Question

If the ratio of the sum of m terms and n terms of an A.P. be $m^2$ : $n^2$, prove that the ratio of $m^{th}$ and $m^{th}$ term is $(2m-1):(2n-1).$

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

According to the question,

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

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

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

$2an-2am=nd-md$

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

$2a=d$

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.