A) 0
B) \[1\]
C) \[\frac{m+n}{2}\]
D) \[\sqrt{mn}\]
Correct Answer: A
Solution :
Let a be the first term and d be the common difference of the given AP, men \[\Rightarrow \] \[{{S}_{m}}-{{S}_{n}}\]a \[\Rightarrow \]\[\frac{m}{2}[2a+(m-1)d]=\frac{n}{2}[2a+(n-1)d]\] \[\Rightarrow \]\[2a(m-n)+\{m(m-1)-n(n-1)\}d=0\] \[\Rightarrow \]\[2a(m-n)+\{{{m}^{2}}-{{n}^{2}}-(m-n)\}d=0\] \[\Rightarrow \]\[(m-n)[2a+(m+n-1)d]=0\] \[\Rightarrow \]\[2a+(m+n-1)d=0\] ?(i) \[(\because \,m-n\ne 0)\] Now, \[{{S}_{m+n}}=\frac{m+n}{2}[2a+(m+n-1)d]\] \[=\frac{m+n}{2}.0\] [using Eq.(i)] \[=0\]
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