Sequences and Series 1 Question 4
4.
If $T _r$ is the $r$ th term of an $AP$, for $r=1,2,3, \ldots$. . If for some positive integers $m$ and $n$, we have $T _m=\frac{1}{n}$ and $T _n=\frac{1}{m}$, then $T _{m n}$ equals
$(1998,2 M)$
(a) $\frac{1}{m n}$
(b) $\frac{1}{m}+\frac{1}{n}$
(c) 1
(d) 0
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Answer:
Correct Answer: 4. (c)
Solution:
- Let
$T _m=a+(m-1) d=\frac{1}{n} $
$T _n=a+(n-1) d=\frac{1}{m}$
and, On subtracting Eq. (ii) from Eq. (i), we get
$(m-n) d=\frac{1}{n}-\frac{1}{m} =\frac{m-n}{m n} $
$\Rightarrow \quad d =\frac{1}{m n}$
Again, $T _{m n}=a+(m n-1) d=a+(m n-n+n-1) d$
$ \begin{aligned} & =a+(n-1) d+(m n-n) d \\ & =T _n+n(m-1) \frac{1}{m n}=\frac{1}{m}+\frac{(m-1)}{m}=1 \end{aligned} $
