Suppose the sum of the first m terms of an ar | vedclass

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(C) Given,$S_m = n$ and $S_n = m$.The formula for the sum of the first $k$ terms is $S_k = \frac{k}{2}[2a + (k-1)d]$.Thus,$S_m = \frac{m}{2}[2a + (m-1

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$-(m+n)$

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(C) Given,$S_m = n$ and $S_n = m$.The formula for the sum of the first $k$ terms is $S_k = \frac{k}{2}[2a + (k-1)d]$.Thus,$S_m = \frac{m}{2}[2a + (m-1)d] = n \implies 2a + (m-1)d = \frac{2n}{m} \quad (i)$.Similarly,$S_n = \frac{n}{2}[2a + (n-1)d] = m \implies 2a + (n-1)d = \frac{2m}{n} \quad (ii)$.Subtracting $(ii)$ from $(i)$:$(m-1)d - (n-1)d = \frac{2n}{m} - \frac{2m}{n}$$(m-n)d = \frac{2(n^2 - m^2)}{mn} = \frac{-2(m-n)(m+n)}{mn}$$d = \frac{-2(m+n)}{mn}$.Substituting $d$ into $(i)$:$2a = \frac{2n}{m} - (m-1)d = \frac{2n}{m} + (m-1) \frac{2(m+n)}{mn} = \frac{2n^2 + 2(m-1)(m+n)}{mn} = \frac{2n^2 + 2(m^2 + mn - m - n)}{mn}$.The sum $S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d]$.Substituting $2a$ and $d$:$S_{m+n} = \frac{m+n}{2} [\frac{2n^2 + 2m^2 + 2mn - 2m - 2n}{mn} + (m+n-1)(\frac{-2(m+n)}{mn})]$.Simplifying the expression inside the bracket:$= \frac{m+n}{2} [\frac{2n^2 + 2m^2 + 2mn - 2m - 2n - 2(m+n)^2 + 2(m+n)}{mn}]$$= \frac{m+n}{2} [\frac{2n^2 + 2m^2 + 2mn - 2m - 2n - 2(m^2 + n^2 + 2mn) + 2m + 2n}{mn}]$$= \frac{m+n}{2} [\frac{2n^2 + 2m^2 + 2mn - 2m^2 - 2n^2 - 4mn}{mn}] = \frac{m+n}{2} [\frac{-2mn}{mn}] = \frac{m+n}{2} (-2) = -(m+n)$.

Suppose the sum of the first $m$ terms of an arithmetic progression is $n$ and the sum of its first $n$ terms is $m$,where $m \neq n$. Then,the sum of the first $(m+n)$ terms of the arithmetic progression is

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