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)

$t_{m} =a + \left(m-1\right)d = \frac{1}{n} \quad...\left(i\right)$
$ t_{n} = a + \left(n-1\right)d= \frac{1}{m}\quad...\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)\quad...\left(iii\right)$
$t_{mn} = a +\left(mn -1\right)d = a + \left(mn-1\right) \times \frac{1}{mn}$
$= a - \frac{1}{mn} +1\quad ...\left(iv\right)$
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$