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

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(C) Given that $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.We know that $S_m = \frac{m}{2}[2a + (m-1)d]$.So,$\frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a +

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

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(C) Given that $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.We know that $S_m = \frac{m}{2}[2a + (m-1)d]$.So,$\frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}$.$\Rightarrow \frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n}$.Cross-multiplying,we get $n[2a + (m-1)d] = m[2a + (n-1)d]$.$2an + n(m-1)d = 2am + m(n-1)d$.$2an - 2am = m(n-1)d - n(m-1)d$.$2a(n-m) = d[mn - m - mn + n] = d(n-m)$.Thus,$d = 2a$.The ratio of the $m^{th}$ and $n^{th}$ term is $\frac{T_m}{T_n} = \frac{a + (m-1)d}{a + (n-1)d}$.Substituting $d = 2a$,we get $\frac{a + (m-1)2a}{a + (n-1)2a} = \frac{a(1 + 2m - 2)}{a(1 + 2n - 2)} = \frac{2m-1}{2n-1}$.

The ratio of the sum of $m$ and $n$ terms of an $A.P.$ is $m^2:n^2$. Then the ratio of the $m^{th}$ and $n^{th}$ term will be:

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