- Home
- Standard 11
- Mathematics
The ratio of sum of $m$ and $n$ terms of an $A.P.$ is ${m^2}:{n^2}$, then the ratio of ${m^{th}}$ and ${n^{th}}$ term will be
$\frac{{m - 1}}{{n - 1}}$
$\frac{{n - 1}}{{m - 1}}$
$\frac{{2m - 1}}{{2n - 1}}$
$\frac{{2n - 1}}{{2m - 1}}$
Solution
(c) Given that $\frac{{\frac{m}{2}[2a + (m – 1)d]}}{{\frac{n}{2}[2a + (n – 1)d]}} = \frac{{{m^2}}}{{{n^2}}}$
$ \Rightarrow $ $\frac{{2a + (m – 1)d}}{{2a + (n – 1)d}} = \frac{m}{n}$
$ \Rightarrow $ $\frac{{a + \frac{1}{2}(m – 1)d}}{{a + \frac{1}{2}(n – 1)d}} = \frac{m}{n}$
$ \Rightarrow $ $an + \frac{1}{2}(m – 1)nd = am + \frac{1}{2}(n – 1)md$
$ \Rightarrow $ $a(n – m) + \frac{d}{2}[mn – n – mn + m] = 0$
$ \Rightarrow $ $a(n – m) + \frac{d}{2}(m – n) = 0$
$ \Rightarrow $ $ a = \frac{d}{2}$ or $d = 2a$
So, required ratio, $\frac{{{T_m}}}{{{T_n}}} = \frac{{a + (m – 1)d}}{{a + (n – 1)d}} = \frac{{a + (m – 1)2a}}{{a + (n – 1)2a}}$
$ = \frac{{1 + 2m – 2}}{{1 + 2n – 2}} = \frac{{2m – 1}}{{2n – 1}}$.
Trick : Replace $m$ by $2m – 1$ and $n$ by $2n – 1$.
Obviously if ${S_m}$ is of degree $2$, then ${T_m}$ is of $1$. $i.e.$ linear.
