A) \[2m+1:2n+1\]
B) \[2m-1:2n-1\]
C) \[2m:n\]
D) \[m:n\]
Correct Answer: B
Solution :
We have, given that, \[\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{\frac{m}{2}\left[ 2a+(m-1)d \right]}{\frac{n}{2}\left[ 2a+(n-1)d \right]}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}\] Replacing m with \[2m-1\] and n with \[2n-1,\] we get \[\Rightarrow \] \[\frac{2a+2(m-1)d}{2a+2(n-1)d}=\frac{2m-1}{2n-1}\] \[\Rightarrow \] \[\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\Rightarrow \frac{{{a}_{m}}}{{{a}_{n}}}=\frac{2m-1}{2n-1}\]
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