Question

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

• A. None of the above
• B. $\frac{(2m–1)}{(2n–1)}$.
• C. $\frac{(m–1)}{(n–1)}$.
• D. $\frac{(2m+1)}{(2n+1)}$.

Answer 1

The correct option is B $\frac{(2m–1)}{(2n–1)}$.

Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by

$S_m$ = ($\frac{m}{2}$) [2a + (m – 1) d], and

$S_n$ = ($\frac{n}{2}$) [2a + (n – 1) d]

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

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

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

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

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

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

$\Rightarrow d=2a$

$\frac{T_m}{T_n}$ = $\frac{a+(m-1)\times 2a}{a+(n-1)\times 2a}=\frac{a+(m-1)\times 2a}{a+(n-1)\times 2a}=\frac{2m-1}{2n-1}$
$(\because d=2a)$