If the sum of first n terms of an A.P. be equ | vedclass

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Question

(a) As given $\frac{n}{2}\left\{ {2a + (n - 1)d} \right\} = \frac{m}{2}\left\{ {a + (m - 1)d} \right\}$

$ \Rightarrow $$2a(m - n) + d({m^2} - m - {

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(a) As given $\frac{n}{2}\left\{ {2a + (n - 1)d} ight\} = \frac{m}{2}\left\{ {a + (m - 1)d} ight\}$

$ \Rightarrow $$2a(m - n) + d({m^2} - m - {n^2} + n) = 0$

$ \Rightarrow $ $(m - n)\left\{ {2a + d(m + n - 1)} ight\} = 0$

$ \Rightarrow $ $2a + (m + n - 1)d = 0$,$(\because \;m 
e n)$

$	herefore $ ${S_{m + n}} = \frac{{m + n}}{2}\left\{ {2a + (m + n - 1)d} ight\} $

$= \frac{{m + n}}{2}\left\{ 0 ight\} = 0$.

If the sum of first $n$ terms of an $A.P.$ be 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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