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If $m t h$ term of an AP is $1 n$ and $n t h$ term is $1 m$ then the sum of its first mn terms is [[1]].

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answer is M N + 1 2.

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Detailed Solution

m t h

term of an AP is

1 n

and

n t h

term is

1 m

then the sum of its first mn terms is

m n + 1 2

.
Given

m t h

term of an AP is

1 n

and

n t h

term is

1 m

.
Let a be the first term and d be the common difference of the given AP.

∴ a m = 1 n ⇒ a + (m − 1) d = 1 n … … … (1)

∴ a n = 1 m ⇒ a + (n − 1) d = 1 m … … (2)

Subtracting equation (2) from equation (1):

⇒ 1 n − 1 m = (m − n) d ⇒ m − n m n = (m − n) d ⇒ d = 1 m n

Putting the value of d in equation (1):

∴ a + (m − 1) 1 m n = 1 n ⇒ a + 1 n − 1 m n = 1 n ⇒ a = 1 m n

Formula used for sum of first term of A.P.:

S n = n 2 [2 a + (n − 1) d]

Therefore, the sum of mn terms:

∴ S m n = m n 2 [2 a + (m n − 1) d] ⇒ S m n = m n 2 2 m n + (m n − 1) 1 m n = m n 2 1 + m n m n = m n + 1 2

Therefore, the sum of the first mn terms of the given AP is

m n + 1 2

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