Question

The ratio of the sums of m and n terms of an A.P. is $m^2:n^2$, then the ratio of the $m^{th}$ and $n^{th}$ terms is

Answer 1

Sum of $m$ terms of an $A.P.=\frac{m}{2}[2a+(m-1\left)d\right]$

Sum of $n$ terms of an $A.P.=\frac{n}{2}[2a+(n-1\left)d\right]$

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

$\Rightarrow \frac{[2a+md-d]}{[2a+nd-d]}=\frac{m}{n}$

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

$\Rightarrow 2an-2am=nd-md$

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

$\Rightarrow 2a=d$

Ratio of $m^{th}$ term to $n^{th}$ term:

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

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

$=\frac{a[1+2m-2]}{a[1+2n-2]}$

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

So, the ratio of $m^{th}$ term and the $n^{th}$ term of the arithmetic series is

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

Hence, this is the answer.