{ 2a + \left( n - 1 \right)d \right\}} = \frac{m}{n} \]
\[ \Rightarrow 2an + ndm - nd = 2am + nmd - md\]
\[ \Rightarrow 2an - 2am - nd + md = 0\]
\[ \Rightarrow 2a\left( n - m \right) - d\left( n - m \right) = 0\]
\[ \Rightarrow 2a\left( n - m \right) = d\left( n - m \right)\]
\[ \Rightarrow d = 2a . . . . . \left( 1 \right)\]
\[\text { Ratio of } \frac{a_m}{a_n} = \frac{a + \left( m - 1 \right)d}{a + \left( n - 1 \right)d}\]
\[ \Rightarrow \frac{a_m}{a_n} = \frac{a + \left( m - \right)2a}{a + \left( n - \right)2a}\text { From } (1)\]
\[ = \frac{a + 2am - 2a}{a + 2an - 2a} \]
\[ = \frac{2am - a}{2an - a}\]
\[ = \frac{2m - 1}{2n - 1}\]
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