Let Tr be the rth term of an AP, for r=1, 2…. . If for some positive integers m and n, we have Tm=1n and Tn=1m, the Tm+n equal

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

Solution:

Let a be the first term and d be the com-mon difference of the given A.P. Then according to the

hypothesis,

$\begin{array}{l} T_{m} - T_{n} = \frac{1}{n} - \frac{1}{m} \\ \Rightarrow (m - n) d = \frac{m - n}{m n} \Rightarrow d = \frac{1}{m n} . \\ \Rightarrow T_{m + n} - T_{m} = (m + n - m) \frac{1}{m n} = \frac{1}{m} \end{array}$

Let $T_{r}$ be the rth term of an AP, for $r = 1,$ 2…. . If for some positive integers m and n, we have $T_{m} = \frac{1}{n}$ and
$T_{n} = \frac{1}{m},$ the $T_{m + n}$ equal

Solution:

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