# Problem Study : Arithmetic Progression

If the ratio of the sum of $\displaystyle m$ terms and $\displaystyle n$ terms of an $\displaystyle A.P.$ is $\displaystyle m^2:n^2$ then show that the ratio of its $\displaystyle m^{\text{th}}$ and $\displaystyle n^{\text{th}}$ term is $\displaystyle (2m-1):(2n-1)$.

$\displaystyle A.P.$ ကိန္းစဥ္ တစ္ခု၏ $\displaystyle m$ အႀကိမ္ေျမာက္အထိ ေပါင္းလဒ္ ($\displaystyle {{S}_{m}}$) ႏွင့္ $\displaystyle n$ အႀကိမ္ေျမာက္အထိ ေပါင္းလဒ္ ($\displaystyle {{S}_{n}}$) တို႔၏ အခ်ိဳးသည္ $\displaystyle m^2:n^2$ ျဖစ္လွ်င္ $\displaystyle m$ အႀကိမ္ေျမာက္ ကိန္းလံုး ($\displaystyle {{u}_{m}}$) ႏွင့္ $\displaystyle m$ အႀကိမ္ေျမာက္ ကိန္းလံုး ($\displaystyle {{u}_{n}}$) တို႔၏ အခ်ိဳးသည္ $\displaystyle (2m-1):(2n-1)$ ျဖစ္ေၾကာင္း သက္ေသျပပါ။ (2012 - ေမးခြန္းေဟာင္း)
Solution

Let the first term and the common difference of the given $\displaystyle A.P.$ be $\displaystyle a$ and $\displaystyle d$ respectively.

By the problem,

$\displaystyle \ \ \ \ \frac{{{{S}_{m}}}}{{{{S}_{n}}}}=\frac{{{{m}^{2}}}}{{{{n}^{2}}}}$

$\displaystyle \therefore \frac{{\frac{m}{2}\left\{ {2a+(m-1)d} \right\}}}{{\frac{n}{2}\left\{ {2a+(n-1)d} \right\}}}=\frac{{{{m}^{2}}}}{{{{n}^{2}}}}$

$\displaystyle \therefore \frac{{m\left\{ {2a+(m-1)d} \right\}}}{{n\left\{ {2a+(n-1)d} \right\}}}=\frac{{{{m}^{2}}}}{{{{n}^{2}}}}$

$\displaystyle \therefore \frac{{2a+(m-1)d}}{{2a+(n-1)d}}=\frac{m}{n}$

$\displaystyle \therefore 2am+mnd-md=2an+mnd-nd$

$\displaystyle \therefore 2am-md=2an-nd$

$\displaystyle \therefore 2am-2an=md-nd$

$\displaystyle \therefore 2a(m-n)=d(m-n)$

$\displaystyle \therefore 2a=d$

$\displaystyle \therefore \frac{{{{u}_{m}}}}{{{{u}_{n}}}}=\frac{{a+(m-1)d}}{{a+(n-1)d}}$

$\displaystyle \ \ \ \ \ \ \ \ \ =\frac{{a+(m-1)(2a)}}{{a+(n-1)(2a)}}$

$\displaystyle \ \ \ \ \ \ \ \ \ =\frac{{a+2am-2a}}{{a+2am-2a}}$

$\displaystyle \ \ \ \ \ \ \ \ \ =\frac{{a(2m-1)}}{{a(2n-1)}}$

$\displaystyle \ \ \ \ \ \ \ \ \ =\frac{{2m-1}}{{2n-1}}$