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

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(C) Given that the ratio of the sum of $m$ and $n$ terms is $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.We know that $S_n = \frac{n}{2}[2a + (n - 1)d]$.Theref

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

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(C) Given that the ratio of the sum of $m$ and $n$ terms is $\frac{S_m}{S_n} = \frac{m^2}{n^2}$.We know that $S_n = \frac{n}{2}[2a + (n - 1)d]$.Therefore,$\frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}$.$\Rightarrow \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$.Cross-multiplying,we get $n[2a + (m - 1)d] = m[2a + (n - 1)d]$.$2an + n(m - 1)d = 2am + m(n - 1)d$.$2an + mnd - nd = 2am + mnd - md$.$2an - 2am = nd - md$.$2a(n - m) = d(n - m)$.Thus,$d = 2a$.The $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.The ratio of $m^{th}$ 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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