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Sum of n Terms

Let Tr be t...

Question

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

\frac{1}{m n}

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\frac{1}{m} + \frac{1}{n}

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Solution

The correct option is C $1$
$T_{m} = a + (m - 1) d = \frac{1}{n}$

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

$⟹ d = \frac{1}{m n}, a = \frac{1}{m} - \frac{n - 1}{m n}$
$T_{m n} = a + (m n - 1) d = \frac{1}{m} - \frac{1}{m} + \frac{1}{m n} + \frac{m n - 1}{m n} = 1$

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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, then Tmn=

The correct option is C 1Tm=a+(m−1)d=1nTn=a+(n−1)d=1m⟹d=1mn,a=1m−n−1mnTmn=a+(mn−1)d=1m−1m+1mn+mn−1mn=1

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

The correct option is C $1$
$T_{m} = a + (m - 1) d = \frac{1}{n}$

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

$⟹ d = \frac{1}{m n}, a = \frac{1}{m} - \frac{n - 1}{m n}$
$T_{m n} = a + (m n - 1) d = \frac{1}{m} - \frac{1}{m} + \frac{1}{m n} + \frac{m n - 1}{m n} = 1$