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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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The ratio of the sums of $m\:\:and\:\:n$ terms of an $A.P.$ is $m^2:n^2$. Show that the ratio of $m^{th}\:and\:n^{th}$ term is $(2m-1):(2n-1)$

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  • $S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
Given: Sum of $m$ terms : Sum of $n$ terms of an $A.P=m^2:n^2$
$\Rightarrow\:\large\frac{S_m}{S_n}=\frac{m^2}{n^2}$
$\Rightarrow\:\large\frac{\large\frac{m}{2}[2a+(m-1)d]}{\large\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}$
$\Rightarrow\:\large\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$
Dividing numerator and denominator by $2$
$\Rightarrow\:\large\frac{a+\large\frac{m-1}{2}d}{a+\large\frac{n-1}{2}d}=\frac{m}{n}$
Let $\large\frac{m-1}{2}$$=M-1 \:\:and\:\:\large\frac{n-1}{2}$$=N-1$
$\Rightarrow\:m=2M-1\:\:and\:\:n=2N-1$
Substituting the values of $m\:\:and\:\:n$
$\Rightarrow\:\large\frac{a+(M-1)d}{a+(N-1)d}=\frac{2M-1}{2N-1}$
But we know that $a+(n-1)d=t_n$
$a+(M-1)d=t_M\:\:and\:\:a+(N-1)d=t_N$
$\therefore\:\large\frac{t_M}{t_N}=\frac{2M-1}{2N-1}$
$i.e.,$ The ratio of $M^{th}$ term and $N^{th}$ term $=(2M-1):(2N-1)$
$\Rightarrow\:$ The ratio of $m^{th}$ and $n^{th}$ term=$(2m-1):(2n-1)$
answered Feb 23, 2014 by rvidyagovindarajan_1
 

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