Question

If the ratio of the sum of $m$ and $n$ term of an A.P is $m^2:n^2$. Find the ratio of $m^{th}$ and $n^{th}$ term

Answer 1

$S_m=\frac{m}{2}(2a+(m-1\left)d\right)$

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$\frac{\frac{m}{2}(4a+(m-1\left)d\right)}{\frac{n}{2}(2a+(n-1\left)d\right)}=\frac{m^2}{n^2}$

$2an+mnd-nd=2am+mnd-md$

$n(2a-d)=m(2a-d)$

$(m-n)(2a-d)=0$

$m\ne n$ as they are different terms

$2a=d....\left(i\right)$

Ratio of $m^{th}$ and $n^{th}$ term (using eq (i))

$\frac{a+(m-1)d}{a+(n-1)d}=\frac{a+(m-1)\left(2a\right)}{a+(n-1)\left(2a\right)}$

$=\frac{a(1+2m-2)}{a(1+2n-2)}=\frac{2m-1}{2n-1}$

Ratio of $m^{th}$ and $n^{th}$ term $(2m-1):(2n-1)$.