Question

Let $T_r$ be the $r$th term of an AP, for $r=1,2,....$ if for some positive integers $m,n$ we have $T_m=1/n$ and $T_n=1/m$, then $T_{m/n}$ equal to ________

• A. $\frac{1}{mn}$
• B. $\frac{1}{m}+\frac{1}{n}$
• C. $\frac{1}{n^2}$
• D. $0$

Answer 1

Answer: (C)

$\frac{1}{n^2}$

We have

$T_m=\frac{1}{n}$ and$T_n=\frac{1}{m}$

$T_m=\frac{1}{n}=a+\left(n-1\right)d---\left(1\right)$

And

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

By $\left(1\right)-\left(2\right)eq{u}^n$ We get,

$T_m-T_n=\frac{1}{n}-\frac{1}{m}=\left(m-n\right)d$

$d=\frac{1}{mn}---\left(3\right)$

By putting value of $d$ in $eq{u}^n\left(1\right)$ we get,

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

$T_{\frac{m}{n}}=\frac{1}{n}-\frac{m-1}{mn}+\frac{1}{mn}.\left(\frac{m}{n}-1\right)$

$=\frac{1}{n}-\frac{1}{n}+\frac{1}{mn}+\frac{1}{n^2}-\frac{1}{mn}$

$=\frac{1}{n^2}$