if the sum of m terms of an A.P to n terms is $m^2$ to$n^2$ then show that the $m$th term to $n$th term is $2m-1$ to $2n-1$

Question

QUESTION: if the sum of m terms of an A.P to n terms is $m^2$ to $n^2$ then show that the $m^{th}$ term to $n^{th}$ term is $2m-1$ to $2n-1$
MY ATTEMPT: using the formula for sum of n terms of an A.P where $a$ is the first term , $d$ is the common difference and $l_m$ and $l_n$ represent the $m^{th}$ term and $n^{th}$ term respectively: $$ \frac{\frac {m}{2} (2a+l_m)}{\frac {n}{2} (2a+l_n)}=\frac {m^2}{n^2} $$ after cancellation: $$ \frac{(2a+l_m)}{(2a+l_n)}=\frac {m}{n} $$ on solving further we get: $$ 2a(n-m)=ml_n-nl_m $$ now i don't know how i can get the value of $\frac{l_m}{l_n}$ from the above equation. any hints or alternate solutions are appreciated. Thanks :)

Answer 1

Given: $$\frac{S_m}{S_n}=\frac{m^2}{n^2} \Rightarrow \frac{\frac{2a_1+d(m-1)}{2}\cdot m}{\frac{2a_1+d(n-1)}{2}\cdot n}=\frac{m^2}{n^2} \Rightarrow \frac{2a_1+d(m-1)}{2a_1+d(n-1)}=\frac{m}{n} \Rightarrow\\ 2a_1n+dn(m-1)=2a_1m+dm(n-1) \Rightarrow \\ 2a_1(n-m)=d(n-m) \Rightarrow d=2a_1.$$ Hence: $$\frac{a_m}{a_n}=\frac{a_1+d(m-1)}{a_1+d(n-1)}=\frac{a_1+2a_1(m-1)}{a_1+2a_1(n-1)}=\frac{2m-1}{2n-1}.$$

Answer 2

$$\begin{align} S_n=\ &An^2+Bn &&&&\scriptsize \left(A=\frac d2; \qquad B=a-\frac d2\right)\\\\ \frac {S_m}{S_n}=\ &\frac {Am^2+Bm}{An^2+Bn}&&=\frac {m^2}{n^2}\\ & B=a-\frac d2&&=0\\ && d &=2a\\ \\ \frac {T_m}{T_n}=\ &\frac {(a-d)+md}{(a-d)+nd} && =\frac {-a+2am}{-a+2an}=\color{red}{\frac {2m-1}{2n-1}} \end{align}$$