Question

The ratio of sum of $m$ and $n$ terms of an A.P. is $m ^ { 2 } : n ^ { 2 }$ , then the ratio of $m ^ { t h }$and $n ^ { t h }$ term will be.

• A. $\frac { m - 1 } { n - 1 }$
• B. $\frac { n - 1 } { m - 1 }$
• C.
• D. $\frac { 2 n - 1 } { 2 m - 1 }$

Answer 1

Answer: (C)

Answer 2

Given that $\frac { \frac { m } { 2 } [ 2 a + ( m - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } = \frac { m ^ { 2 } } { n ^ { 2 } }$

$\Rightarrow$ $\frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m } { n }$ $\Rightarrow$ $\frac { a + \frac { 1 } { 2 } ( m - 1 ) d } { a + \frac { 1 } { 2 } ( n - 1 ) d } = \frac { m } { n }$

$\Rightarrow$ $a n + \frac { 1 } { 2 } ( m - 1 ) n d = a m + \frac { 1 } { 2 } ( n - 1 ) m d$

$\Rightarrow$ $a ( n - m ) + \frac { d } { 2 } [ m n - n - m n + m ] = 0$

$\Rightarrow$ $a ( n - m ) + \frac { d } { 2 } ( m - n ) = 0$ $\Rightarrow$ $a = \frac { d } { 2 }$ or $d = 2 a$

So, required ratio,

$\frac { T _ { m } } { T _ { n } } = \frac { a + ( m - 1 ) d } { a + ( n - 1 ) d } = \frac { a + ( m - 1 ) 2 a } { a + ( n - 1 ) 2 a }$

$= \frac { 1 + 2 m - 2 } { 1 + 2 n - 2 } = \frac { 2 m - 1 } { 2 n - 1 }$.

Trick : Replace $m$ by $2 m - 1$ and $n$ by $2 n - 1$. Obviously if $S _ { m }$ is of degree 2, then $T _ { m }$ is of $1$ linear.