The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term

Let’s consider that a and b to be the first term and the common difference of the A.P. respectively.

Then from the question, we have

\( \dfrac{s_m}{s_n}=\dfrac{m^2}{n^2}\)

\( s_n=\dfrac{\dfrac{m}{2}[2a+(m-1)d]}{\dfrac{n}{2}[2a+(n-1)d]}=\dfrac{m^2}{n^2}\)

\( =\dfrac{2a+(m-1)d}{2a+(m+1)d}\)

= n[2a+(m−1)d]=m[2a+(n−1)d]

⇒2an+mnd−nd+2am+mnd−nd

= md−nd=2am−2an

= (m−n)d=2a(m−n)

=d=2a

Now, the ratio of m^th and n^th terms is 

\( \dfrac{m^2}{n^2} =\dfrac{a+(m-1)d}{a+(m+1)d}\)

= \( \dfrac{a+(m-1)2a}{a+(m+1)2a}\)

= \( \dfrac{a(1+2m-2)}{a(1+2n-2)}\)

= \( \dfrac{2m-1}{2n-1}\)

Therefore, the ratio of m^th and n^th term is (2m -1) : (2n -1)