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Q.
If the ratio of the sum of m and $ n $ terms of an AP is $ {{m}^{2}}:{{n}^{2}}, $ then the ratio of its mth and nth terms is:
Bihar CECEBihar CECE 2005
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) $