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}}$

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!