Mathematics - Sequence & Series Question with Solution | TestHub

Let the first term be $a$ and common difference be $d$. The sum of $n$ terms is $S_n=\frac{n}{2}(2a+(n-1)d)$. Given $\frac{S_m}{S_n}=\frac{m^2}{n^2}$, we substitute the formula for $S_n$: $$\frac{\frac{m}{2}(2a+(m-1)d)}{\frac{n}{2}(2a+(n-1)d)}=\frac{m^2}{n^2}$$ This simplifies to $\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$. Cross-multiplying gives $n(2a+(m-1)d)=m(2a+(n-1)d)$, which expands to $2an+n(m-1)d=2am+m(n-1)d$. Rearranging terms, $2a(n-m)=d(m(n-1)-n(m-1))=d(mn-m-mn+n)=d(n-m)$. Since $m\ne n$, we can divide by $(n-m)$, yielding $2a=d$. The ratio of the $m$ -th term to the $n$ -th term is $\frac{a_m}{a_n}=\frac{a+(m-1)d}{a+(n-1)d}$. Substituting $d=2a$: $$\frac{a+(m-1)(2a)}{a+(n-1)(2a)}=\frac{a(1+2m-2)}{a(1+2n-2)}=\frac{2m-1}{2n-1}$$