The ratio of sum of m and n terms of an A.P. | vedclass

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Question

(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}

Vedclass · Accepted Answer

$\frac{{2m - 1}}{{2n - 1}}$

Vedclass · Answer

(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.

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

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