Question

Let $t_r$ denotes the $r^{th}$ term of an $A.P$. Also, suppose that $t_m=\frac{1}{n}$ and $t_n=\frac{1}{m},\left(m\ne n\right)$, for some positive integers $m$ and $n$, then which of the following is necessarily a root of the equation $(l+m-2n)x^2+(m+n-2l)x+(n+l-2m)=0$?

• A. $t_n$
• B. $t_m$
• C. $t_{m+n}$
• D. $t_{mn}$

Answer 1

Answer: (D)

$t_m=a+\left(m-1\right)d=\frac{1}{n}...\left(i\right)$
$t_n=a+\left(n-1\right)d=\frac{1}{m}...\left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right)$, we get
$\left(m-n\right)d=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow \left(m-n\right)d=\frac{m-n}{mn}$
$\Rightarrow d=\frac{1}{mn}\left(\because m\ne n\right)...\left(iii\right)$
$t_{mn}=a+\left(mn-1\right)d=a+\left(mn-1\right)\times \frac{1}{mn}$
$=a-\frac{1}{mn}+1...\left(iv\right)$
From $\left(i\right)$ and $\left(ii\right)$
$a+\left(m-1\right)\cdot \frac{1}{mn}=\frac{1}{n}$
$\Rightarrow a+\frac{1}{n}-\frac{1}{mn}=\frac{1}{n}$
$\Rightarrow a=\frac{1}{mn}$
$\Rightarrow t_{mn}=1$