A) \[m-1:n-1\]
B) \[2m+1:2n+1\]
C) \[2m-1:2n-1\]
D) none of these
Correct Answer: C
Solution :
Let a and d be the first term and common difference respectively. Given that \[\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{m/2[2a+(m-1)d]}{n/2[2a+(n-1)d]}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}\] Replace \[m\]by \[2m-1\]and\[n\]by \[2n-1\] \[\Rightarrow \] \[\frac{2a+(2m-2)d}{2a+(2n-2)d}=\frac{2m-1}{2n-1}\] \[\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\] \[\Rightarrow \]Ration of \[mth\]and \[nth\]terms \[=(2m-1):(2n-1)\]
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