Question

Let $T_r$ be the $r^{th}$ term of an A.P., for $r=1,2,3,$_______. If for some positive integers m, n we have $T_m=\frac{1}{n}$ and $T_n=\frac{1}{m}$, then $T_{mn}$ equals.

• A. $\frac{1}{mn}$
• B. $\frac{1}{m}+\frac{1}{n}$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$1$

Let, common difference$=d$

$r^{th}$ term $=t_r$

$1^{st}$ term= a

Now $m^{th}$ term= $t_m=a+(m-1)d$

Since, $t_m=\frac{1}{n}$

Therefore,

$\frac{1}{n}=a+(m-1)$ …… (1)

Now $n^{th}$ term $t_n=a+(m-1)d$

Given ,$t_n=\frac{1}{m}$

Therefore,

$\frac{1}{m}=a+(n-1)$ …… (2)

Subtract equation $\left(2\right)$ from equation$\left(1\right)$

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

$\frac{m-n}{mn}=(m-n)d$

$d=\frac{1}{mn}$

Put value d in equation $\left(1\right)$

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

$\frac{1}{n}=a+\frac{m}{mn}+\frac{1}{mn}$

$\frac{1}{n}=a+\frac{1}{n}-\frac{1}{mn}$

$\frac{1}{n}=a+\frac{1}{n}-\frac{1}{mn}$

$a=\frac{1}{mn}$

Now $m{n}^{th}$ term ,

$t_{mn}=\frac{1}{mn}+(mn-1)\frac{1}{mn}$

$=\frac{1}{mn}+1-\frac{1}{mn}$

$t_{mn}=1$