Question

The ratio of sum of m and n term of an AP in $\frac{m^2}{n^2}$ or $m^2-n^2$ Then the ratio of m and n then is

• A. $\frac{2m-1}{2n-1}$
• B. $\frac{2m+1}{2n+1}$
• C. $\frac{2m-1}{2n+1}$
• D. $\frac{2n+1}{2n-1}$

Answer 1

The correct option is A $\frac{2m-1}{2n-1}$

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

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

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

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

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

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

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

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

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

Ratio of $m^{th}$ and $n^{th}$ term is

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

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

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

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

So, the ratio of $m^{th}$ term and the $n^{th}$ term of the arithmetic series is $(2m-1):(2n-1)$.