Hi Sara,
if the ratio of sum of the first m and n terms of an A.P is m² : n², show that the ratio of its mth and nth terms is (2m-1) : (2n-1)
Let's take a look.
The sum of the first m terms of an AP is \(S_m=\frac{m}{2}[2a+(m-1)d]\)
The sum of the first n terms of an AP is \(S_n=\frac{n}{2}[2a+(n-1)d]\)
The mth term is \(T_m=a+(m-1)d\)
The nth term is \(T_n=a+(n-1)d\)
So given:
\(\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2} \)
SHOW THAT:
\(\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\)
Ok lets look at what we know is true:
\(\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}\\~\\ \frac{[2a+(m-1)d]}{[2a+(n-1)d]}=\frac{m}{n}\\~\\ n[2a+dm-d]=m[2a+dn-d]\\ 2an+dmn-dn=2am+dnm-dm\\ 2an-2am=dnm-dm-dmn+dn\\ 2a(n-m)=d(n-m)\\ d=2a\\ \)
Now I need to show the following
\(\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\)
So I am going to sube d=2a into the LHS
\(LHS\\=\frac{a+(m-1)*2a}{a+(n-1)*2a}\\ =\frac{a+2am-2a}{a+2an-2a}\\ =\frac{2am-a}{2an-a}\\ =\frac{a(2m-1)}{a(2n-1)}\\ =\frac{2m-1}{2n-1}\\ =RHS\)
there you go, that is how it is done :))
+2236 
