Question

The ratio of the sums of first $m$ and first $n$ terms of an arithmetic series is $m^2:n^2$ show that the ratio of the $m^{th}$ and $n^{th}$ terms is $(2m-1):(2n-1)$

Answer 1

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

It is given that the ratio of the sums of first $m$ terms and first $n$ terms of an arithmetic series is $m^2:n^2$, therefore,

$\frac{S_m}{S_n}=\frac{m^2}{n^2}$

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

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

$\Rightarrow \frac{[2a+(m-1\left)d\right]}{[2a+(n-1\left)d\right]}=\frac{m^2\times n}{n^2\times m}$

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

$\Rightarrow n[2a+(m-1\left)d\right]=m[2a+(n-1\left)d\right]$

$\Rightarrow 2na+n(m-1)d=2ma+m(n-1)d$

$\Rightarrow 2na+nmd-nd=2ma+mnd-md$

$\Rightarrow 2na-2ma=nd-md$

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

$\Rightarrow 2a=d$

We also know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$.

Now using the value of $d$ as $d=2a$, consider the ratio of the $m$th and $n$th terms as follows:

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

$=(2m-1):(2n-1)$

Hence, the ratio of $m^{th}$ and $n^{th}$ terms is $(2m-1):(2n-1)$.