A) 0
B) \[n\]
C) \[m\]
D) \[m+n\]
Correct Answer: A
Solution :
As given \[\frac{n}{2}\left\{ 2a+(n-1)d \right\}=\frac{m}{2}\left\{ a+(m-1)d \right\}\] \[\Rightarrow \]\[2a(m-n)+d({{m}^{2}}-m-{{n}^{2}}+n)=0\] \[\Rightarrow \] \[(m-n)\left\{ 2a+d(m+n-1) \right\}=0\] \[\Rightarrow \] \[2a+(m+n-1)d=0\], \[(\because \ m\ne n)\] \[\therefore \] \[{{S}_{m+n}}=\frac{m+n}{2}\left\{ 2a+(m+n-1)d \right\}=\frac{m+n}{2}\left\{ 0 \right\}=0\].
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