A) \[\frac{2m+1}{2n+1}\]
B) \[\frac{2m-1}{2n-1}\]
C) \[\frac{m-1}{n-1}\]
D) \[\frac{m+1}{n+1}\]
Correct Answer: B
Solution :
Let a and d be the first term and common difference of an AP respectively, then, \[{{S}_{m}}=\frac{m}{2}[2a+(m-1)d]\] and \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] Given, \[\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{\frac{m}{2}\{2a+(m-1)d\}}{\frac{n}{2}\{2a+(n-1)d\}}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}\] \[\Rightarrow \] \[2an+(mn-n)d=2am+(mn-m)d\] \[\Rightarrow \] \[2a(n-m)+(mn-n-mn+m)d=0\] \[\Rightarrow \] \[2a(n-m)+(m-n)d=0\] \[\Rightarrow \] \[(m-n)(d-2a)=0\] \[\Rightarrow \] \[d=2a\] ?..(iii) \[(\because \,\,\,\,m\ne n)\] Now, \[\frac{{{a}_{m}}}{{{a}_{n}}}=\frac{a+(m-1)d}{a+(n-1)d}=\frac{a+(m-1)2a}{a+(n-1)2a}\] \[=\frac{a(1+2m-2)}{a(2+2n-2)}=\frac{2m-1}{2n-1}\]
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