# 12.   The ratio of the sums of m and n terms of an A.P. is $m^2 : n^2$ . Show that the ratio of mth and nth term is $( 2m-1) : ( 2n- 1 )$.

S seema garhwal

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is $m^2 : n^2$ .

To prove :  the ratio of mth and nth term is $( 2m-1) : ( 2n- 1 )$.

$\therefore \, \frac{sum\, of\, m\, \, terms}{sum\, of\, n\, \, terms }=\frac{m^2}{n^2}$

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

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

Put $m=2m-1\, \, and\, \, n=2n-1$, we get

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

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

$\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{a+(m-1)d}{a+(n-1)d}$

From equation (1) ,we get

$\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{2m-1}{2n-1}$

Hence proved.

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