Question

If S_n denotes the sum of first n terms of an A.P. < a_n > such that

\[\frac{S_m}{S_n} = \frac{m^2}{n^2}, \text { then }\frac{a_m}{a_n} =\]

पर्याय

• \[\frac{2 m + 1}{2 n + 1}\]

• \[\frac{2 m - 1}{2 n - 1}\]

• \[\frac{m - 1}{n - 1}\]

• \[\frac{m + 1}{n + 1}\]

Answer 1

\[\frac{2 m - 1}{2 n - 1}\]

\[\frac{S_m}{S_n} = \frac{m^2}{n^2}\]

\[ \Rightarrow \frac{\frac{m}{2}\left\{ 2a + \left( m - 1 \right)d \right\}}{\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}} = \frac{m^2}{n^2}\]

\[ \Rightarrow \frac{\left\{ 2a + \left( m - 1 \right)d \right\}}{\left\{ 2a + \left( n - 1 \right)d \right\}} = \frac{m}{n} \]

\[ \Rightarrow 2an + ndm - nd = 2am + nmd - md\]

\[ \Rightarrow 2an - 2am - nd + md = 0\]

\[ \Rightarrow 2a\left( n - m \right) - d\left( n - m \right) = 0\]

\[ \Rightarrow 2a\left( n - m \right) = d\left( n - m \right)\]

\[ \Rightarrow d = 2a . . . . . \left( 1 \right)\]

\[\text { Ratio of } \frac{a_m}{a_n} = \frac{a + \left( m - 1 \right)d}{a + \left( n - 1 \right)d}\]

\[ \Rightarrow \frac{a_m}{a_n} = \frac{a + \left( m - \right)2a}{a + \left( n - \right)2a}\text {  From } (1)\]

\[ = \frac{a + 2am - 2a}{a + 2an - 2a} \]

\[ = \frac{2am - a}{2an - a}\]

\[ = \frac{2m - 1}{2n - 1}\]