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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}$$[2a+(n-1)d]$
Given: Sum of $m$ terms : Sum of $n$ terms of an $A.P=m^2:n^2$
Dividing numerator and denominator by $2$
Let $\large\frac{m-1}{2}$$=M-1 \:\:and\:\:\large\frac{n-1}{2}$$=N-1$
Substituting the values of $m\:\:and\:\:n$
But we know that $a+(n-1)d=t_n$
$i.e.,$ 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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