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Question

# If Sn denotes the sum of first n terms of an A.P. < an > such that $\frac{{S}_{m}}{{S}_{n}}=\frac{{m}^{2}}{{n}^{2}},\mathrm{then}\frac{{a}_{m}}{{a}_{n}}=$ (a) $\frac{2m+1}{2n+1}$ (b) $\frac{2m-1}{2n-1}$ (c) $\frac{m-1}{n-1}$ (d) $\frac{m+1}{n+1}$

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Solution

## (b) $\frac{2m-1}{2n-1}$ $\frac{{S}_{m}}{{S}_{n}}=\frac{{m}^{2}}{{n}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{m}{2}\left\{2a+\left(m-1\right)d\right\}}{\frac{n}{2}\left\{2a+\left(n-1\right)d\right\}}=\frac{{m}^{2}}{{n}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{\left\{2a+\left(m-1\right)d\right\}}{\left\{2a+\left(n-1\right)d\right\}}=\frac{m}{n}\phantom{\rule{0ex}{0ex}}⇒2an+ndm-nd=2am+nmd-md\phantom{\rule{0ex}{0ex}}⇒2an-2am-nd+md=0\phantom{\rule{0ex}{0ex}}⇒2a\left(n-m\right)-d\left(n-m\right)=0\phantom{\rule{0ex}{0ex}}⇒2a\left(n-m\right)=d\left(n-m\right)\phantom{\rule{0ex}{0ex}}⇒d=2a.....\left(1\right)$ $\mathrm{Ratio}of\frac{{a}_{m}}{{a}_{n}}=\frac{a+\left(m-1\right)d}{a+\left(n-1\right)d}\phantom{\rule{0ex}{0ex}}⇒\frac{{a}_{m}}{{a}_{n}}=\frac{a+\left(m-\right)2a}{a+\left(n-\right)2a}\mathrm{From}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{a+2am-2a}{a+2an-2a}\phantom{\rule{0ex}{0ex}}=\frac{2am-a}{2an-a}\phantom{\rule{0ex}{0ex}}=\frac{2m-1}{2n-1}$

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