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)

$t_{mn}$

Answer 2

$t_{m} =a + \left(m-1\right)d = \frac{1}{n} \quad...\left(i\right)$ $t_{n} = a + \left(n-1\right)d= \frac{1}{m}\quad...\left(ii\right)$ 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 $\left(ii\right)$ $a +\left(m-1\right)\cdot\frac{1}{mn} = \frac{1}{n}$ $\Rightarrow a+ \frac{1}{n} -\frac{1}{mn} = \frac{1}{n}$ $\Rightarrow a= \frac{1}{mn}$ $\Rightarrow t_{mn} = 1$