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

Q) The ratio of the sum of m and n terms of an A.P. is $m^2:n^2$ . Show that the ratio m^th and n^thterm is (2m-1) : (2n-1).

Ans: Let:

First term of given A.P. = $a$ ,

Common difference = $d$ ,

Then, the sums of m and n terms are given by

$S_{m}=\frac{m}{2}\{2a+(m-1)d\}$

$S_{n}=\frac{n}{2}\{2a+(n-1)d\}$

Then, $\frac{S_{m}}{S_{m}}=\frac{m^{2}}{n^{2}}$

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

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

$\implies \{2a+(m-1)d\}n=\{2a+(n-1)d\}m$

$\implies 2a(n-m)=d\{(n-1)m-(m-1)n\}$

$\implies 2a(n-m)=d(n-m)$

$\implies d=2a$

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

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