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(A) Let the first term be $a$ and the common difference be $d$.Given that the sum of the first $n$ terms $S_n$ is equal to the sum of the first $m$ te

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(A) Let the first term be $a$ and the common difference be $d$.Given that the sum of the first $n$ terms $S_n$ is equal to the sum of the first $m$ terms $S_m$:$\frac{n}{2}[2a + (n - 1)d] = \frac{m}{2}[2a + (m - 1)d]$$n[2a + (n - 1)d] = m[2a + (m - 1)d]$$2an + n(n - 1)d = 2am + m(m - 1)d$$2a(n - m) + d(n^2 - n - m^2 + m) = 0$$2a(n - m) + d[(n^2 - m^2) - (n - m)] = 0$$2a(n - m) + d[(n - m)(n + m) - (n - m)] = 0$Since $m 
e n$,we can divide by $(n - m)$:$2a + d(n + m - 1) = 0$Now,the sum of the first $(m + n)$ terms is:$S_{m+n} = \frac{m + n}{2}[2a + (m + n - 1)d]$Substituting $2a + (m + n - 1)d = 0$:$S_{m+n} = \frac{m + n}{2} 	imes 0 = 0$

If the sum of first $n$ terms of an $A.P.$ is equal to the sum of its first $m$ terms,$(m \ne n)$,then the sum of its first $(m + n)$ terms will be

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