The ratio of the sum of …

Let a be the first term and d be the common difference of the given AP. Then,
S_m = sum of first m terms of the given AP;
S_n = sum of first n terms of the given AP.
{tex}\frac { S _ { m } } { S _ { n } } = \frac { m ^ { 2 } } { n ^ { 2 } } \Rightarrow \frac { \frac { m } { 2 } [ 2 a + ( m - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } = \frac { m ^ { 2 } } { n ^ { 2 } }{/tex}
{tex} \Rightarrow \frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m } { n }{/tex}
{tex}\Rightarrow{/tex} 2an+mnd-nd= 2am+mnd-md
{tex}\Rightarrow{/tex} 2an-2am=nd-md
{tex}\Rightarrow{/tex} 2a(n - m) = d(n - m) {tex}\Rightarrow{/tex}2a=d...(i)
{tex}\therefore \quad \frac { T _ { m } } { T _ { n } } = \frac { a + ( m - 1 ) d } { a + ( n - 1 ) d } = \frac { a + ( m - 1 ) \cdot 2 a } { a + ( n - 1 ) \cdot 2 a }{/tex} [from (i)]
{tex}= \frac { a + 2 a m - 2 a } { a + 2 a n - 2 a } = \frac { 2 a m - a } { 2 a n - a } = \frac { a ( 2 m - 1 ) } { a ( 2 n - 1 ) } = \frac { 2 m - 1 } { 2 n - 1 }{/tex}.