A) \[\frac{m-1}{n-1}\]
B) \[\frac{n-1}{m-1}\]
C) \[\frac{2m-1}{2n-1}\]
D) \[\frac{2n-1}{2m-1}\]
Correct Answer: C
Solution :
Given that \[\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \]\[\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}\]\[\Rightarrow \]\[\frac{a+\frac{1}{2}(m-1)d}{a+\frac{1}{2}(n-1)d}=\frac{m}{n}\] \[\Rightarrow \] \[an+\frac{1}{2}(m-1)nd=am+\frac{1}{2}(n-1)md\] \[\Rightarrow \]\[i.e.\] \[\Rightarrow \] \[a(n-m)+\frac{d}{2}(m-n)=0\]\[\Rightarrow \]\[a=\frac{d}{2}\] or \[d=2a\] So, required ratio, \[\frac{{{T}_{m}}}{{{T}_{n}}}=\frac{a+(m-1)d}{a+(n-1)d}=\frac{a+(m-1)2a}{a+(n-1)2a}\] \[=\frac{1+2m-2}{1+2n-2}=\frac{2m-1}{2n-1}\]. Trick: Replace \[m\] by \[2m-1\] and \[n\] by\[2n-1\]. Obviously if \[{{S}_{m}}\] is of degree 2, then \[{{T}_{m}}\] is of \[a\] \[i.e.\] linear.
You need to login to perform this action.
You will be redirected in
3 sec
